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{<i, x) liE I & x E H(i)}. Then (1) provides us with a function Fe:::; R such that dom F = dom R = I. Then since <i, F(i) E F e:::; R, we must have F(i) E H(i). Thus the conclusion of (2) holds. (2) = (4) Let d be a set meeting conditions (a) and (b) of (4). Let H be the identity function on d; then (\I BEd) H(B) #- 0. Hence by (2) there is a function f with domain d such that (\I BE d)f (B) E H(B) = B. Let C = ran f. Then for BEd we have B n C = {f(B)}. (Nothing else could belong to B n C by condition (b).) (4) = (3) Given a set A, define d = {{B} x BIB is a nonempty subset of A}. Then each member of d is nonempty, and any two distinct members are disjoint (if <x, y) E ({B} x B) n ({Bf} x Bf), then x = B = Bf). Let C be a set (provided by (4)) whose intersection with each member of d is a singleton: en ({B} x B) = {<B, x)}, where x E B. It is a priori possible that C contains extraneous elements not belonging to any member of d. So discard them by letting F = C n (Ud). We claim that F is a choice function for A. Any member of F belongs to some {B} x B, and hence is of the form <B, x) for x E B. For anyone nonempty set Be:::; A, there is a unique x such that <B, x) E F, because F n ({B} x B) is a singleton. This x is of course F(B) and it is a member of B. (3) = (1) Consider any relation R. Then (3) provides us with a choice function G for ran R; thus G(B) E B for any nonempty subset B of ran R. Then define a function F with dom F = dom R by F(x) = G |
({y I xRy}). Then F(x) E {y I xRy}, i.e., <x, F(x) E R. Hence Fe:::; R. It remains to include parts (5) and (6) of the theorem. (6)= (1) The strategy behind this application (and others) of Zorn's lemma is to form a collection d of pieces of the desired object, and then to show that a maximal piece serves the intended purpose. In the present case, we are given a relation R and we choose to define d = {fe:::; R Ifis a function}. Axiom of Choice 153 Before we can appeal to Zorn's lemma, we must check that d is closed under unions of chains. So consider any chain fJI e:::; d. Since every member of fJI is a subset of R, UfJI is a subset of R. To see that UfJI is a function, we use the fact that fJI is a chain. If <x, y) and <x, z) belong to UfJI, then and <x, z) E H E fJI <x, y) E G E fJI for some functions G and H in d. Either G e:::; H or H e:::; G; in either event both <x, y) and <x, z) belong to a single function, so y = z. Hence UfJI is in d. Now we can appeal to (6), which provides us with a maximal function F in d. We claim that dom F = dom R. For otherwise take any x E dom R - dom F. Since x E dom R, there is some y with xRy. Define F' = F u {<x, y)}. Then F' Ed, contradicting the maximality of F. Hence dom F = dom R. (6) = (5) Let C and D be any sets; we will show that either C ~ D or D ~ C. In order to utilize (6), define d = {Ilfis a one-to-one function & domfe:::; C & ranfe:::; D}. Consider any chain fJI e:::; d. As in the preceding paragraph, UfJI is a function, and a similar argument shows that UfJI |
is one-to-one. Next consider <x, y) E UfJI; then <x, y) EfE d. Consequently x E C and y ED. Thus dom UfJI e:::; C and ran UfJI e:::; D. Hence UfJI E d and we can apply (6) to obtain a maximal JEd. We claim that either domJ = C (in which case C ~ D) or ranJ = D (in which case D ~ C since J-l is then a one-to-one function from D into C). Suppose to the contrary that neither condition holds, so that there exist elements C E C - domJ and dE D - ranJ Then f' =Ju {<c, d)} is in d, contradicting the maximality of J (You will observe that the strategy underlying this application of Zorn's lemma is the same one as in the preceding paragraph.) At this point we have proved that part of Theorem 6M shown in Fig. 40. The proof will be completed in Chapter 7. Remarks Zorn's lemma first appeared in a 1922 paper by Kuratowski. Earlier maximality principles that were similar in spirit had been published by Felix Hausdorff. The importance of Zorn's lemma lies in the fact that it is well suited to many applications. in mathematics. For example, to prove in linear algebra that every vector space has a basis requires some form of choice, and Zorn's lemma is a convenient form to use here. (Take d to be the collection of all linearly independent sets; then a maximal element will 154 6. Cardinal Numbers and the Axiom of Choice be a basis.) Similarly in proving that there exist maximal proper ideals in a ring with identity or in proving that there exist maximal Abelian subgroups in a group, Zorn's lemma provides a suitable tool. We can give a plausibility argument for Zorn's lemma as follows. d cannot be empty, because 0 is a chain and so 0 = U 0 E d. Probably o is not maximal, so we can choose a larger set. If that larger set is not maximal, we can choose a still larger one. After infinitely many steps, even if we have not found a maximal set we have at least formed a chain. So we can take its union and continue. The procedure can stop only when we finally reach a maximal set. So |
if only we are patient enough, we should reach such a set... 2 I 4 3 II Fig. 40. This much of Theorem 6M we prove now; the rest is postponed. We now formally add the axiom of choice to our list of axioms. And we do it without a marginal stripe. This is because historically the axiom of choice has had a unique status. Initially some mathematicians objected to the axiom, because it asserts the existence of a set without specifying exactly what is in it. To this extent it is less "constructive" than the other axioms. Gradually the axiom has won acceptance (at least acceptance by most mathematicians willing to accept classical logic). But it retains a slightly tarnished image, from the days when it was not quite respectable. Consequently it has become wide-spread practice, each time the axiom of choice is used, to make explicit mention of the fact. (No such gesture is extended to the other axioms, which are used extensively and mentioned rarely.) In Chapter 7 we will complete our axiomatization by adding the replacement axioms and the regularity axiom. I Without the axiom of choice, we can still prove that for a finite set I, if H(i) #- 0 for all i E I, then there is a function f with domain I such that f (i) E H(i) for all i E I (Exercise 19). But for an infinite set I, the axiom of choice is indispensable. (If we are making seven choices, we can explicitly mention each one; for ~o choices this is no longer possible.) In particular, if we are choosing one thing, then we do not need the Axiom of Choice 155 axiom of choice. (We mention this because people sometimes overreact when first hearing of the axiom of choice.) For example, assume that A #- 0. Then there exists some Yo in A; we can use Yo in further arguments (to show, e.g., that some singleton {Yo} is a subset of A). This makes no use of the axiom of choice, even when phrased as "we can choose some fixed Yo in A." For a second example, assume we are given a relation R. For any particular x in dom R, there exists some Yo for which xRyo. And we can conclude that for any x in dom R, there is a singleton {Yo} included in |
{t I xRt}. We have not yet used the axiom of choice. What does require the axiom of choice is saying that for any x in dom R there is some Yx for which xRyx' and then putting all these Yx's together into a set, e.g., {Yx I x E dom R}. This in effect is making many choices, one for each x in dom R, which may be an infinite set. (A human being cannot perform infinitely many actions. But the axiom of choice asserts that, despite our practical limitations, there exists, in theory, a set produced by making infinitely many choices.) Another situation in which the axiom of choice can be avoided is where we can specify exactly which object we want to choose. For example, we can show, without using the axiom of choice, that there is a choice function for OJ. Namely, define F: (9OJ - {0}) --+ OJ by F(A) = the least member of A for 0 #- A s OJ. The crucial point here is that we can write down an expression defining the selected member of A. If OJ is replaced by IR, for example, there is no longer the possibility of defining a way to select numbers from arbitrary nonempty subsets of IR. Since there is a choice function on OJ, it follows that there is a choice function on any set equinumerous to OJ. (Why?) And similarly any finite set has a choice function. The following example is due to Russell. If we have ~o pairs of shoes, then we can select one shoe from each pair without using the axiom of choice. We simply select the left shoe from each pair. But if we have ~o pairs of socks, then we must use the axiom of choice if we are to select one sock from each pair. For there is no definable difference between the two socks in a pair. Example Assume thatfis a function from A onto B. We claim that B ~ A. To verify this, recall that by Theorem 3J(b), the proof of which used choice, there is a right inverse g: B --+ A such that fog = lB. And 9 is one-to-one since it has a left inverse. The function 9 shows that B ~ A. 156 6. Cardinal Numbers and the Axiom of Choice Conversely assume that B ~ A and B is nonempty. Then |
(since B ~ A) there is a one-to-one function g: B -+ A. By Theorem 3J(a) there is a left inverse f: A -+ B such that fog = lB. And f maps A onto B since it has a right inverse. Hence we can conclude: If B is non empty, then B ~ A iff there is a function from A onto B. As a special case, we can conclude that a nonempty set B is dominated by OJ iff there is a function from OJ onto B. (This special case can be proved without the axiom of choice.) We now utilize the axiom of choice to prove that ~o is the least infinite cardinal number. Theorem 6N (b) ~o S K for any infinite cardinal K. (a) For any infinite set A, we have OJ ~ A. Proof Part (b) is merely a restatement of part (a) in terms of cardinals; it suffices to prove (a). So consider an infinite set A. The idea is to select ~o things from A: a first, then a second, then a third,.... Let F be a choice function for A (as provided by the axiom of choice, III). We must somehow employ F repeatedly; in fact ~o times. As a first try, we could attempt to define a function g: OJ -+ A by recursion. We know A is nonempty; fix some element a E A. Maybe we can take g(O) = a, g(n+) = the chosen member of A - {g(O),..., g(n)} = F(A - g[n+]). No, we cannot. This would require a stronger recursion theorem than the one proved in Chapter 4. We are attempting to define g(n+) not just from g(n), but from the entire list g(O), g(1),..., g(n). But our second try will succeed. We will define by recursion a function h from OJ into the set of finite subsets of A. The guiding idea is to have, for example, h(7) = {g(O),..., g(6)}. Formally, h(O) = 0, h(n+) = h(n) u {F(A - h(n))}. Thus we start with 0, and successively add chosen |
new elements from A. A - h(n) is non empty because A is infinite and h(n) is a finite subset. And h(n +) is again a finite subset of A. Now we can go back and recover g. Define g(n) = F(A - h(n)) Axiom of Choice 157 so that g(n) is the thing we added to the set h(n) to make h(n+). Thus g: OJ --+ A; we must check that 9 is one-to-one. Suppose then that m #- n. One number is less than the other; say mEn. Then m+ ~ n and so But g(n) ¢ h(n), since Hence g{m) #- g(n). g(m) E h(m+) S h(n). g(n) = F(A - h(n)) E A - h(n). We can now list some consequences of the foregoing theorem. 1. Any infinite subset of OJ is equinumerous to OJ. In this special case we can avoid using the axiom of choice, since we have a choice function for OJ anyway. 2. We have by part (b) of the theorem that if K < ~o' then K is finite. Since the converse is clear, we have K < ~o iff K is finite. 3. We get another proof that subsets of finite sets are finite (Corollary 6G). For if card A ::; card n < ~o, then by the Schroder-Bernstein theorem card A < ~o' and hence A is finite. Another consequence of the preceding theorem is the following char acterization of the infinite sets (proposed by Dedekind in his 1888 book as a definition of" infinite "). Corollary 6P A set is infinite iff it is equinumerous to a proper subset of itself. Proof Half of this result is contained in Corollary 60, where we showed that if a set was equinumerous to a proper subset of itself, then it was infinite. Conversely, consider an infinite set A. Then by the above theorem, there is a one-to-one function f from OJ into A. Define a function 9 from A into A by g(f(n)) = f(n+) for n E OJ, g(x) = x for x � |
� ranf (see Fig. 41). Then 9 is a one-to-one function from A onto A - {f(O)}. -l 158 6. Cardinal Numbers and the Axiom of Choice o • • 2 • 3 • \\\\ • • • • 9 Fig. 41. /(0) is not in ran g. Exercises 18. Prove that the following statement is equivalent to the axiom of choice: For any set d whose members are nonempty sets, there is a functionfwith domain d such thatf(X) E X for all X in d. 19. Assume that H is a function with finite domain I and that H(i) is nonempty for each i E I. Without using the axiom of choice, show that there is a functionfwith domain I such thatf(i) E H(i) for each i E I. [Suggestion: Use induction on card I.] 20. Assume that A is a non empty set and R is a relation such that (\Ix E A)(3y E A) yRx. Show that there is a functionf: OJ --+ A withf(n+ )Rf(n) for all n in OJ. (Teichmiiller-Tukey lemma) Assume that d 21. that for every set B, is a nonempty set such BEd ¢> every finite subset of B is a member of d. Show that d has a maximal element, i.e., an element that is not a subset of any other element of d. 22. Show that the following statement is another equivalent version of the axiom of choice: For any set A there is a function F with dom F = UA and such that x E F(x) E A for all x E UA. 23. Show that in the proof to Theorem 6N, we have g[n] = h(n). 24. How would you define the sum of infinitely many cardinal numbers? Infinite products? 25. Assume that S is a function with domain OJ such that S(n) $; S(n+) for each n E OJ. (Thus S is an increasing sequence of sets.) Assume that B is a subset of the union Une co S(n) such that for every infinite subset Bf of B there is some n for which Bf n S(n) is infinite. Show that |
B is a subset of some S(n). Countable Sets 159 COUNTABLE SETS The definition below applies the word "countable" to those sets whose elements can, in a sense, be counted by means of the natural numbers. A "counting" ofa set can be taken to be a one-to-one correspondence between the members of the set and the natural numbers (or the natural numbers less than some number n). This requires that the set be no larger than OJ. Definition A set A is countable iff A ~ OJ, i.e., iff card A ::; ~o. Since we have recently found that K < ~o ¢> K is finite, we can also formulate the definition as follows: A set A is countable iff either A is finite or A has cardinality ~o. For example, the set OJ of natural numbers, the set 71. of integers, and the set IQ of rational numbers are all infinite countable sets. But the set IR of real numbers is uncountable (Theorem 6B). Any subset of a countable set is obviously countable. The union of two countable sets is countable, as is their Cartesian product. (The union has at most cardinality ~o + ~o' the product at most ~o. ~o. But both of these numbers equal ~o.) On the other hand, 9A is uncountable for any infinite set A. (If 2" ::; ~o' then K < ~o.) Recall (from the example preceding Theorem 6N) that a nonempty set B is countable iff there is a function from OJ onto B. This fact is used in the proof of the next theorem. Theorem 6Q A countable union of countable sets is countable. That is, if d is countable and if every member of d is a countable set, then Ud is countable. Proof We may suppose that 0 ¢ d, for otherwise we could simply remove it without affecting Ud. We may further suppose that d #- 0, since U0 is certainly countable. Thus d is a countable (but nonempty) from OJ x OJ onto Ud. We already know of functions from OJ onto OJ x OJ, and the composition will map OJ onto Ud, thereby showing that Ud is countable. Since d is countable but nonempty, there is a function G from |
OJ onto d. Informally, we may write d = {G(O), G(l),... }. (Here G might not be one-to-one, so there may be repetitions in this enumeration.) We are given that each set G(m) is countable and nonempty. 160 6. Cardinal Numbers and the Axiom of Choice Hence for each m there is a function from w onto G(m). We must use the axiom of choice to select such a function for each m. Because the axiom of choice is a recent addition to our repertoire, we will describe its use here in some detail. Let H: w -> "'(Ud) be defined by H(m) = {g I 9 is a function from w onto G(m)}. We know that H(m) is nonempty for each m. Hence there is a function F with domain w such that for each m, F(m) is a function from w onto G(m). To conclude the proof we have only to let f(m, n) = F(m)(n). Then f -1 is a function from w x w onto Ud. Example F or any set A, define a sequence in A to be a function from some natural number into A. Let Sq(A) be the set of all sequences in A: Sq(A) = {f I (3n E w) f maps n into A} =oAu 1 Au 2 Au.. ·. The length of a sequence is simply its domain. In order to verify that Sq(A) is a legal set, note that iff: n -> A, then so thatf E glI(w x A). Hence Sq(A) s:: glI(w x A). f s:: n x A s:: w x A, We now list some observations that establish the existence of tran scendental real numbers. 1. Sq(w) has cardinality ~o. This can be proved by using primarily the fact that w x w ~ w. Another very direct proof is the following. Consider any f E Sq(w); say the length off is n. Then define H(f) = 2J (0)+1 • 3J (1)+1 • '" • r;;~11)+1, where Pi is the (i + l)st prime. (If the length of f is 0, then |
H(f) = 1.) Thus H: Sq(w) -> wand by the fundamental theorem of arithmetic (which states that prime factorizations are unique) H is one-to-one. Hence card Sq(w) :::;; ~O, and the opposite inequality is clear. (In Chapter 4 we did not actually develop the theory of prime numbers. But there are no difficulties in embedding any standard development of the subject into set theory.) 2. Sq(A) is countable for any countable set A. By the countability of A there is a one-to-one function 9 from A into w. This function naturally induces a one-to-one map from Sq(A) into Sq(w). Hence card Sq(A) :::;; card Sq(w) = ~o. (An alternative proof writes Sq(A) = U{n A I nEw}, a countable union of countable sets.) We can think of this set A as an alphabet, and the elements of Sq(A) as being words on the alphabet A. In this terminology, the present example Countable Sets 161 3. There are ~o algebraic numbers. can be stated: On any countable alphabet, there are countably many words. (Recall that an algebraic number is a real number that is the root of some polynomial with integer coefficients. For this purpose we exclude from the polynomials the function that is identically equal to 0.) As a first step in counting the algebraic numbers, note that the set 7L of integers has cardinality ~o + ~o = ~o. Next we calculate the cardinality of the set P of polynomials with integer coefficients. We can assign to each polynomial (of degree n) its sequence (of length n + 1) of coefficients; e.g., 1 + 7x - 5x 2 + 3x4 is assigned the sequence of length 5 whose successive values are 1, 7, - 5, 0, 3. This defines a one-to-one map from Pinto Sq(7L), a countable set. Hence P is countable. Since each polynomial in P has only finitely many roots, the set of algebraic numbers is a countable union of finite sets. Hence it is countable, by Theorem 6Q. Since the set of algebraic numbers is certainly infinite, it |
has cardinality ~o. 4. There are uncountably many transcendental numbers. (Recall that a transcendental number is defined to be a real number that is not algebraic.) Since the set of algebraic numbers is countable, the set of transcendental numbers cannot also be countable lest the set ~ be countable. (Soon we will be able to show that the set of transcendental numbers has cardinality 2No.) Exercises 26. Prove the following generalization of Theorem 6Q: If every member of a set d has cardinality K or less, then card Ud ::; (card d). K. 27. (a) Let A be a collection of circular disks in the plane, no two of which intersect. Show that A is countable. (b) Let B be a collection of circles in the plane, no two of which intersect. Need B be countable? (c) Let C be a collection of figure eights in the plane, no two of which intersect. Need C be countable? 28. Find a set d of open intervals in ~ such that every rational number belongs to one of those intervals, but Ud #- ~. [Suggestion: Limit the sum of the lengths of the intervals.] 29. Let A be a set of positive real numbers. Assume that there is a bound b such that the sum of any finite subset of A is less than b. Show that A is countable. 30. Assume that A is a set with at least two elements. Show that Sq(A) ~ W A. 162 6. Cardinal Numbers and the Axiom of Choice ARITHMETIC OF INFINITE CARDINALS We can use the axiom of choice to show that adding or multiplying two infinite cardinals is a more trivial maHer than it first appeared to be. Lemma 6R For any infinite cardinal K, we have K' K = K. Proof Let B be a set of cardinality K. It would suffice to show that B x B ~ B, and we will almost do this. Define.Ye = {f I f = 0 or for some infinite A ~ B,J is a one-to-one correspondence between A x A and A}. Our strategy is to use Zorn's lemma to obtain a maximal function fo in.Ye. Although fo might not quite show that B x B ~ B, it will come close enough. (The reason for including 0 as a |
member of.Ye is that our statement of Zorn's lemma requires that.Ye contain the union of any chain, even the empty chain.) Before applying Zorn's lemma to.Ye, we must check for closure under unions of chains. Let Cfj be a chain included in.Ye. We may suppose that Cfj contains some nonempty function, since otherwise UCfj = 0 E.Ye. As we have seen before, UCfj is again a one-to-one function. Define A = U{ran f If E Cfj} = ran UCfj (compare Exercise 8 of Chapter 3). A is infinite since Cfj contains some nonempty function. We claim that UCfj is a one-to-one correspondence between A x A and A. The only part of this cla.im not yet verified is that dom UCfj = A x A. First consider any <a1, a2) E A x A. Then a1 E ranf1 and a2 E ranf2 for some f1 and f2 in Cfj. Either f1 ~f2 or f2 ~f1; by symmetry we may suppose that f1 ~ f 2. Then <a1, a2) E ranf2 x ranf2 = domf2 ~ U{domf If E Cfj} = dom UC§'. Conversely any member of dom UCfj belongs to domf for some f E Cfj. But domf = ranf x ranf ~ A x A. Thus dom UCfj = A x A. So UCfj is a one-to-one correspondence between A x A and A; hence UCfj E.Ye. Zorn's lemma now provides us with a maximal fo E.Ye. First, we must check that fo #- 0. Since B is infinite, it has a subset A of cardinality ~o' Because ~o' ~o = ~o, there is a one-to-one correspondence 9 between A x A and A. Thus 9 E.Ye; since 9 properly extends 0, it is impossible for 0 to be maximal is.Ye. Hence fo #- 0. So by the definition of.Ye, fo is a one-to-one correspondence between Ao x Ao and Ao, where Ao is some infinite subset of B. Let A. = card Ao ; then A. is infinite and A. • A. = A |
.. One might hope that by virtue of the maximality, Ao would be all of B. This may not be true, but we will show that A. = K and that B - Ao has smaller cardinality. Arithmetic of Infinite Cardinals 163 In order to show that card (B - Ao) < A., suppose that to the contrary A. :::; card(B - Ao). Then B - Ao has a subset D of cardinality A.. We will show that this contradicts the maximality of fo by finding a proper extension of fo that is a one-to-one correspondence between the sets (Ao u D) x (Ao u D) and Ao u D. We have (Ao u D) x (Ao u D) = (Ao x Ao) u (Ao x D) u (D x Ao) u (D x D) ~ D Fig. 42. The set B x B (in Lemma 6R). (see Fig. 42). Of the four sets on the right side of this equation, Ao x Ao is already paired with Ao by f o' The remainder (Ao x D) u (D x Ao) u (D x D) (shaded in Fig. 42) clearly has cardinality A.'A.+A.'A.+A.'A.=A.+A.+A. = 3. A. :::; A. • A. = A. •. Hence there is a one-to-one correspondence 9 between the shaded sets and D. Thenfo u 9 E.YC and properly extendsfo'contradicting the maximality of fo' Hence our supposition that A. :::; card(B - Ao) is false. By cardinal comparability, card(B - Ao) < A.. 164 6. Cardinal Numbers and the Axiom of Choice Finally we have K = card Ao + card(B - Ao) ::; A. + A. = 2. A. ::; A.. A. = A. ::; K, whence A. = K. Hence K. K = K. Absorption Law of Cardinal Arithmetic Let K and A. be cardinal numbers, the larger of which is infinite and the smaller of which is nonzero. Then K + A. = K. A. = max(K, A.). Proof By the symmetry we may suppose that A. ::; K. Then K ::; K + A. :: |
; K + K = K. 2 ::; K • K = K and Hence equality holds throughout. K ::; K. A. ::; K. K = K. Example We do not have a well-defined subtraction operation for infinite cardinal numbers. If you start with ~o things and take away ~o things, then the number of remaining things can be anywhere from 0 to ~o' depending on which items were removed. But if you start with K things (where K is infinite) and remove A. things where A. is strictly less than K, then exactly K things remain. To prove this, let J1. be the cardinality of the remaining items. Then K = A. + J1. = max (A., J1.), so that we must have J1. = K. Example There are exactly 2No transcendental numbers. This follows from the preceding example. If from the 2No real numbers we remove the countable set of algebraic numbers, then 2No numbers remain. Example For any infinite cardinal K, we have KK = 2K. To prove this, observe that whence equality holds throughout. We conclude this section with two counting problems: 1. How many functions from ~ into ~ are there? 2. How many of these are continuous? The first question asks for the cardinality of the set Ihl~. The cardinal number of this set is Continuum Hypothesis 165 This last expression cannot be further simplified; it provides the answer to the first question. (As with finite numbers, K A • means KW ) and not (KAt) Now consider the second question; let C(~) be the set of continuous functions in Ihl~. It is easy to see that 2No :::; card C(~) :::; 22"0, but we need an exact answer. We claim that card C(~) = 2No. To prove this, we will consider the restriction of the continuous functions to the set o of rational numbers (where 0 is regarded as a subset of ~). Iff and 9 are two distinct continuous functions, then f - 9 is not identically zero. (Here f - 9 is the result of subtracting 9 from f, not the relative complement.) Hence by continuity, there is an open interval throughout whichf - 9 is nonzero. In this interval lies some rational, so f ~ 0 #- 9 ~ 0. Hence the map from C(~) into Q~ assigning to |
each continuous function f its restriction f ~ 0 is a one-to-one map. Thus C(~) :::; Q~ and so card C(~) :::; card Q~ = (2 No)No = 2No. Exercises In the proof of Lemma 6R we utilized a certain set.Ye. How do we 31. show from the axioms that such a set exists? 32. Let fJ' A be the collection of all finite subsets of A. Show that if A is infinite, then A ~ fJ' A. 33. Assume that A is an infinite set. Prove that A ~ Sq(A). 34. Assume that 2 :::; K :::; A. and A. is infinite. Show that KA = 2A. 35. Find a collection d of 2No sets of natural numbers such that any two distinct members of d have finite intersection. [Suggestion: Start with the collection of infinite sets of primes.] 36. Show that for an infinite cardinal K, we have K! = 2\ where K! is defined as in Exercise 14. CONTINUUM HYPOTHESIS We have in this chapter given some examples of countable sets and uncountable sets. But every uncountable set examined thus far has had cardinality 2No or more. This raises the question: Are there any sets with cardinali ty between ~ 0 and 2 NO? The" con tin u um hypothesis" is the assertion that the answer is negative, i.e., that there is no K with ~o < K < 2No. Or equivalently, the continuum hypothesis can be stated: Every uncountable set of real numbers is equinumerous to the set of all real numbers. 166 6. Cardinal Numbers and the Axiom of Choice Cantor conjectured that the continuum hypothesis was true. And David Hilbert later published a purported proof. But the proof was incorrect, and more recent work has cast doubt on the continuum hypothesis. In 1939 Godel proved that on the basis of our axioms for set theory (which we here assume to be consistent) the continuum hypothesis could not be disproved. Then in 1963 Paul Cohen showed that the continuum hypothesis could not be proved from our axioms either. But since the continuum hypothesis is neither provable nor refutable from our axioms, what can we say about its truth or falsity? We have some informal ideas about what sets are like, |
but our intuition might not assign a definite answer to the continuum hypothesis. Indeed, one might well question whether there is any meaningful sense in which one can say that the continuum hypothesis is either true or false for the "real" sets. Among those set-theorists nowadays who feel that there is such a meaningful sense, the majority seems to feel that the continuum hypothesis is false. The "generalized continuum hypothesis" is the assertion that for every infinite cardinal K, there is no cardinal number between K and 2K. Godel's 1939 work shows that even the generalized continuum hypothesis cannot be disproved from our axioms. And of course Cohen's result shows that it cannot be proved from our axioms (even in the special case K = ~o). There is the possibility of extending the list of axioms beyond those in this book. And the new axioms might conceivably allow us to prove or to refute the continuum hypothesis. But to be acceptable as an axiom, a statement must be in clear accord with our informal ideas of the concepts being axiomatized. It would not do, for example, simply to adopt the generalized continuum hypothesis as a new axiom. It remains to be seen whether any acceptable axioms will be found that settle satisfactorily the correctness or incorrectness of the continuum hypothesis. The work of Godel and Cohen also shows that the axiom of choice can neither be proved nor refuted from the other axioms (which we continue to assume are consistent). But unlike the continuum hypothesis, the axiom of choice conforms to our informal view of how sets should behave. For this reason, we have adopted it as an axiom. Results such as those by Godel and Cohen belong to the metamathematics of set theory. That is, they are results that speak of set theory itself, in contrast to theorems within set theory that speak of sets. ORDERINGS AND ORDINALS CHAPTER 7 In this chapter we will begin by discussing both linear orderings (mentioned briefly in Chapter 3) and, more generally, partial orderings. But we will soon focus our attention on a special case of the linear orderings, namely the so-called well orderings. The well orderings will lead us to the study of ordinal numbers and to the fulfillment of promises (e.g., the definition of cardinal numbers) that have been made. PARTIAL ORDERI NGS A partial ordering is a special sort of relation. Before making any |
definitions, it will be helpful to consider a few examples. 1. Let S be any fixed set, and let c s be the relation of strict inclusion on subsets of S: c s = {<A, B) I A ~ B ~ S & A -:f. B}. Of course we write" A c s B" in place of the more awkward" <A, B) E c s'" 2. Let P be the set of positive integers. The strict divisibility relation on P is {<a, b) E P X P I a' q = b for some q -:f. I}. 167 168 7. Orderings and Ordinals For example, 5 divides 60 (here q = 12), but 2 does not divide 3, nor does 3 divide 2. No number in P strictly divides itself. 3. For the set ~ of real numbers, we have the usual ordering relation <. For any distinct real numbers x and y, either x < y or y < x. Definition A partial ordering is a relation R meeting the following two conditions: (a) R is a transitive relation: xRy & yRz = xRz. (b) R is irreflexive: It is never the case that xRx. In the foregoing examples, it is easy to see that c s' strict divisibility, and < are all partial ordering relations. The preferred symbols for partial ordering relations are < and similar symbols, e.g., -<, c, and the like. If < is such a relation, then we can define: x :::; y iff either x < y or x = y. {O, I, 2} {O. I} {O} {1,2} {2} o Fig. 43. The inclusion ordering on {O, 1, 2}. Partial Orderings 169 4 6 10 9 14 15 25 21 35 49 Fig. 44. A finite part of an infinite picture. It is easy to see that, for example, x ::; y < z = x < z. At the end of Chapter 3 we drew some pictures (Fig. 15) of orderings. For partial orderings the pictures show more variety. Again we represent the members of A by dots, placing the dot for x below the dot for y whenever x < y, and drawing lines to connect the dots. For example, Fig. 43 is a picture of the partial ordering c s on the |
set S = {O, 1, 2}. There is no need to add a direct nonstop line from 0 to {O, I}, because there are lines from 0 to {O} and from {O} to {O, I}. Figure 44 is a picture of a small piece of the divisibility ordering on the positive integers; the full picture would have to be infinite. The following theorem lists some easy consequences of our definitions. Theorem 7A Assume that < is a partial ordering. Then for any x, y, and z: (a) At most one of the three alternatives, x < y, x = y, y < x, can hold. (b) x::;y::;x=x=y. Proof In part (a), if we had both x < y and x = y, then we would have x < x, contradicting irrefiexivity. And if both x < y and y < x, then by transitivity x < x, again contradicting irrefiexivity. 170 7. Orderings and Ordinals In part (b) if x#- y then we would have x < y < x, contradicting part -1 (a). We are particularly interested in cases where part (a) of this theorem can be strengthened to read "exactly one," i.e., trichotomy holds. Definition R is a linear ordering on A iff R is a binary relation on A that is a transitive relation and satisfies trichotomy on A, i.e., for any x and y in A exactly one of the three alternatives, holds. xRy, x = y, yRx, Note that we have to specify the set A in this definition. We can also speak of a partial ordering on A, this being just a binary relation on A that is a partial ordering. Examples The usual ordering relation on ~ is linear. Strict divisibility is a partial ordering, but is not a linear ordering on the positive integers. And inclusion c s is nonlinear if S has two or more members. Notice that if R is a linear ordering on A, then R is also a partial ordering (since trichotomy implies irreflexivity-recall Theorem 3R). Digression In the study of partial orderings, there is always the question whether to use strict orderings «) or weak orderings (:::;) as the basic concept. We have taken the" <" |
course, by demanding that a partial ordering be irreflexive. For the" :::; " version, one demands that a partial ordering on A be reflexive on A. Yet a third course is to specify neither extreme, but to allow any pairs <x, x) to belong to the ordering or not, as they please. Each alternative has its own minor advantages and its own drawbacks. One feature of demanding reflexivity is that whenever :::; is a partial ordering on A, then A = fld :::;. Consequently we can treat just the ordering:::; ; the set A is then encoded inside this relation. For strict orderings this feature is lost; for example, 0 is a partial ordering on any set. Often we will want to refer to a set A together with an ordering < on A. Definition A structure is a pair <A, R) consisting of a set A and a binary relation R on A (i.e., R ~ A x A). In particular, we can speak of a partially (or linearly) ordered structure if R is a partial (or linear) ordering relation on A. (The terms poset and loset are sometimes used here.) There is a certain amount of terminology for orderings that has proved sufficiently useful to have become standard. Let < be a partial ordering; let D be a set. An element m of D is said to be a minimal element of D Partial Orderings 171 iff there is no x in D with x < m (or equivalently, iff for all x in D, x 1- m). And m is a least (or smallest or minimum) element of D iff m :::; x for all x in D. A least element is also minimal. For a linear ordering on a set that includes D the two concepts coincide, since x1-m = m:::;x for a linear ordering. But in the nonlinear case, minimality is weaker than leastness. Example Consider the strict divisibility relation on the set P of positive integers. Then 1 is the least element of P. But let D be {a E P I a #- I}. Then every prime is a minimal element of D, and D has no least element. A least element of a set (if one exists) is automatically unique. (If m 1 and m 2 are both least, then m 1 :::; m 2 and m2 :::; m 1, whence equality holds.) In |
this event, the least element is the only minimal element. But a set can have many minimal elements, as the above example shows. It is also possible for a set to have no minimal elements at all. (What is an example of such a set?) If R is a partial ordering, then R - 1 Similarly an element m of D is a maximal element of D iff there is no x in D with m < x (or equivalently, iff for all x in D, m 1- x). And m is a greatest (or largest or maximum) element of D iff x:::; m for all x in D. The remarks we have made concerning minimal and least elements carry over to maximal and greatest elements. That is, a greatest element is also maximal. For a linear ordering (on a set that includes D) the two concepts coincide. A set can have only one greatest element. But if a set has no greatest element,it can have many maximal elements (or it might have none). is also a partial ordering (Exercise 2). And for a set D, the minimal elements of D with respect to R -1 are exactly the maximal elements with respect to R. When it is necessary to specify the ordering, we can speak of "R-minimal" elements. Suppose that < is a partial ordering on A and consider a subset C of A. An uppe~ bound of C is an element b E A such that x :::; b for all x E C. Here b mayor may not belong to C; if it belongs to C, then it is clearly the greatest element of C. If b is the least element of the set of all upper bounds for C, then b is the least upper bound (or supremum) of C. Lower bound and greatest lower bound (or infimum) are defined analogously. Example Consider a fixed set S and the partial ordering c s on &JS. For A and B in &JS, the set {A, B} has a least upper bound (with respect to c s)' namely A u B. Similarly A n B is the greatest lower bound of {A, B}. If d £; &JS, then Ud is the least upper bound of d. And nd (if d #- 0) is the greatest lower bound. 172 7. Orderings and Ordinals Example Next consider the set 0 of rational numbers with its usual linear ordering relation <. The set of positive rational numbers has no upper |
bounds at all. The set {x E 0 I x 2 < 2} has upper bounds in 0, but no least upper bound in 0. Example In the set of positive integers ordered by divisibility, greatest lower bounds are called "greatest common divisors" (g.c.d.'s). And least upper bounds are called" least common multi pIes" (I.c.m.'s). In this ordering, any finite set has an I.c.m., but infinite sets have no upper bounds. Any nonempty set of positive integers has a g.c.d. Exercises 1. Assume that < A and < B are partial orderings on A and B, respectively, and thatf is a function from A into B satisfying x < AY = f(x) < Bf(Y) for all x and Y in A. (a) Can we conclude that f is one-to-one? (b) Can we conclude that x < AY ¢> f(x) < Bf(Y)? 2. Assume that R is a partial ordering. Show that R -1 is also a partial ordering. 3. Assume that S is a finite set having n elements. Show that a linear ordering on S contains!n(n - 1) pairs. WELL ORDERINGS Definition A well ordering 1 on A is a linear ordering on A with the further property that every nonempty subset of A has a least element. For example, we proved in Chapter 4 that the usual ordering on (J) is a well ordering. But the usual ordering on the set 7L of integers is not a well ordering; 7L has no least element. Well orderings are significant because they can be used to index constructions that proceed "from the bottom up," where at every stage in the construction (except the last) there is a unique next step. This idea is 1 This is a grammatically unfortunate phrase, in that it uses "well" as an adjective. Some authors insert a hyphen, which helps a little, but not much. Well Orderings 173 made precise in the transfinite recursion theorem, which we will encounter presently. We can very informally describe what well orderings are like; later we will give exact statements and proofs. Assume that < is a well ordering on A. Then the set A itself, if nonempty, has a least element to' And then what is left, A - |
{to}, has (if it is nonempty) a least element t1. Next consider A - {to, t1}, and so forth. We obtain to < t1 < t2 <.... There may be still more in A - {to, tl'... }, in which case there is a least element tro' And this might continue: Eventually we use up all the elements of A, and things grind to a halt. We can use the axiom of choice to obtain another way of characterizing the well orderings. Theorem 7B Let < be a linear ordering on A. Then it is a well ordering iff there does not exist any functionf: w->A withf(n+) < f(n) for every nE w. A functionf: w -> A for whichf(n+) <f(n) (for all nEw) is sometimes called a descending chain, but this terminology should not be confused with other uses we make of the word "chain." The theorem asserts that, for a linear ordering <, it is a well ordering iff there are no descending chains. Proof If there is a descending chain f, then ran f is a nonem pty subset of A. And clearly ranf has no least element; for each elementf(n) there is a smaller elementf(n+). Hence the existence of a descending chain implies that < is not a well ordering. Conversely, assume that < is not a well ordering on A, so that some nonempty B ~ A lacks a least element. Then (\fx E B)(3y E B) y < x. By Exercise 20 of Chapter 6, there is a descending chain f: w -> B with f(n+) < f(n) for each n in w. -1 If < is some sort of ordering on A (at least a partial ordering) and tEA, then the set seg t = {x I x < t} is called the initial segment up to t. (A less ambiguous notation would be seg< t, but in practice the simpler notation suffices.) For example, w was ordered by E, and consequently we had for nEW, seg n = {x I x E n} = n. 174 7. Orderings and Ordinals Transfinite Induction Principle Assume that < is a well ordering on A. Assume that B is a subset of A with the special property that for every |
t in A, seg t s B = t E B. Then B coincides with A. Before proving this, let us define a subset B of A to be a < -inductive subset of A iff it has this special property, i.e., for every t in A, seg t s B = t E B. This condition, in words, states that t's membership in B is guaranteed once it is known that all things less than t are already in B. The transfinite induction principle can be restated: If < is a well ordering on A, then any < -inductive subset of A must actually coincide with A. Proof If B is a proper subset of A, then A - B has a least element m. By the leastness, y E B for any y < m. But this is to say that seg m S B, so by assumption mE B after all. -1 The above proof is so short that we might take a second glance at the situation. First take the least element to of A. Then seg to = 0, and so automatically seg to S B. The assumption on B then tells us that to E B. Next we proceed to the least element t1 of A - {to}' Then seg t1 = {to} S B, so we obtain t1 E B. And so forth and so forth. But to make the "and so forth" part secure, we have the foregoing actual proof of the principle. The transfinite induction principle is a generalization of the strong induction principle for (0, encountered at the end of Chapter 4. The following theorem is somewhat of a converse to transfinite induction. It asserts that the only linear orderings for which the transfinite induction principle is valid are the well orderings. Although we make no later use of this theorem, it is of interest in showing why one might choose to study well orderings. Theorem 7C Assume that < is a linear ordering on A. Further assume that the only < -inductive subset of A is A itself. That is, assume that for any B S A satisfying the condition e:d we have B = A. Then < is a well ordering on A. (\it E A)(seg t s B = t E B), Proof Let C be any subset of A; we will show that either C has a least element or C is empty. We decide (for reasons that may seem mysterious at |
the moment) to consider the set B of "strict lower bounds" ofC: B = {t E A I t < x for every x E C}. Well Orderings 175 Note that B n C = 0, lest we have t < t. We ask ourselves whether condition (u) holds for B. Case I Condition (*) fails. Then there exists some tEA with seg t ~ B but t ¢ B. We claim that t is a least element of C. Since t ¢ B, there is some x E C with x :::; t. But x cannot belong to seg t, which is disjoint from C. Thus x = t and t E C. And t is least in C, since anything smaller than t is in seg t and hence not in C. Case II Condition (*) holds. Then by the hypothesis of the theorem, -1 A = B. Consequently in this case C is empty. Next we turn to the important business of defining a function on a well-ordered structure by transfinite recursion. Assume that < is a well ordering on A. Conceivably we might possess some rule for defining a function value F(t) at tEA, where the rule requires knowing first all values F(x) for x < t. Then we can start with the least element to of A, apply our rule to find F(to), go on the next element, and so forth. That phrase "and so forth" has to cover a great deal of ground. But, because < is a well ordering, after defining F on any proper subset B of A, we always have a unique next element (the least element of A - B) to which we can apply our rule so as to continue. Now let us try to make these ideas more concrete. The "rule" of the preceding paragraph might be provided by a function G. Then F(t) for tEA is to be found by applying G to the values F(x) for x < t: F(t) = G(F ~ seg t). We will say that F is "G-constructed" if the above equation holds for every t E dom F. For the right side of this equation to succeed, the domain of G must contain all functions of the form F ~ seg t. This leads us to define, for a set B, the set < AB of all functions from initial segments of < into B: < A B |
= {f I for some t E A,J is a function from seg t into B}. To check that there is indeed such a set, observe that if!: seg t --> B, then! ~ A x B. Hence <AB is obtainable by applying a subset axiom to.9(A x B). Transfinite Recursion Theorem, Preliminary Form Assume that < is a well ordering on A, and that G: < AB --> B. Then there is a unique function F: A --> B such that for any tEA, F(t) = G(F ~ seg t). 176 7. Orderings and Ordinals Example In the case of the well-ordered set w, we have for each n in W the equation seg n = {x I x En} = n. Hence the above theorem asserts the existence of a unique F: w --> B satisfying for every n in w In particular, we have F(n) = G(F ~ n). F(O) = G(F ~ 0) = G(0), F(1) = G(F ~ 1) = G({<O, F(O)}), F(2) = G(F ~ 2) = G({<O, F(O), (1, F(1)}). This can be compared with the recursion theorem of Chapter 4. There is the difference that in Chapter 4 the value of F(n) was constructed by using only the one immediately preceding value of the function. But now in constructing F(n) we can use all of the previous values, which are given to us by F ~ n. (This alteration is not made solely for reasons of generosity. It is forced on us by the fact that in an arbitrary well ordering, there may not always be an "immediately preceding" element.) Before proving the transfinite recursion theorem, we want to state it in a stronger form. In our informal comments we spoke of a "rule" for forming F(t) from the restriction F ~ seg t. We want to be broad-minded and to allow the case in which the rule is not given by a function. We have in mind such rules as F(t) = {F(x) I x < t} = ran(F ~ seg t). There is no function G such that G(a) = ran a for every set a; such a function would have to have everything |
in its domain, but its domain is merely a set. To be sure, for any fixed set B there is indeed a function G with G(a) = ran a for every a E B. But our desire is to avoid having to produce in advance the set B. In terms of proper classes, we can formulate the improved version of transfinite recursion as follows. Let < be a well ordering on A and let G be a "function-class," i.e., a class of ordered pairs that satisfies the definition of a function except for not being a set. Further suppose that the domain of G includes < AV, where < AV is the class {f I for some t in A,f is a function with dom f = seg t}. Then there exists a unique function F with domain A that is "G-constructed" in the sense that for all tEA. (Here F is a set.) F(t) = G(F ~ seg t) Well Orderings 177 Because we are working in Zermelo-Fraenkel set theory, we must reword the above formulation of transfinite recursion so as to avoid reference to the class G. Instead we allow ourselves only a formula y(x, y) that defines G: G = {<x, y) I y(x, y)}. Applying this rewording (the standard rewording for these predicaments), we obtain the statement below. Transfinite Recursion Theorem Schema For any formula y(x, y) the following is a theorem: Assume that < is a well ordering on a set A. Assume that for any f there is a unique y such that y(j, y). Then there exists a unique function F with domain A such that for all t in A. y(F ~ seg t, F(t)) This is a theorem schema; that is, it is an infinite package of theorems, one for each formula y(x, y). (The formula y(x, y) is allowed to mention other fixed sets in addition to x and y.) We will say that F is y-constructed if the condition y(F ~ seg t, F(t)) holds for every t in dom F. Example We obtain one instance of transfinite recursion by taking y(x, y) to be: y = ran x. Now it is automatically true that for any f there is a unique y such |
that y = ranf. So we are left with the theorem: Assume that < is a well ordering on A. Then there exists a unique function with domain A such that for all t in A. F(t) = ran(F ~ seg t) This is the instance of transfinite recursion we will need in order to make ordinal numbers. Example The preliminary form of transfinite recursion is obtained by choosing the formula y(x, y): y(x, y) ¢> either or (i) x E < AB and y = G(x) (ii) x ¢ < AB and y = 0. Then for any f there is a unique y such that y(j, y). We can conclude from the transfinite recursion theorem that there is a unique function F with domain A that is y-constructed: For tEA, either (i) (F ~ seg t) E < AB 178 7. Orderings and Ordinals and F(t) = G(F ~ seg t) or (ii) (F ~ seg t) ¢ < AB and F(t) = 0. Since G: < AB -> B, we can see (by transfinite induction) that we have alternative (i) for every t in A. Thus F is G-constructed. One momentary drawback to the above statement of transfinite recursion is that we are unable to prove it with the axioms now at our disposal. Actually this drawback is shared by a number of other statements that are intuitively true under our informal ideas about sets. For example, we are unable to prove that o u (JJJ 0 u (JJJ(JJJ 0 u (JJJ(JJJ (JJJ 0 u... is a set. In fact we cannot prove the existence of any inductive set other than ro. These deficiencies will be eliminated in the next section by the replacement axioms. Exercises 4. Let < be the usual ordering on the set P of positive integers. For n in P, let f(n) be the number of distinct prime factors of n. Define the binary relation R on P by mRn ¢> either f(m) <f(n) or [J(m) = f(n) & m < n]. Show that R is a well ordering on P. Does <P, R) resemble any of the pictures in Fig. 45 ( |
p. 185)? 5. Assume that < is a well ordering on A, and thatf: A -> A satisfies the condition x < y = f(x) <f(y) for all x and y in A. Show that x:::; f(x) for all x III A. [Suggestion: Consider f(f(x)).] 6. Assume that S is a subset of the real numbers that is well ordered (under the usual ordering on reals). Show that S is countable. [Suggestion: For each x in S, choose a rational number between x and the next member of S, if any.] 7. Let C be some fixed set. Apply transfinite recursion to ro (with its usual well ordering), using for y(x, y) the formula Let F be the y-constructed function on ro. y = C u UU ran x. (a) Calculate F(O), F(I), and F(2). Make a good guess as to what F(n) is. (b) Show that if a E F(n), then a ~ F(n+). (c) Let C = U ran F. Show that C is a transitive set and that C ~ C. (The set C is called the transitive closure of C, denoted TC C.) Replacement Axioms 179 REPLACEMENT AXIOMS If H is a function and A is a set, then H[A] is a set, simply because it is included in ran H. (Lemma 3D guarantees that ran H is a set.) But now consider a "function-class" H, i.e., a class of ordered pairs that satisfies the definition of being a function, except that it may not be a set. Is it still true that H[A] is a set? With the axioms stated thus far, we are unable to prove that it is. (The un provability of this can actually be proved; we will return to this point in Chapter 9.) But certainly H[ A] cannot be too big to be a set, for it is no larger than the set A. So if we adopt the principle that what distinguishes the sets from the proper classes is the property of being limited in size, then it is eminently reasonable to adopt an axiom that, in a way, asserts that H[A] is a set. Now the axiom in Z |
ermelo-Fraenkel set theory cannot legally refer to a class H; it must instead involve a formula <p that defines H. (You will recall that there was a similar situation in the case of the subset axioms.) For each formula that might define a function-class, we get an axiom. Replacement Axioms For any formula <p(x, y) not containing the letter B, the following is an axiom: \iA[(\ix E A) \iYl \iY2(<P(X, Yl) & <p(x, Y2) = Yl = Y2) = 3B \iy(y E B ¢> (3x E A) <p(x, y))]. Here the formula <p(x, y) is permitted to name sets other than x and y; we could have emphasized this fact by writing <p(x, y, t l'..., tk) and inserting the phrase "\it 1 ••• \itk" at the beginning of the axiom. But <p(x, y) must not mention the set B whose existence is being asserted by the axiom. To translate the replacement axioms into words, define the class H = {<x, y) I x E_A & <p(x, y)}. Then the hypothesis of the axiom, (\ix E A) \iYl \iY2(<P(X, Yl) & <p(x, Y2) = Yl = Y2)' asserts that H is a function-class. And the second line, 3B \iy(y E B ¢> (3x E A) <p(x, y)), asserts that if we let B = {y I (3x E A) <p(x, y)} = H[A], then B is a set. (It then follows that H, being included in A x B, is also a set.) 180 7. Orderings and Ordinals An alternative way of translating replacement into words is as follows. Read <p(x, y) as "x nominates y." Then the hypothesis of the axiom says, "Each member of A nominates at most one object." And the conclusion says, "The collection of all nominees is a set." The name "replacement" reflects the idea of replacing each x in the set A by |
its nominee (if any) to obtain the set B. Example If A is a set, then {£?lIa I a E A} is also a set, by Exercise 10 of Chapter 2. But we now have an easy proof of this fact. Take <p(x, y) to' be y = £?lIx. That is, let each x nominate its own power set. Then replacement tells us that the collection of all power sets of members of A forms a set. Example Let S be any set. Then replacement tells us that {card x I XES} is a set. We take the formula <p(x, y) to be y = card x. Proof of the Transfinite Recursion Theorem The proof is similar, in its general outline, to the proof of the recursion theorem on ro. Again we construct the desired function F as the union of many approximating functions. For t in A, say that a function v is y-constructed up to t iff dom v = {x I x ~. t} and for any x in dom v, y(v ~ seg x, v(x)). 1. First we claim that if t1 ~ t2, v1 is y-constructed up to t1 and v2 is y-constructed up to t 2, then v1(x) = v2 (x) for all x ~ t 1, Should this fail, there is a least x ~ t1 with v1 (x) #- v2(x). By the leastness of x, we have v1 ~ seg x = v2 ~ seg x. Also we have y(v1 ~ seg x, v1 (x)) and whence by our assumption on y we conclude that v1(x) = v2 (x) after all. This establishes the claim.'In particular, by taking t1 = t2 we see that for any tEA there is at most one function v that is y-constructed up to t. We next want to form the set % of all functions v that are, for some t in A, y-constructed up to t: % = {v I (3t E A) v is a function y-constructed up to t}. % is provided by a replacement axiom. Take for <p(t, v) the formula v is a function that is y-constructed up |
to t. We have shown that tEA & <p(t, v1) & <p(t, v2) = v1 = v2. Replacement Axioms 181 Hence by replacement there is a set % such that for any v, VE% ¢> (3tE A) <p(t, v) ¢> (3t E A) v is y-constructed up to t. Now let F be U%, the union of all the v's. Thus <x, y) E F ¢> v(x) = y for some v in %. First observe that F is a function. For suppose that <x, Yl) and <x, Y2) belong to F. By (u) there exist v!' t1' v2' and t2 such that vi(x) = Yi and Vi is y-constructed up to ti for i = 1, 2. Either x:::; tl :::; t2 or x:::; t2 :::; t 1, and in either event we have by our earlier claim Yl = V1 (x) = v2(x) = Y2' 2. Next we claim that for any x E dom F, y(F ~ seg x, F(x)). For if x E dom F, there exists v in % with x E dom v. Then we have y(v ~ seg x, v(x)) since v E %, v ~ seg x = F ~ seg x by h~") and part 1, v(x) = F(x) by h;:{), from which we conclude that y(F ~ seg x, F(x)). 3. We now claim that dom F = A. If this fails, then there is a least tEA - dom F. Then seg t ~ dom F; in fact seg t = dom F. Take the unique Y such that y(F, y) and let v = F u {<t, y)}. We want to show that v is y-constructed up to t. Clearly v is a function and dom v = {x I x :::; t}. For any x < t we have v ~ seg x = F ~ seg x and v(x) = F(x), and so by part 2 we conclude that y(v ~ seg x, v( |
x)). For the case x = t we have v ~ seg t = F and v(t) = y, and so by our choice of y we obtain y(v) ~ seg t, v(t)). Hence v is y-constructed up to t. But this implies that t E dom F after all. 4. Finally we claim that F is unique. For suppose that Fl and F2 both satisfy the conclusion of the theorem. We can apply transfinite induc tion; let B be the set on which Fl and F2 agree: B = {tE A I F1(t) = F2(t)}. It suffices to show that for any tEA, seg t ~ B = t E B. But this is easy. If seg t ~ B, then we have F 1 ~ seg t = F 2 ~ seg t. Also we have and 182 7. Orderings and Ordinals whence by our assumption on y we conclude that F 1{t) = F 2 {t). Hence t E B and we are done. -1 Exercises 8. Show that the subset axioms are provable from the other axioms. 9. Show that the pairing axiom is provable from the other axioms. EPSI LO N-I MAGES2 Our first application of transfinite recursion will be to the construction of the ordinal numbers. Assume that < is a well ordering on A and take for y{x, y) the formula: y = ran x. The transfinite recursion theorem then presents us with a unique function E with domain A such that for any tEA: E{t) = ran{E ~ seg t) = E[seg t] = {E{x) I x < t}. Let rx = ran E; we will call rx the E-image of the well-ordered structure <A, <). (Later rx will also be called an ordinal number, but we want to postpone introducing that terminology.) Example To get some idea of what the E-image might look like, take the three-element set A = {r, s, t}, where r < s < t. Then we calculate E{r) = {E{x) I x < r} = 0, E{s) = {E{x) I x < s} = {E{r)} = {0}, E |
{t) = {E{x) I x < t} = {E{r), E{s)} = {0, {0}}· Thus E{r) = 0, E{s) = 1, E{t) = 2, and the E-image rx of <A, <) is 3. You are invited to contemplate whether natural numbers will always appear in rx like this. In the above example, rx was a set equinumerous with A. Moreover, while < was a well ordering on A, the membership relation Ea = {<x, y) E rx x rx I x E y} provided a well ordering on rx. We will show that this always happens, whence the name "E-image." 2 The membership symbol (E) is not typographically the letter epsilon but originally it was, and the name "epsilon" persists. Epsilon-Images 183 Theorem 7D Let < be a well ordering on A and let E and rx be as described as above. (a) E(t) ¢. E(t) for any tEA. (b) E maps A one-to-one onto rx. (c) For any 8 and t in A, 8 < t iff E(8) E E(t). (d) rx is a transitive set. Proof We prove that E(t) ¢. E(t) by the "least counterexample" method. That is, let S be the set of counterexamples: S = {t E A I E(t) E E(t)}. We hope that S is empty. But if not, then there is a least i E S. Since E(i) E E(i), there is (by the definition of E(i)) some 8 < i with E(i) = E(8). But then E(8) E E(8), contradicting the leastness of i. Hence S = 0, which proves part (a). For part (b), it is obvious that E maps A onto rx; we must prove that E is one-to-one. If 8 and t are distinct members of A, then one is smaller than the other; assume 8 < t. Then E(8) E E(t), but, by part (a), E(t) ¢. E( |
t). Hence E(8) "# E(t). In part (c) we have 8 < t = E(8) E E(t) by definition. Conversely if E(8) E E(t), then (by the definition of E(t)) there is some x < t with E(8) = E(x). Since E is one-to-one, we must have 8 = x and hence 8 < t. Finally for part (d) it is easy to see that rx is a transitive set. If -l U E E(t) E rx, then there is some x < t with u = E(x); consequently u E rx. Define the binary relation on rx: Ea = {<x, y) E rx x rx I x E y}. Under the assumptions of the preceding theorem, Ea is a well ordering on rx. We will postpone verifying this fact until we can do so in a more general setting (see Corollary 7H). Since rx is a transitive set, we can characterize Ea by the condition X Ea Y ¢> X EyE rx. 184 Exercises 7. Orderings and Ordinals 10. For any set S, we can define the relation Es by the equation: Es = {<x, y) E S X S I x E y} (a) Show that for any natural number n, the E-image of <n, En) is n. (b) Find the E-image of <w, E",). (a) Although the set 7l. of integers is not well ordered by its normal ordering, show that the ordering 11. 0,1,2,..., -1, -2, -3,... is a well ordering on 7l.. (b) Suppose that we define the usual function Eon 7l., using the well ordering of part (a). Calculate E(3), E( -1), and E( - 2). Describe ran E. ISOMORPHISMS Theorem 7D told us that a well-ordered structure <A, <) looks a great deal like its E-image <IX, Ea)' We had a one-to-one correspondence E between A and IX with the order-preserving property:. s < t iff E(s) E E(t) for sand t in A. |
In formalizing this concept of" looking alike," we need not restrict attention to well orderings. Instead we might as well formulate our definition in broader terms. Definition Consider structures <A, R) and <B, S). An isomorphism from <A, R) onto <B, S) is a one-to-one function f from A onto B such that xRy iff f(x)Sf(y) for x and y in A. If such an isomorphism exists, then we say that <A, R) is isomorphic to <B, S), written: <A, R) ~ <B, S). Example Let < be a well ordering on A and let IX be the E-image of <A, <). Then Theorem 7D states that <A, <) is isomorphic to <IX, Ea), and that E is an isomorphism. Example Isomorphic structures really do look alike. We can draw pictures (at least in the simple cases) as in Figs. 43 and 44. Figure 45 consists of pictures of four well-ordered structures. Figure 45a shows a well ordering on {r, s, t}, and Fig. 45b shows a well ordering on {x, y, z}. The two structures are isomorphic, and the two pictures are obviously very much alike. Figures 45c and 45d provide pictures of infinite structures. Figure 45c Isomorphisms 185 • t I • s I • r (ab) 3 2 • 0 (c) (3, I) (3,0) (2, I) (2,0) (I, I) (1,0) · • I I (0 I) • (0,0) (d) Fig, 45. Four well orderings, of which two are isomorphic. shows the usual ordering on w. Figure 45d is a picture of w x w with the "lexicographic" (dictionary) ordering <m, n) <L <p, q) iff mE p or (m = p & n E q). Although w is equinumerous to w x w, the pictures do not look alike. For example, in Fig. 45c the set seg n is always finite, whereas in Fig. 45d the set seg<m, n) is infinite for m "# O. And the two structures are not isomorphic. Actually we have used the word "is |
omorphic" before, e.g., in Theorem 4H. Both there and here we are using special cases of the general concept of 186 7. Orderings and Ordinals isomorphism of structures. In Theorem 4H we had "structures" <N, S, e) consisting of a set N together with a function S: N -> N and an element e EN. Now we have struct~res <A, R), where R is a binary relation on A. We will have no need here for greater generality, but you are nonetheless invited to devise general concepts of structure and isomorphism. We have the easy theorem stating that isomorphism is an "equivalence concept." That is, it obeys laws of reflexivity, symmetry, and transitivity. Theorem 7E Let <A, R), <B, S), and <C, T) be any structures. Then <A, R) ~ <A, R), <A, R) ~ <B, S) = <B, S) ~ <A, R), <A, R) ~ <B, S) ~ <C, T) = <A, R) ~ <C, T). Proof For the first of these we can use the identity function on A; it is an isomorphism of <A, R) onto itself. To prove the second we take the inverse of the given isomorphism; to prove the third we take the composition of the two given isomorphisms. -l If two structures are isomorphic, then for any statement that is true of one, there is a corresponding statement that is true of the other. This is because the structures look exactly alike. But to nail down this general principle in precise terms, we would have to say formally what we meant by "true statement" and "corresponding statement." This would lead us away from set theory into logic. Instead we will just list (in Theorem 7G) some particular instances of the general principle. Lemma 7F Assume that f is a one-to-one function from A into B, and that < B is a partial ordering on B. Define the binary relation < A on A by X<AY iff f(x) <Bf(y) for x and y in A. (a) The relation < A is a partial ordering on A. (b) (c) If < B is a linear ordering on B, then < |
A is a linear ordering on A. If < B is a well ordering on B, then < A is a well ordering on A. Proof The proof is straightforward and is left, for the most part, to the exercises. We will give details here for only the following part: If < B is a well ordering on B, then any nonempty subset S of A has a least (with respect to <) element. First note that f[S] is a nonempty subset of B. Hence it has a least element, which must bef(m) for some unique m E S. We claim that m is least in S. Let x be any other member of S. Then Ordinal Numbers 187 f{x) E f[S], and by the leastness of f{m) we have f{m) ::; B f{x). In fact, f{m) < Bf{x) since f is one-to-one. Hence m < AX, which shows that m is indeed least in S. -l Theorem 7G Assume that structures <A, < A) and <B, < B) are isomorphic. If one is a partially (or linearly or well) ordered structure, so also is the other. Proof Suppose that one is a partially ordered structure; by symmetry we may suppose that it is <B, < B)' We are given an isomorphism f of <A, < A) onto <B, < B)' Thus for X and y in A. Hence we can apply the above lemma. X < AY iff f{x) < Bf{Y) Theorem 7D told us that well orderings were isomorphic to their E-images. In light of the above theorem, we can now conclude that the E-images are well ordered by epsilon. Corollary 78 Assume that < is a well ordering on A and let rx be the E-image of <A, <). Then rx is a transitive set and Ea is a well ordering on rx. Exercises. 12. Complete the proof of Lemma 7F. 13. Assume that two well-ordered structures are isomorphic. Show that there can be only one isomorphism from the first onto the second. 14. Assume that <A, <) is a partially ordered structure. Define the function F on A by the equation F{a) = {x |
E A I x ::; a}, and let S = ran F. Show that F is an isomorphism from <A, <) onto <S, c s )' ORDINAL NUMBERS It will be extremely useful in our discussion of well orderings if we can assign a "number" to each well-ordered structure that measures its "length." Two well-ordered structures should receive the same number iff they look alike, i.e., iff they are isomorphic. The situation is much like that in Chapter 6, where we assigned cardinal numbers to sets so as to measure their size. Two sets received the same cardinal number iff they were equinumerous. The key to the present situation lies in the following theorem. 188 7. Orderings and Ordinals Theorem 71 Two well-ordered structures are isomorphic iff they have the same E-image. That is, if < 1 and < 2 are well orderings on Al and A2, respectively, then <AI' < 1) ~ <A2'< 2) iff the E-image of <AI' < 1) is the same as the E-image of <A 2, < 2)' Proof In one direction, this follows at once from the fact that well ordered structures are isomorphic to their E-images. That is, if <AI' < 1) and <A 2, < 2) have the same E-image (x, then <AI' < 1) ~ <(X, Ea) ~ <A2, < 2)' For the other direction, assume thatfis an isomorphism from <AI' < 1) onto <A2, < 2)' LetE 1 : Al -> (Xl and E2: A2 -> (X2 be the usual isomorphisms of the well orderings onto their E-images: El(S) = {El(X) I X <1 s} E2(t) = {E2(Y) I Y <2 t}. and Then we can use transfinite induction to show that El (s) = E2(f(S)) for all sEAl' For let B be the set on which this equation holds: B = {SE Al I El(S) = E2(f(S))}. Suppose that seg S ~ B. Then we calculate El(S) = {El(X) I X <1 S} = {E 2 |
(f(x)) I x < 1 S} = {E2(Y) I Y <2f(s)} = E2(f(S)). by definition since seg S ~ B (see below) The third line on this calculation requires some thought. Clearly {E2(f(X)) I x < 1 S} ~ {E2(Y) I Y < 2 f(s)} because we can take Y = f(x). Conversely, the other inclusion holds because fmaps onto A2 and therefore any y less thanf(s) must equalf(x) for some (unique) x less than s. Thus if seg s ~ B, then El (s) = E2(f(S)) and so s E B. By the transfinite induction principle, B = AI' i.e., for all sEAl' Consequently, (Xl = {E 1 (s) I sEA I} = {E2(f(S)) I SEAl} = {E2(t) I tE A2} = (X2' Thus the E-images coincide. sincefmaps onto A2 Ordinal Numbers 189 The solution to our problem of how to assign "numbers" to well orderings is provided by the foregoing theorem. We just use E-images. Definition Let < be a well ordering on A. The ordinal number of <A, <) is its E-image. An ordinal number is a set that is the ordinal number of some well-ordered structure. (More generally, we could seek to define an "order type" ot<A, R) whenever R was a partial ordering on A. The order type should measure the "shape" of R, in the sense that ot<A, R) = ot<B, S) iff <A, R) ~ <B, S). This can be done; see Exercise 32. But the methods that are required are quite unlike the methods used for ordinal numbers.) We can generate examples of ordinal numbers by calculating E-images. The example preceding Theorem 7D shows that 3 is an ordinal number. And study of Fig. 45c will show that w is an ordinal number. More examples will appear presently. First we need some additional information about well orderings. If < is any ordering on A and C is a subset of A, then the elements of C are, of course, still |
ordered by <. Now it is not strictly true that < is an ordering on C, because (if C "# A) < is not a binary relation on C. But < n (C x C) is an ordering on C. This notation is excessively cumber some for the simplicity of the ideas involved. And so we define <C, <0) = <C, < n (C x C). The symbol ° reminds us that the relation < can be cut down to fit the set C. In particular, if C = seg c, then we have the ordered structure <seg c, <0). Theorem 7J Assume that < is a partial ordering on A, and that C ~ A. Then <0 is a partial ordering on C. This continues to hold with "partial" replaced by "linear" or "well." Proof This is an immediate consequence of Lemma 7F, wherein we -l take f to be the identity function on C. The following theorem is a sort of trichotomy law for well-ordered structures. Theorem 7K For any two well-ordered structures, either they are isomorphic or one is isomorphic to a segment of the other. More precisely, let < A and < B be well orderings on A and B. Then one of the following 190 7. Orderings and Ordinals alternatives holds: <A, <,4) ~ <B, <B)' <A, <,4) ~ <seg b, <~) for some b E B, <seg a, <~) ~ <B, < B) for some a E A. Proof The idea is to start pairing elements of A with elements of B in the natural way: We pair the least elements of the two sets, then the next smallest, and so forth. Eventually we run out of elements on one side or the other. If the two sets run out simultaneously, then <A, <,4) ~ <B, < B). Otherwise one set runs out first; in this case it is isomorphic to a segment of the other. The pairing is obtained by transfinite recursion. Let e be some extraneous object belonging neither to A nor to B. By transfinite recursion there is a unique function F: A -> B u {e} such that for each tEA, F(t) = lethe least element of B - F[seg t] \ ifany, if B - F[ |
seg t] is empty. Case leE ran F. Let a be the least element of A for which F(a) = e. We claim that F ~ seg a is an isomorphism from <seg a, <~) onto <B, <B). Let r = F ~ seg a. Clearly FO: seg a -> B, and its range is all of B since B - F[seg a] = 0. We have x:::;,4 y <,4 a = F[seg x] ~ F[seg y] = B - F[ seg y] ~ B - F[ seg x] = F(x):::; B F(y). Furthermore if x <,4 y <,4 a then F(x) "# F(y), because F(x) E F[seg y] but by construction F(y) ¢. F[seg y]. Therefore FO is one-to-one, and in fact x <,4 y <,4 a = F(x) < B F(y). Consequently, FO preserves order, since F(x) < B F(y) = F(y) $. B F(x) = y$.,4x = x <,4y. Thus r is an isomorphism as claimed. Case II ran F = B. Then F: A -> B, and just as in Case I (but without the a) F is one-to-one and preserves order. Thus F is an isomorphism from <A, <,4) onto <B, < B)· Ordinal Numbers 191 Case III Otherwise ran F is a proper subset of B. Let b be the least ran F. We claim that ran F = seg b. By the leastness of b, we element of B have seg b ~ ran F. Conversely, we cannot have b < B F(x) for any x, because F(x) is least in B - F[seg x], a set to which b belongs. Hence ran F = seg b. As in Cases I and II, F is one-to-one and preserves order. Hence F is -l an isomorphism from <A, <,4) onto <seg b, < ~). We now return to the study of ordinal numbers. The next theorem will show that any ordinal number |
is its own E-image. To extract the maximum of information from the proof, the following definition will help. Definition A set A is well ordered by epsilon iff the relation E,4 = {<x, y) E A x A I x E y} is a well ordering on A. We have shown that any ordinal number is a transitive set that is well ordered by epsilon (Corollary 7H). The converse is also true; the ordinal numbers are exactly the transitive sets that are well ordered by epsilon: Theorem 7L Let rx be any transitive set that is well ordered by epsilon. Then rx is an ordinal number; in fact rx is the E-image of <rx, Ea)' Proof Let E be the usual function from rx onto its E-image. We can use transfinite induction to show that E is just the identity function on rx. Note that for t E rx, because rx is a transitive set. As a consequence we have seg t = t. If the equation E(x) = x holds for all x in seg t, then E(t) = {E(x) I XEa t} = {x I x Ea t} = seg t = t. Hence by transfinite induction, E is the identity function on rx, so the E-image of < rx, Ea) is just rx itself.'-l The class of all ordinal numbers is (as we will prove shortly) too large to be a set. But except for not being a set, it has many of the properties of the ordinal numbers themselves. The next theorem shows that the class of all ordinal numbers is a transitive class well ordered by epsilon. 192 7. Orderings and Ordinals Theorem 7M The following are valid for any ordinal numbers rx, /3, and y. (a) (b) (c) (d) (transitive class) Any member of rx is itself an ordinal number. (transitivity) (irreflexivity) (trichotomy) Exactly one of the alternatives, rx E /3 E Y = rx E y. rx ¢. rx. rx E /3, rx = /3, holds. (e) Any nonempty set S of ordinal numbers has a |
least element 11 (i.e., 11 ~ rx for every rx E S). Proof For part (a), consider any x in rx. Now rx is the E-image of some well-ordered structure <A, <). So x = E(t) where E is the usual isomorphism and t is some member of A. We will prove that x is an ordinal by showing that it is the E-image of <seg t, <0). By Tqeorem 7J the restriction < ° is a well ordering on seg t. The E-image of <seg t, < 0) is easily seen to be E[seg t], but this is just E(t), which is x. Hence x is an ordinal. Part (b) is true because the ordinal y is a transitive set. Part (c) follows from Theorem 7D(a). If rx E rx, then rx = E(t) for some t (in the notation used above in proving part (a)). But then E(t) E E(t), which contradicts Theorem 7D(a). In trichotomy, the "at most one" part follows from transitivity and irreflexivity (as in Theorem 7A). The "at least one" halfis a consequence of Theorem 7K, which tells us that either <rx, Ea) and </3, Ep) are isomorphic, or else one is isomorphic to a segment of the other. Case I <rx, Ea) ~ </3, Ep). Then both have the same E-image, by Theorem 71. But ordinals are their own E-images by Theorem 7L, so rx = /3. Case II By symmetry we may suppose that <rx, Ea) is isomorphic to <seg b, E~), where bE /3. But then b is an ordinal, seg b = b, and Ep = Ed. Thus we have <rx, Ea) ~ <b, Ed). Now by Case J, rx = b. Hence rx E /3. For part (e), first consider some arbitrary /3 E S. If /3 n S = 0, then we claim that fJ is least in S. This is because rxES = rx¢'/3 = /3~rx by trichotomy. |
There remains the possibility that /3 n S "# 0. Then we have here a nonempty subset of /3. Hence /3 n S has a least (with respect to Ep) Ordinal Numbers 193 member 11. We claim that 11 is least III S. To verify this, consider any (1.E S: (1.¢f3 => f3~(1. => f3t;;(1. => Il E (1., (1.Ef3 => (1.Ef3nS => Il~~ Either way, 11 ~ (1.. Corollary 7N ordinal number. (a) Any transitive set of ordinal numbers is itself an (b) 0 is an ordinal number. (c) (d) number. If (1. is any ordinal number, then (1.+ is also an ordinal number. If A is any set of ordinal numbers, then U A is also an ordinal Proof (a) We can conclude from the preceding theorem that any set of ordinals is well ordered by epsilon. If, in addition, the set is a transitive set, then it is an ordinal number by Theorem 7L. (b) 0 is a transitive set of ordinals. (It is the E-image of the well ordered structure <0, 0)·) (c) Recall that (1.+ = (1. U {(1.}. The members of (1.+ are all ordinals (by Theorem 7M(a)). And (1.+ is a transitive set by Theorem 4E (compare Exercise 2 of Chapter 4). (d) We can also apply part (a) to prove part (d). Any member of UA is a member of some ordinal, and so is itself an ordinal. To show that U A is a transitive set, we calculate ()EUA => => => ()E(1.EA ()t;;(1.EA ()t;;UA. for some (1. since (1. is transitive Thus by part (a), UA is an ordinal. In fact more is true here. If A is any set of ordinals, then U A is an ordinal that is the least upper bound of A. To verify this, first note that the epsilon ordering on the class of all ordinals is also defined by c. That is, for ordinals (1. and |
13 we have: (1.Ef3 => (1.cf3 since 13 is a transitive set and (1. "# 13 => => f3i(1. (1.Ef3 lest 13 E 13 by trichotomy. Thus (1. E 13 iff (1. c 13. Since U A is the least upper bound for A in the inclusion ordering, it is the least upper bound for the epsilon ordering. 194 7. Orderings and Ordinals In the same vein, we can assert that IX+ is the least ordinal larger than IX. Clearly IX E IX+, so IX+ is larger than IX. Suppose IX E 13 for some ordinal 13. Then IX ~ 13 as well, and hence IX+ ~ 13. As noted above, this yields the fact that IX+ § 13. This shows that IX+ is the least ordinal larger than IX. Observe also that any ordinal is just the set of all smaller ordinals. That is: IX = {x I x E IX} = {x I x is an ordinal & x E IX}. Since our ordering relation is E, this can be read as, "IX equals the set of ordinals less than IX." At long last we can get a better picture of what the ordinals are like. The least ordinal is O. Next come 0+, 0++, •.•, i.e., the natural numbers. The least ordinal greater than these is the least upper bound of the set of natural numbers. This is just,Uw, which equals w. (Or more directly: w is an ordinal and the ordinals less than ware exactly its members.) Then come w +, W + +, •••• All of these are countable ordinals (i.e., ordinals that are countable sets). The least uncountable ordinal n is the set of all smaller ordinals, i.e., is the set of all countable ordinals. But here we are on thin ice, because we have not shown that there are any uncountable ordinals. That defect will be corrected in the next section. BuraH-Forti Theorem There is no set to which every ordinal number belongs. Proof Theorem 7M told us that the class of all ordinals was a transitive class well ordered by epsilon. If it were a set, then by Theorem 7L it would be an ordinal itself |
. But then it would be a member of itself, and no ordinal has that property. -l This theorem is also called the "Burali-Forti paradox." Like the later paradox of Russell, it showed the inconsistency of Cantor's casual use of the abstraction principle, which permitted speaking of "the set of all ordinals." The theorem was published in 1897 by Burali-Forti, and was the first of the set-theoretic paradoxes to be published. Exercises 15. (a) Assume that < is a well ordering on A and that tEA. Show that <A, <) is never isomorphic to <seg t, < 0). (b) Show that in Theorem 7K, at most one of the three alternatives holds. 16. Assume that IX and 13 are ordinal numbers with IX E 13. Show that IX+ E 13+. Conclude that whenever IX"# 13, then IX+ "# 13+. Debts Paid 195 17. Assume that <A, <) is a well-ordered structure with ordinal number rx and that B ~ A. Let 13 be the ordinal number of the well-ordered structure <B, < 0) and show that 13 § rx. 18. Assume that S is a set of ordinal numbers. Show that either (i) S has no greatest element, Us ¢. S, and Us is not the successor of another ordinal; or (ii) Us E S and Us is the greatest element of S. 19. Assume that A is a finite set and that < and -< are linear orderings on A. Show that <A, <) and <A, -<) are isomorphic. 20. Show that if Rand R -1 are both well orderings on the same set S, then S is finite. 21. Prove the following version of Zorn's lemma. Assume that < is a partial ordering on A. Assume that whenever C is a subset of A for which < ° is a linear ordering on C, then C has an upper bound in A. Then there exists a maximal element of A. DEBTS PAID In this section we will payoff the debts incurred in Chapter 6. We will define cardinal numbers as promised, and we will complete the proof of Theorem 6M. The first theorem assures an adequate supply of ordinal numbers. Recall that B is dominated by A (B ~ A |
) iff B is equinumerous to a subset of A. Hartogs'Theorem Forany set A, there is an ordinal not dominated by A. Proof There is a systematic way to construct the least such ordinal rx. Any ordinal that is dominated by A is smaller than (i.e., is a member of) rx. And conversely, any ordinal smaller than (i.e., a member of) rx should be dominated by A, if rx is to be least. Hence we decide to try defining rx = {f3 I 13 is an ordinal & 13 ~ A}. The essential part of the proof is showing that rx is a set. After all, the theorem is equivalent to the assertion that rx is not the class of all ordinals. We begin by defining W = {<B, <) I B ~ A & < is a well ordering on B}. We claim that any member of rx is obtainable as the E-image of something in w. (Thus rx is, in a sense, no larger than W.) Consider any ordinal 13 in rx. Then some function f maps 13 one-to-one onto a subset B of A. There is a well ordering < on B such that f is an isomorphism from <13, Ep) onto <B, <), namely < = {<f(x),J(y) I x EyE f3}. 196 7. Orderings and Ordinals Thus <B, <) E Wand the E-image of <B, <) is the same as the E-image of <13, Ep), which is just 13· Thus 13 is indeed the E-image of a member of W. W is a set, because if <B, <) E W, then <B, <) E gilA x £!l!(A x A). We can then use a replacement axiom to construct the set tff' of E-images of members of W. The preceding paragraph shows that rx s tff'. (In fact it is not hard to see that rx = tff'.) Since rx is a set, we can conclude from the Burali-Forti theorem that some ordinals do not belong to rx. This proves the theorem, but we will continue and show that rx itself is the least such ordinal. Clearly rx is a set of ordinals |
, and since yEf3Erx = ysf3~A = yE~.rx is transitive. Hence rx is an ordinal. We could not have rx ~ A, lest rx belong to itself. And rx is the least such ordinal, since 13 E rx = 13 ~ A by construction. -l Hartogs' theorem does not require the axiom of choice. But the following theorem does. Well-Ordering Theorem For any set A, there is a well ordering on A. This theorem is often stated more informally: Any set can be well ordered. As we will observe presently, the well-ordering theorem is equivalent to the axiom of choice. We can prove the well-ordering theorem from Zorn's lemma (Exercise 22). But in order to complete the proof of Theorem 6M, we will instead use the axiom of choice, III. Proof Let G be a choice function for A, and let rx be an ordinal not dominated by A (furnished by Hartogs' theorem). Our strategy is to order A by first choosing a least element, then a next-to-least, and so forth. For the "and so forth" part, we use rx as a base; rx is large enough that we will exhaust A before coming to the end of rx. Let e be an extraneous object not belonging to A. We can use transfinite recursion to obtain a function F: rx -> A u {e} such that for any y E rx, F(y) = {~(A - F[y]) if A - F[y]"# 0, if A - F[ y] = 0. In other words, F(y) is the chosen member of A - F[y], until A is exhausted. Suppose y E 13 E rx. If neither F(y) nor F(f3) equals e, then F(y) "# F(f3), since F(y) E F[f3] but F(f3) ¢. F[f3]. That is, F is one-to-one until e appears. It follows that e E ran F, lest F be a one-to-one function from rx into A, contradicting the fact that A does not dominate rx. Debts Paid 197 Let b be the least element of rx for which F( |
b) = e. Then F[b] = A. And by the preceding paragraph, F ~ b is a one-to-one map from b onto A. This produces a well ordering on A, namely < = {<F(P), F(y) I P EyE b}. Then F ~ b is an isomorphism from <b, Ed) onto <A, <). Example The well-ordering theorem claims that there exists a well ordering W on the set IR of real numbers. Now the usual ordering on IR is definitely not a well ordering, so W is some unusual ordering. What is it? We have a theorem asserting the existence of W, but its proof (which used choice) gives no clues as to just what W is like. It is entirely possible that there is no formula of the language of set theory that will explicitly define a well ordering on IR. The following variant of the well-ordering theorem follows easily from it. Numeration Theorem Any set is equinumerous to some ordinal number. I Proof Consider any set A; by the preceding theorem there is a well ordering < on A. Then A is equinumerous to the E-image rx of <A, <). The name "numeration theorem" reflects the possibility of "counting" the members of A by using the ordinal numbers in rx, where a "counting" is a one-to-one correspondence. The numeration theorem produces a satisfactory definition of cardinal number. Definition For any set A, define the cardinal number of A (card A) to be the least ordinal equinumerous to A. The following theorem makes good our promise of Chapter 6. Theorem 7P (a) For any sets A and B, card A = card B iff A:::::: B. (b) For a finite set A, card A is the unique natural number equi numerous to A. Proof First, note that by our definition, A :::::: card A for any set A. Hence if card A = card B, then we have A :::::: card A = card B :::::: B. Conversely if A :::::: B, then A and B are equinumerous to exactly the same ordinals. Hence the least such ordinal is the same in both cases. 198 7. Orderings and Ordinals Part (b) follows from the fact that the natural numbers are also ordinal numbers. If A is finite, then ( |
by definition) A:::::: n for a unique natural number n. A is not equinumerous to any smaller ordinal, since the smaller ordinals are just the natural numbers in n. Hence card A = n. -l Proof of Theorem 6M, concluded Figure 46 shows how Fig. 40 is to be extended. Here "WO" is the well-ordering theorem, which we proved from (3). It remains to establish (5) = WO and WO = (6). Proof of (5) = WO Given any set A, we obtain from Hartogs' theorem an ordinal rx not dominated by A. By (5), we have A ~ rx; i.e., there is a 3.. 4 / WO Fig. 46. The proof of Theorem 6M. one-to-one function f from A into rx. This induces a well ordering of A; define 8 < t ¢> f(8) Ef(t) for 8 and t in A. Then < is a well ordering of A (by Lemma 7F). Proof of WO= (6) Consider any d that is closed under unions of chains. By WO there exists a well ordering < of d. We want to make a very large chain C(} S d such that UC(} is a maximal element. The idea is to go through the members of d in order, adding to C(} any member of d that includes each set previously placed in C(}. This construction uses trans finite recursion. That is, the transfinite recursion theorem gives us a function F: d --> 2 such that for any A E d, F(A) = l~ if A includes every set B < A for which F(B) = 1, otherwise. Then let C(} = {A E dl F(A) = I}. Thus F is the characteristic function of C(}, and for A E d, A E C(} ¢> A includes every B < A for which BE C(}. We claim that UC(} is a maximal element of d. Debts Paid 199 First of all, C(} is a chain. For if A and B are in C(}, then and A5,B = AsB is a chain, UC(} E d. To prove maximality, suppose that Since C(} UC(} s DE d. Then DEC(}, since D includes every |
member of C(}. Hence D S UC(}, and so D = UC(}. Thus UC(} is not a proper subset of any member of d. -l B 5, A = Bs A. There are several easy consequences of our definition of cardinal numbers. Say that an ordinal number is an initial ordinal iff it is not equinumerous to any smaller ordinal number. Then any initial ordinal is its own cardinal number. Conversely any cardinal number must, by the definition, be an initial ordinal. Thus the cardinal numbers and the initial ordinals are exactly the same things. Next consider any cardinal number, say K = card S. Then K ~ S (and K is the least ordinal with this property). Also card K = K, sinceK is an initial ordinal. That is, K is itself a set of cardinality K. The cardinal numbers, like any other ordinal numbers, are well ordered by the epsilon relation as described in Theorem 7M. It is now easy to verify that this ordering agrees with the ordering of cardinal numbers defined in Chapter 6. For any cardinals K and A, Conversely, KEA = KCA = K~A&K#A = K5,A&K#A = K < A. Kt-A = AEK = A5,K = K 1- A. by the above In particular, any nonempty set of cardinal numbers has a smallest element. And the word "smallest" is used unambiguously here, since the epsilon ordering and the cardinality ordering agree. The smallest cardinal after ~o is of course denoted as ~ l' We can also characterize ~ 1 as the least uncountable ordinal. Next in order of magnitude comes ~2' And this continues. In Chapter 8, we define ~a for every ordinal rx. Exercises 22. Prove the well-ordering theorem from the version of Zorn's lemma given in Exercise 21. 200 7. Orderings and Ordinals 23. Assume that A is a set and define (as in Hartogs' theorem) rx to be the set of ordinals dominated by A. Show that (i) rx is a cardinal number, (ii) card A < rx, and (iii) rx is the least cardinal greater than card A. 24. Show that for any ordinal number rx, there is a cardinal number K that is (as an |
ordinal) larger than rx. 25. the following holds: (Transfinite induction schema) Let <p(x) be a formula and show that Assume that for any ordinal rx, we have (Vx E rx)<p(x) => <p(rx). Then <p(rx) for any ordinal rx. RANK In this final section of the chapter we want to return to the informal view of sets mentioned in Chapter I-and especially to Fig. 3. The description that was somewhat vague at the start of this book can now be made quite precise. We want to define for every ordinal number rx the set ~. Vo is to be empty, and, in general, ~ is to contain those sets whose members are all in some "P for f3 less than rx. Thus we want aE~ ¢> ac;;"P ¢> a E &"P for some f3 E rx for some f3 E rx, or equivalently, Theorem 7U will show that in particular cases, this equation can be simplified. If rx = {) +, the equation will become whereas if rx is not the successor of another ordinal, then the equation will become ~ = U{Vp I f3 E rx} = U Vp' pea Thus V3 = &&&0 and VO) = Vo U VI U V2 U.... But we are getting ahead of ourselves. We have yet to show how to define ~. The equation Rank 201 defines ~ in terms of Vp for f3 less than rx. So we need to define ~ by transfinite recursion. The class of all ordinals is well ordered, so transfinite recursion over this class should give us a function-class F such that F(rx) = U{&F(f3) I f3 E rx} for each ordinal rx. But there is an obvious hitch here. We cannot work directly with classes that are not sets. Instead we will sneak up gradually on F. Lemma 7Q For any ordinal number b there is a function Fd with domain b such that for every rx E b. Proof We apply transfinite recursion. On b we have the well ordering Ed' Take for y(x, y) the formula y = U{&z I ZE ran x}. We can show that {&z I Z E ran x |
} is a set by showing that it is included in &&U(ran x). But an easier proof uses a replacement axiom: Let the formula w = &Z. Then because ran x <p(z, w) be is a set, {&z I Z E ran x} is also a set. Obviously for any f there is a unique y such that y(f, y), namely y = U{&z I Z E ranf}· Transfinite recursion then gives us a function Fd such that for every rx E b. Now then, seg rx = {t I t Ed rx} = {t I t E rx} = rx. Thus Z E ran(Fd ~ seg rx) ¢> ¢> Z E ran(Fd ~ rx) Z = Fd(f3) for some f3 E rx. So h~") becomes as desired. Fd(rx) = U{&z I Z = Fd(f3) for some f3 E rx} = U{g;> Fd(f3) I f3 E rx} 202 7. Orderings and Ordinals Now consider any two ordinals band 8. The smaller of the two is b n 8. (Why?) We have two functions Fd and F. from the preceding lemma. These functions agree on b n 8: Lemma 7R Let band 8 be ordinal numbers; let Fd and F. be functions from Lemma 7Q. Then for all rx E b n 8. Proof By the symmetry, we suppose that b ~ 8. Hence b £; 8 and b n 8 = b. We will establish the equation Fd(rx) = F.(rx) by using transfinite induction in <b, Ed). Define B = {rx E b I Fd(rx) = F.(rx)}. In order to ~how that B = b, it suffices to show that B is "Ed-inductive," i.e., that seg rx ~ B = rx E B for each rx E b. We calculate: seg rx £; B = Fd(P) = F.(P) for P E rx = U{&Fd(P) I P E rx} = U{&F.(P) I P E rx} = Fd(rx |
) = F.(rx) = rxEB for rx E b. And so we are done. In particular (by taking b = 8) we see that the function Fd from Lemma 7Q is unique. We can now unambiguously define ~. Definition Let rx be an ordinal number. Define ~ to be the set Fd(rx), where b is any ordinal greater than rx (e.g., b = rx+). Theorem 78 For any ordinal number rx, ~ = U{&Yp I P E rx}. Proof Let b = rx+. Then ~ = Fd(rx) and Yp = Fd(P) for P E rx. Hence the -l desired equation reduces to Lemma 7Q. Lemma 7T For any ordinal number rx, ~ is a transitive set. Proof We would like to prove this by transfinite induction over the class of all ordinals. This can be done by utilizing Exercise 25. But we can also avoid that exercise by proving for each ordinal b that ~ is a transitive set whenever rx E b. This requires only transfinite induction over <b, Ed). Rank 203 Let B be the set on which the conclusion holds: B = {IX E b I ~ is a transitive set}. We want to show that B = b; this will follow from the transfinite induction theorem if we show that B is "Ed-inductive," i.e., that IXs:B = IXEB for all IX E b. (Recall that seg IX = IX.) Suppose then that IX S: B. Then for any 13 E IX, "P is a transitive set, as is &"P (by Exercise 3 of Chapter 4). It follows that ~ is a transitive set, because XE ~ = XE&"P = x S: &"P = x S: ~. for some 13 E IX for some 13 E IX Thus IX E B, and we are done. There are three sorts of ordinal number. First, there is 0, called zero. Second, there are the ordinals of the form IX + for a smaller ordinal IX. These are called the successor ordinals. Third, there are all the others, called limit ordinals. For example, the least limit ordinal is w. If A. is a limit ordinal and 13 E A., then 13 + E |
A.. This is because p+, being the least ordinal larger than 13, must satisfy p+ ~ A.. And p+ "# A. since A. is a limit ordinal. The following theorem describes ~ for each of the three sorts of ordinal number. (a) For ordinals 13 E IX we have "P S: ~. Theorem 7U (b) Vo = 0· (c) ~+ = &~ for any ordinal number IX. (d) V;. = Up e;' "P for any limit ordinal A.. Proof For part (a), first observe that "P E &Vp S: ~, so that Vp E ~. Since ~ is a transitive set, we also have "P S: ~. Part (b) is clear. For part (c), we first obtain from part (a) 13 E IX = "P S: ~ = &"P S: &~. Now ~+ = U{&"P I 13 ~ IX} and this union is equal to the largest set, &~. Finally for part (d) we have one inclusion: X E V;. = X E &"P = XE"P+ = XE U Vp pe;' for some 13 E A. by part (c) since 13 + E A.. 204 7. Orderings and Ordinals The other inclusion is similar: XE UYp = XEYpSYp+=&Yp PEA. for some 13 E A. = XE VA.. Thus equality holds. We will say that a set A is grounded iff A S ~ for some ordinal number rx. In this event, we define the rank of A (rank A) to be the least such ordinal rx. Thus for a grounded set A: A S Yrank A and A E "(rank A)+ • The number rank A indicates how many times the power set operation must be used to obtain all the members of A. For example, every ordinal rx is grounded and rank rx = rx (Exercise 26). There is one methodological objection that can be raised concerning the definition of rank. We want rank A to be the least member of {rx I A S ~}. But (for a grounded set A) this is not a set of ordinals, but a proper class of ordinals. Nonetheless, it has a least member. Being nonempty, it has some member 13 |
. Either 13 is its least member, or {rx E 13 I A S ~} is a nonempty set of ordinals. In the latter case, we can apply Theorem 7M(e) to obtain a least element. Theorem 7V (a) If a E A and A is grounded, then a is grounded and rank a E rank A. (b) If every member of A is grounded, then A is also grounded and rank A = U{(rank a)+ I a E A}. Proof For the first part, assume that A is grounded and that a E A. Then As Yrank A = a E Yrank A = a E &Yp = as Yp = rank a E rank A. for some 13 E rank A for some 13 E rank A For part (b), assume that every member of A is grounded. Define rx = U{(rank at I a E A}. To show that rx is a set, we use a replacement axiom first to conclude that {(rank at I a E A} is a set. Then by the union axiom, rx is a set. Rank 205 By Corollary 7N(d), rx is an ordinal. First we claim that A ~ ~. (This will show that A is grounded and that rank A ~ rx.) We calculate a E A = a ~ ~ank a = a E "(rank a) + = a E ~ by Theorem 7U (a), since (rank at E rx for each a E A. Thus A ~ ~. To show that rx ~ rank A, we use part (a). a E A = = rank a E rank A (rank a)+ ~ rank A. Hence rank A is an upper bound for the ordinals (rank a)+, and so is at least as large as their least upper bound rx. -l Part (b) of the foregoing theorem can be translated into words as: The rank of a set is the least ordinal that is strictly greater than the ranks of all members of the set. Theorem 7W The following two statements are equivalent. (a) Every set is grounded. (b) (Regularity) Every nonempty set A has a member m with mn A=0. Proof First assume that every set is grounded, and let A be a nonempty set. The idea is to take mEA having the least possible rank 11. Thus from {rank |
alaEA}, we take the least member 11, and then we select mEA having rank m = 11· If x E m, then by Theorem 7V(a), rank x Ell. We cannot have x E A, due to the leastness of 11. Hence m n A = 0 as desired. For the converse, assume that (b) holds. Suppose that, contrary to our expectations, some set c is not grounded. Then some set (e.g., {c}) has a nongrounded member. And hence some transitive set B has a nongrounded member; we can take B to be the transitive closure of {c} (see Exercise 7). Let A = {x E B I x is not grounded}. Since A is nonempty, by (b) there is some mEA with m n A = 0. We claim that every member of m is grounded. If x Em, then x E B since B is transitive. But x ¢. A because m n A = 0, so x must be grounded. Since every member of m is grounded, we can conclude from Theorem -l 7V(b) that m is also grounded. This contradicts the fact that mEA. 206 7. Orderings and Ordinals Near the bottom of p. 8, we stated as "a fundamental principle" that every set is grounded. So on an informal level at least, we must accept both parts of the above theorem as true. But neither part is provable from the axioms adopted up to now. To correct this defect, we will now adopt the regularity axiom. For this axiom we could use either (a) or (b) from the above theorem. We select (b) because it has a more elementary formulation. Regularity Axiom Every nonempty set A has a member m with mn A=0. The regularity axiom is also known as the foundation axiom or the Fundierungsaxiom. The idea first appeared in a 1917 paper by Mirimanoff; the axiom was listed explicitly in von Neumann's 1925 paper. The concept of rank, as well as the concept of regularity, appeared in Mirimanoff's 1917 paper. The idea of rank is a descendant of Russell's concept of type. The following theorem lists some basic consequences of regularity. (a) No set is a member of itself. Theorem 7X (b) There do not exist any sets a |
and b with a E band bE a. (c) There is no function f with domain w such that... E f(2) E f(l) E f(O). Proof For part (c), suppose to the contrary thatf(n+) E f(n) for each nEW. Let A = ranf. Any member m of A must equal f(n) for some n. Thenf(n+) E m n A, so that m n A is always nonempty. This contradicts regularity. Parts (a) and (b) can be proved either by similar arguments or as -l consequences of part (c). We leave the details as an exercise. Since every set is grounded, the sets are arranged in an orderly hierarchy according their rank. This is the situation that Fig. 3 attempts to illustrate. Thus the universe of all sets is, in a sense, determined by two factors: the extent of the class of all ordinals, and the variety of subsets that are assigned to a set by the power set 1. 2. operation. Rank Exercises 207 26. Show that every ordinal number rx is grounded, and that rank rx = rx. 27. Show that the set IR of real numbers, as constructed in Chapter 5, has rank w + + + + + 28. Show that ~ = {X I rank X E rx}. 29. Prove parts (a) and (b) of Theorem 7X. 30. Show that for any sets rank{a, b} = max (rank a, rank bt, rank!!J>a = (rank at, rank Ua ~ rank a. 31. Define kard A to be the collection of all sets B such that (i) A is equinumerous to B, and (ii) nothing of rank less than rank B is equinumerous to B. (a) Show that kard A is a set. (b) Show that kard A is nonempty. (c) Show that for any sets A and B, kard A = kard B iff A:::::: B. 32. Let <A, R) be a structure. Define the isomorphism type it<A, R) of this structure to be the set of all structures <B, S) such that (i) <A, R) is isomorphic to <B, S), and (ii) nothing of rank less |
than rank<B, S) IS isomorphic to <B, S). (a) Show that it<A, R) is a set (and not a proper class). (b) Show that it<A, R) is nonempty. (c) Show that it<A, R) = it<B, S) iff <A, R) ~ <B, S). 33. Assume that D is a transitive set. Let B be a set with the property that for any a in D, a ~ B = aE B. Show that D ~ B. 34. Assume that and show that x = u and y = v. {x, {x, y}} = {u, {u, v}} 208 7. Orderings and Ordinals 35. Show that if a+ = b+, then a = b. 36. Show that the rank of any set S is the same as the rank of its transitive closure TC S (as defined in Exercise 7). 37. Show that a set rx is an ordinal number iff it is a transitive set with the property that for any distinct x and y in rx, either x E y or y E x. 38. Show that whenever A. is a limit ordinal, then A. = UA.. 39. Prove that a set is an ordinal number iff it is a transitive set of transitive sets. ORDINALS AND ORDER TYPES CHAPTER 8 This chapter ends with a discussion of the arithmetic of ordinal numbers. That topic is preceded by material that is useful for ordinal arithmetic, and that also has independent interest. The logical dependencies among the sections ofthis chapter are as follows. The section on alephs depends on the preceding section on transfinite recursion. The section on ordinal operations can be independent ofthe others, but it refers to the section on alephs for examples. The section on isomorphism types is independent of the earlier sections, and in turn forms the basis for the following section on the arithmetic of order types. The section on ordinal arithmetic refers to the preceding section for addition and multiplication, and to transfinite recursion for exponentiation. But it is indicated how all of the operations can be based on transfinite recursion, thereby avoiding order types. In either event, the section on ordinal operations is utiliz'ed here. TRANSFINITE RECURSION AGAIN In Chapter 7 we |
defined a set ~ for each ordinal rY.. That definition proceeded in a somewhat roundabout manner, defining first a function 209 210 8. Ordinals and Order Types Fd for each b, and then proving that Fd(lX) was actually independent of b (as long as b was greater than IX). We could then define ~ to be Fd(lX) for any large b. It is interesting to note that the goal of the construction was not a function but a definition. We could not hope to have a function F with F(IX) = ~ for every IX, because the domain of F would have to be the proper class of all ordinals. We now want a general transfinite recursion theorem that will yield directly the definition of ~, and will be applicable to other cases as well. In terms of proper classes, we can state this theorem as follows: Let G be afunction-class whose domain is the class V of all sets. Then we claim that there is a function-class F whose domain is the class of all ordinals and such that for every IX. F(IX) = G(F ~ IX) Now we must apply the standard rewording. In place of G we have a formula y(x, y) that defines G. The assumption that G is a function-class with domain V is reworded to state that for any x there is a unique y such that y(x, y) holds. In place of F we also have a formula <p(x, y) defining F. For every ordinal IX there must be a unique y such that <p(IX, y) holds. The equation is reworded to state that whenever f is a function with domain IX such that <p(f3, f(f3)) holds for all 13 E IX (i.e., whenever f = F ~ IX) then the unique y such that <p(IX, y) equals the unique y such that y(j, y). Transfinite Recursion Schema on the Ordinals Let y(x, y) be any formula. Then we can construct another formula <p(u, v) such that the following is a theorem. Assume that for every f there is a unique set y such that y(j, y). Then for every ordinal number IX there is a unique y such that <p(IX, y). Furthermore whenever f is a function whose domain is an ord |
inal number IX and such that <p(f3, f(f3)) for all 13 less than IX, we then have for every y. <p(IX, y) iff y(j, y) We will give the proof presently. The significance of the theorem is that it justifies making a certain definition. We select some available symbol, say" t," and define the term ta = the unique y such that <p(IX, y) Transfinite Recursion Again 211 for every ordinal rx. The operation of going from rx to ta is y-constructed in the following sense. Whenever f is the restriction of the operation to some ordinal rx (i.e., domf = rx and f(f3) = tp for each 13 less than rx), then we can conclude from the above theorem that y(j, ta). We might abbreviate this by writing y(t ~ rx, ta)' Example We will apply the foregoing theorem to construct ~. For y use the formula we used before (in Lemma 7Q), so that y(j, y) is y = U{&z I z E ran f}· Now plug this into the theorem and the subsequent discussion, but select the symbol" V" instead of" t." We get a set ~ for each rx, but we must use the fact that the operation is y-constructed to be sure that we have the right ~. So consider any fixed rx and let f be the restriction of the operation to rx: f(f3) = Vp for 13 less than rx. Then we can conclude that y(j, ~), which when expanded becomes ~ = U{&z I z E ranf} = U{&Vp I 13 E rx}. Comparing this equation with Theorem 7S, we see (by transfinite induction) that we have the same ~ as in Chapter 7. Proof of the theorem The formula <p(rx, y) is the following: There exists an ordinal b greater than rx and there exists a y-constructed function F6 with domain b such that F6(rx) = y. As in Chapter 7, F6 is said to be y-constructed iff y(F6 ~ 13, FAf3)) for every 13 E b. First we will show that these F6 functions always exist. We are given that for every f |
there is a unique y such that y(j, y). Hence for any ordinal b we can apply transfinite recursion (from Chapter 7) to the well ordered structure <b, E6). By doing so, we obtain the unique y-constructed function F6 with domain b. Now suppose we are given some ordinal rx and we want to find some y such that <p(rx, y). We choose b to be any larger ordinal (such as rx+) and take y = F6(rx). Then clearly <p(rx, y) holds. Next we will prove a uniqueness result. Namely, we claim that if Fd and F. are y-constructed functions with domains band 8, respectively, then F6(rx) = F.(rx) for all rx in b n 8. (This corresponds to Lemma 7R in the special case of ~.) The claim is proved by transfinite induction. For the 212 8. Ordinals and Order Types inductive step, suppose that FAP) = F.(P) for all P E rx. Then F6 ~ rx = F. ~ 0:. Since both y(F6 ~ rx, FArx)) and y(F. ~ rx, F.(rx)), we can conclude that F6(rx) = F.(rx). It now follows that for any ordinal rx there is a unique y such that <p(~, y), and we know what this unique y is. Finally assume that f is a function whose domain is rx and that <p(P, f(P)) holds for all P E rx. We must show that <p(rx, y) ~ y(j, y) for each y. But what is f? It must be the function Fa' by the above. The only y for which y(Fa' y) is FArx), where {) is some larger ordinal. And it is also the only y for which <p(rx, y). -l In the next section we will see further examples of the use of transfinite recursion on the ordinals. ALEPHS Suppose that we have defined some class A of ordinals. Now A might or might not be a set. If A is bounded, i.e., if there is an ordinal P such that rx E P for each rx in A, then A ~ P+ and |
consequently A is a set. On the -:-other hand, if A is unbounded, then UA is the class of all ordinals. (This holds because any p, failing to bound A, is less than some rx in A. Thus P E rx E A and P E UA.) By the Burali-Forti theorem, A cannot be a set. Thus we can conclude that A is a set iff it is bounded. Now focus attention on the case where we have defined an unbounded (and hence proper) class of ordinals. As an example we will take the class of infinite cardinal numbers, but later we can apply our work to other classes. The class of infinite cardinal numbers is unbounded. (One way to prove this fact is to observe that for any ordinal P we obtain from Hartogs' theorem the least ordinal rx not dominated by p. Then rx is an initial ordinal and is larger than p. An alternative proof uses 2card p.) The class of infinite cardinal numbers, like any class of ordinals, inherits the well ordering by epsilon on the class of all ordinals, given by Theorem 7M. We want to enumerate its members in ascending order. There is a least infinite cardinal, which we have always called ~o. Then there is a next one (the least infinite cardinal greater than ~o), which we call ~ l'And so forth. We will now expand that "and so forth." Suppose we have worked our way up to rx, i.e., we have defined ~P for all P less than rx and we are ready to try defining ~a' (In the preceding paragraph rx = 2.) Naturally we define ~a = the least infinite cardinal different from ~P for every P less than rx. Alephs 213 Such a cardinal must exist, because {~p I 13 E IX} is merely a set, whereas the class of infinite cardinals is unbounded. Now that we know how to construct ~a from the smaller alephs, we can apply transfinite recursion on the ordinals. Choose for y(j, y) the formula y is the least infinite cardinal not in ranf. Then for any set f, there is a unique y such that y(j, y), again because ran f is merely a set whereas the class of infinite cardinals is unbounded. Then transfinite recursion lets us pick a symbol ( |
we pick "~") and define ~a for each IX in a way that is },-constructed. That is, if f is the function with domain IX defined by f(f3) = ~p for 13 E IX, then y(j, ~J And y(j, ~a)' when written out, becomes ~a is the least infinite cardinal not in ran f, which by the definition off becomes ~a is the least infinite cardinal different from ~p for every 13 less than IX. The following theorem verifies that this construction enumerates the infinite cardinals in ascending order. Theorem 8A (b) Every infinite cardinal is of the form ~a for some IX. If IX E 13, then ~a < ~p. (a) Proof (a) This is a consequence of the fact that ~a is the least cardinal meeting certain conditions that become more stringent as IX increases. Both ~a and ~p meet the condition of being different from ~y for all y less than IX. Since ~a was the least such candidate (and ~p "# ~a)' we have ~a < ~p' (b) We use transfinite induction on the class of infinite cardinal numbers (a subclass of the ordinals). Suppose, as the inductive hypothesis, that K is an infinite cardinal for which all smaller infinite cardinals are in the range of the aleph operation. Consider the corresponding set {f3l ~p < K} of ordinals. This is a set (and not a proper class), being no larger than K. And it is a transitive set by part (a). So it is itself an ordinal; call it IX. By construction, ~a is the least infinite cardinal different from ~p whenever ~p < K. By the inductive hypothesis, this is the least infinite cardinal different from all those less than K, which is just K itself. -l The construction up to here is applicable to any unbounded class A of ordinals that we might have defined. We take for y(j, y) the formula y is the least member of A not in ran f. 214 8. Ordinals and Order Types Transfinite recursion then lets us define ta for every rx in such a way that ta is the least member of A different from tp for every 13 less than rx. The analogue of Theorem 8A holds (and with the same proof), showing that we have constructed an enumeration of A in ascending order |
. Now that we have a legal definition of ~a' we can try to see what that definition can give us. There are three possibilities for rx: zero, a successor ordinal, or a limit ordinal. We already know about ~o, so consider the case where rx is a successor ordinal, say rx = 13+. By the construction and Theorem 8A, ~p+ is the least infinite cardinal greater than ~y for every y less than 13 +. We can simplify this since the largest candidate for ~y is ~p. Thus ~p+ = the least cardinal greater than ~p. By Exercise 23 of Chapter 7, the least cardinal greater than ~p, as a set, is the set of all ordinals dominated by the set ~p. Now consider the case of limit ordinal A.. As before, ~). is the least infinite cardinal greater than ~p for every 13 less than A.. We claim that in fact ~).=U~p· pe). From Chapter 7 we know that Up e). ~p is the least ordinal greater than ~p for every 13 in A.. The following lemma shows that it is actually a cardinal number. Lemma 8B The union of any set of cardinal numbers is itself a cardinal number. Proof Let A be any set of cardinals. Then UA is an ordinal by Corollary 7N(d). We must show that it is an initial ordinal. So assume that rx E UA; we must show that rx *' UA. We know that rx EKE A for some cardinal number K. Thus rx ~ K ~ UA, so that if rx ~ UA, then rx ~ K. But it is impossible to have rx ~ K, since K is an initial ordinal. Hence rx *' UA. -1 As another application of transfinite recursion on the ordinals, we can define the beth numbers :Ja. (The letter :J, called beth, is the second letter of the Hebrew alphabet.) The following three equations describe the term :Ja that we want to define: :Jo=~o, :Ja + = 2J :J). = U :Ja, ", ae). Ordinal Operations 215 where A. is a limit ordinal. The transfinite recursion machinery lets us convert these three equations into a legal definition. To operate the machinery, let y(f, y) be the formula: Either or (i |
) (ii) or or (iii) (iv) f is a function with domain 0 and y = ~o'f is a function whose domain is a successor ordinal IX + and y = 2f (a), f is a function whose domain is a limit ordinal A., and y = U(ran f), none of the above and y = 0. Then transfinite recursion gives us a formula qJ with the usual properties. We select the symbol :J and define :Ja = the unique y such that q>(IX, y). Then the machinery tells us that "y(:J ~ IX, :JJ" Writing out this condition with IX = 0 produces the equation :Jo = ~o' Similarly, by using a successor ordinal and then a limit ordinal we get the other two equations o, :Ja + = 2J :J,\= U :Ja · ilEA. The continuum hypothesis (mentioned in Chapter 6) can now be stated by the equation ~1 = :J 1, and the generalized continuum hypothesis is the assertion that ~a = :Ja for every IX. (We are merely stating these hypotheses as objects for consideration; we are not claiming that they are true.) Exercises 1. Show how to define a term ta (for each ordinal IX) so that to = 5, ta+ = (tat, and t,\ = Uae,\ ta for a limit ordinal A.. 2. that if w § IX, then ta = IX. In the preceding exercise, show that if IX E w, then ta = 5 + IX. Show ORDINAL OPERATIONS There are several operations on the class of ordinal numbers that are of interest to us. A few of these operations have already been defined: the successor operation assigning to each ordinal p its successor p+, and the aleph and beth operations assigning to p the numbers ~p and :Jp. More examples will be encountered when we study ordinal arithmetic. 216 8, Ordinals and Order Types These operations do not correspond to functions (which are sets of ordered pairs), because the class of all ordinals fails to be a set. Instead for each ordinal p we define ap ordinal tp as the unique ordinal meeting certain specified conditions. We will say that the operation is monotone iff the condition 0( E P = ta E tp always holds. We will say that the operation is continuous iff the equation tA = U |
tp peA holds for every limit ordinal A.. Finally we will say that the operation is normal iff it is both monotone and continuous. Example If 0( E p, then 0(+ E p+ (by Exercise 16 of Chapter 7). Thus the successor operation on the ordinals is monotone. But it lacks continuity, because w+ =f. Unewn+ = w. Example The aleph operation is normal by Theorem 8A (for mono tonicity) and Lemma 8B with the accompanying discussion (for continuity). More generally, whenever the ta's enumerate an unbounded class A of ordinals in ascending order, then monotonicity holds (by the analogue of Theorem 8A). Pursuing the analogy, we can assert that continuity holds if A is closed in the sense that the union of any nonempty set of members of A is itself a member of A. Conversely whenever we have defined some normal ordinal operation ta, then its range-class {ta 10( is an ordinal number} is a closed unbounded class of ordinal numbers (Exercise 6). Recall (from the discussion following Corollary 7N) that whenever S is a set of ordinal numbers, then Us is an ordinal that is the least upper bound of S. It is therefore natural to define the supremum of S (sup S) to be simply US, with the understanding that we will use this notation only when S is a set of ordinal numbers. For example, w = sup{O, 2, 4,... }. The condition for an operation taking p to tp to be continuous can be stated for limit ordinals A.. The following result shows that for a continuous operation, the condition for monotonicity can be replaced by a more "local" version. Ordinal Operations 217 Theorem Schema 8C Assume that we have defined a continuous operation assigning an ordinal tp to each ordinal number f3. Then the operation is monotone provided that ty E ty+ for every ordinal y. Proof We consider a fixed ordinal IX and prove by transfinite induction on p that IX E P = ta E tp' Case I p is zero. Then the above condition is vacuously true, since IX f/: p. Case II P is a successor ordinal y +. Then IXEP = a§y = ta § ty = ta E ty+ = Case III P is a limit ordinal. |
Then ta E tp' by the inductive hypothesis SInce ty E ty+ IX E P = = IX E IX+ E P taEta+§tp by continuity. Hence ta E tp. -l Example The beth operation (assigning::1p to P) is normal. The continuity is obvious from its definition. And since ::1y is always less than 2:1." we have (oy the above theorem) monotonicity. The next results will have useful consequences for the arithmetic of ordinal numbers. Theorem Schema 8D Assume that we have defined a normal operation assigning an ordinal ta to each ordinal IX. Then for any given ordinal P that is at least as large as to, there exists a greatest ordinal y such that ty § p. We know that any nonempty set of ordinals has a least member, but the above theorem asserts that the set {y I ty § P} has a greatest member. Proof Consider the set {IX I ta E P}. This is a set of ordinals and it is (by monotonicity) a transitive set:-So it is itself an ordinal. It is not 0, because to § p. Could it be a limit ordinal A.? If so, then t;, = sup{ta IIX E A.} §p, 218 8. Ordinals and Order Types whence A. E A.. This is impossible, so our set must in fact be a successor ordinal y +. Thus y is the largest member of the set, and so is the largest ordinal for which ty.§ p. -l Theorem Schema 8E Assume that we have defined a normal operation assigning ta to each ordinal number IX. Let S be a nonempty set of ordinal numbers. Then Proof We get the " ~" half by monotonicity: tsup S = sup{ta IIX E S}. IX E S = = IX.§ sup S ta.§tsuPS' whence sup{ ta IIX E S}.§ tsup s· For the other inequality, there are two cases. If S has a largest member 15, then sup S = 15 and so tsups = to.§ sup{ta IIX E S}. If S has no largest member, then sup S must be a limit ordinal (since S #- 0). So by continuity, tsups = sup{tp I p E sup |
S}. If P E sup S, then P EyE S for some y. Consequently tp E t y'and so Since P was an arbitrary member of sup S, tp E sup{ta IIX E S}. sup{tp I PE sup S}.§ sup{ta IIX E S} as desired. -l For a monotone operation, we always have P.§ tp'by Exercise 5. Is it possible to have P = tp? Yes; the identity operation (assigning P to P) is normal. But even for normal operations with rapid growth (such as the beth operation), there are "fixed points" P with P = t p' The following theorem, although not essential to our later work, shows that in fact such fixed points form a proper class. Veblen Fixed-Point Theorem Schema (1907) Assume that we have defined a normal operation assigning ta to each ordinal number IX. Then the operation has arbitrarily large fixed points, i.e., for every ordinal number P we can find an ordinal number y with ty = Y and P.§ y. Proof From the monotonicity we have P.§ tp' If P = tp' we are done just take y = p. So we may assume that p E tp' Then by monotonicity We claim that the supremum of these ordinals is the desired fixed point. PEt p E tIp E.. '. Ordinal Operations 219 More formally, we define by recursion a function f from w into the ordinals such that f(O) = f3 and Thus f(l) = t p'f(2) = tIp' an?.so forth.+ We will write tp for f(n). As mentioned above, by monotomclty tp E tp, so that f3 E tpE t~ E.... Define A. = sup ranf = sup{tp I nEw}. We claim that t;, = A.. Clearly A. has no largest element, and hence is a limit ordinal. Let S = ranf = {tp I nEw}, so that A. = sup S. By Theorem 8E, t;, = sup{ta IIX E S} = sup{tp+ I nEw} = A.. Thus we have a fixed point. -l This proof actually gives us the least fixed |
point that is at least as large as f3 (by Exercise 7). For example, consider the aleph operation with f3 = O. Then o E ~oE ~NoE'" and the supremum of these numbers is the least fixed point. Since the class of fixed points of the operation t is an unbounded class of ordinals, we can define the derived operation t' enumerating the fixed points in ascending order. The definition of t' is produced by the transfinite recursion machinery; the crucial equation is t~ = the least fixed point of t different from tp for every f3 E IX. It turns out (Exercise 8) that t' is again a normal operation. Exercises 3. Assume that we have defined a monotone operation assigning ta to each ordinal IX. Show that and for any ordinals f3 and y. 4. Assume that we have defined a normal operation assigning ta to each IX. Show that whenever A. is a limit ordinal, then t;, is also a limit ordinal. 220 8. Ordinals and Order Types 5. Assume that we have defined a monotone operation assigning ta to each IX. Show that p § tp for every ordinal number p. 6. Assume that we have defined a normal operation assigning ta to each IX. Show that the range-class {ta IIX is an ordinal number} is a closed unbounded class of ordinal numbers. 7. Show that the proof of Veblen's fixed-point theorem produces the least fixed point that is at least as large as p. 8. Assume that we have defined a normal operation assigning ta to each IX. Let t' be the derived operation, enumerating the fixed points of t in order. Show that t' is again a normal operation. ISOMORPHISM TYPES We have ordinal numbers to measure the length of well orderings. If you know the ordinal number of a well-ordered structure, then you know everything there is to know about that structure, "to within isomorphism." This is an informal way of saying that two well-ordered structures receive the same ordinal iff they are isomorphic (Theorem 7I). Our next undertaking will be to extend these ideas to handle structures that are not well ordered. Although our intent is to apply the extended ideas to linearly ordered structures, the initial definitions will be quite general. Consider then a structure <A, R |
). The "shape" of this structure is to be measured to within isomorphism by it<A, R), the isomorphism type of <A, R). This is to be defined in such a way that two structures receive the same isomorphism type iff they are isomorphic: it<A, R) = it<B, S) iff <A, R) ~ <B, S). As a first guess, we could argue as follows: Isomorphism,is an equivalence concept, so why not take it<A, R) to be the equivalence class of <A, R)? Then as in Lemma 3N, two structures should have the same equivalence class iff they are isomorphic. In one way, this first guess is a very bad idea. The "equivalence classes" here will fail to be sets. If A is nonempty, then no set contains every structure that is isomorphic to <A, R); this is Exercise 9. With but one modification, this first guess will serve our needs perfectly. Do not take all structures isomorphic to <A, R), just those of least possible rank IX. Then these structures will be in ~+, and so it<A, R) will be a set. We now will write this down officially. Isomorphism Types 221 Definition Let R be a binary relation on A. The isomorphism type it<A, R) of the structure <A, R) is the set of all structures <B, S) such that (i) <A, R) ~ <B, S), and (ii) no structure of rank less than rank<B, S) is isomorphic to <B, S). It will help to bring in (but only for very temporary use) some terminology. Call a structure pioneering iff there is no structure of smaller rank isomorphic to it. If two pioneering structures are isomorphic, then clearly they have the same rank. We claim that any structure <A, R) is isomorphic to some (not necessarily unique) pioneering structure. The class {O( 10( is the rank of a structure isomorphic to <A, R)} is nonempty (rank<A, R) is in it). So it has a least element 0(0' which we can call the pioneer ordinal for <A, R). Some structure of rank 0(0 is isomorphic to <A, R), and it is pioneering |
by the leastness of 0(0 • The definition of it<A, R) can now be phrased: It is the set of all pioneering structures that are isomorphic to <A, R). This set has been seen to be nonempty. I Note that it<A, R) is indeed a set, being a subset of ~+, where 0( is the pioneer ordinal for <A, R). Theorem SF Structures <A, R) and <B, S) have the same isomorphism type iff they are isomorphic: it<A, R) = it<B, S) iff <A, R) ~ <B, S). Proof First assume that it<A, R) = it<B, S). Let <C, T) be any structure in this common isomorphism type. Then <A, R) ~ <C, T) ~ <B, S). For the converse, assume that <A, R) ~ <B, S). Then the same structures are isomorphic to each of these two; in particular, the same pioneering structures are isomorphic to each. Hence it<A, R) = it<B, S). -l Digression Concerning Cardinal Numbers The crucial property of cardinal numbers, card A = card B iff A ~ B, looks a lot like Theorem 8F, simplified by omission of the binary relations. This indicates the possibility of an alternative definition of cardinal numbers. 222 8. Ordinals and Order Types Specifically, define kard A to be the set of all sets B equinumerous to A and having the least possible rank. Then (as in Theorem 8F), kard A = kard B iff A ~ B. In comparing card A with kard A, we see that the definition of card A uses the axiom of choice (but not regularity). The definition of kard A relies on regularity but does not require the axiom of choice. (This is a point in favor of kard.) For a finite nonempty set A, kard A fails to be a natural number. (This is a point in favor of card.) Exercises 9. Assume that <A, R) is a structure with A =f. 0. Show that no set contains every structure isomorphic to <A, R). to. (a) Show that kard n = {n} for n = 0, 1, and |
2. (b) Calculate kard 3. ARITHMETIC OF ORDER TYPES We now focus attention solely on the order types. Definition An order type is the isomorphism type of some linearly ordered structure. We will use Greek letters p, (1, • • • for order types. Any member of an order type p is said to be a linearly ordered structure of type p. First we want to define addition of order types. The basic idea is that p + (1 should be the order type "first p and then (1." For the real definition of p + (1, first select <A, R) of type p and <B, S) of type (1 with An B = 0. (This is possible, by Exercise 11.) Then define the relation R tB S by R tB S = R u S u (A x B). (We reject the cumbersome R AtB BS notation.) Finally we define p + (1 by p + (1 = it<A u B, R tB S). The idea is that we want to order A u B in such a way that any member of A is less than any member of B. But within A and B, we order according to Rand S. The next lemma verifies that p + (1 is a well-defined order type. Lemma 8G Assume that <A, R) and <B, S) are linearly ordered structures with A and B disjoint. Arithmetic of Order Types 223 (a) R EB S is a linear ordering on A u B. (b) If <A, R) ~ <A', R'), <B, S) ~ <B', S'), and A' n B' = 0, then <A u B, R EB S) ~ <A' u B', R' EB S'). Proof (a) First of all, REB S is irreflexive because a pair of the form <x, x) cannot belong to R or S, nor can it belong to A x B since A and B are disjoint. To check that REB S is transitive, suppose that <x, y) and <y, z) belong to REB S. If both of these pairs come from R (or from S), then of course xRz (or xSz). Otherwise one of the pairs comes from A x B. We may suppose that <x, |
y) E A x B; the other case is similar. Then x E A and the pair <y, z) comes from S (because y E B). Hence z E B and <x, z) E A x B. Finally, it is easy to see that REB S is connected on A u B. Consider any x and y in A u B. There are three cases: both in A, both in B, or mixed. But all three are trivial. Part (b) is sufficiently straightforward so that nothing more need be 1 said. The role of this lemma is to assure us that when we define P + (J = it<A u B, R EB S), the end result will be independent of just which structures of type p and (J are utilized. Example For each ordinal IX, we have the order type it<lX, Ea)' Distinct ordinals yield distinct order types, since it<lX, Ea) = it</3, Ep) = <IX, Ea) ~ </3, Ep) = 1X=/3 for ordinals, by Theorems 71 and 7L. The order type it<lX, Ea) will be denoted as iX. In particular, we have the order types I, 3, and w (which are itO, E 1) and so forth). And I + 3 = 4. (Please explain why.) Also I + w = w, whereas w + 1= w+. Example It is traditional to use '1 and A to denote the order type of the '1 = it<lQl, <Q) rationals and reals, respectively (in their usual ordering): A = it<lR, <R)' and Then 1+'1 is an order type with a least element but no greatest element. (Or more pedantically, the ordered structures of type 1+'1 have least elements but not greatest elements.) But '1 + I has a greatest element but no least element; hence 1+'1 =I- '1 + 1. Also A + A =I- A, because in type A + A there is a bounded nonempty set without a least upper bound. On the other hand '1 + '1 = '1. This is not obvious, but see Exercise 19. 224 8. Ordinals and Order Types Any order type p can be run backwards to yield an order type p*. More specifically, we |
can select a linearly ordered structure <A, R) of type p. Then we define p* = it<A, R - 1). It is routine to check that p* is a well-defined order type (compare Exercise 43 of Chapter 3). For any finite ordinal IX, we have a* = a by Exercise 19 of Chapter 7. But w* =f. w; in fact w* is the order type of the negative integers, which are not well ordered. It is easy to see that '1* = '1 and A.* = A.. Example The sum w* + w is the order type of the set Z of integers. But w + w* is different; it has both a least element and a greatest element, and infinitely many points between the two. Now consider the multiplication of order types. The product p. (J can be described informally as "p, (J times." More formally, we select structures <A, R) and <B, S) of types p and (J, respectively. Then define R * S to be "Hebrew lexicographic order" on A x B: <ai' bi)(R * S)<a 2, b2 ) iff either b i Sb 2 or (b i = b2 and a i Ra 2 ). This orders the pairs in A x B according to their second coordinates, and then by their first coordinates when the second coordinates coincide. We can now define the product p. (J = it<A x B, R * S). The next lemma verifies that p. (J is a well-defined order type. This lemma does for multiplication what Lemma 8G does for addition. Lemma 88 Assume that <A, R) and <B, S) are linearly ordered structures. (a) R * S is a linear ordering on A x B. (b) If <A, R) ~ <A', R') and <B, S) ~ <B', S'), then <A x B, R * S) ~ <A' x B', R' * S'). Proof (a) It is easy to see from the definition of R * S that it is irreflexive (because both Rand S are) and is connected on A x B. It remains to verify that it is a transitive relation. So assume that <ai' bi )(R * S)<a 2, b2 |
) and <a 2, b2 )(R * S)<a 3, b3 ). This assumption breaks down into four cases, as illustrated in Fig. 47. But in each case, we have <ai' bi)(R * S)<a 3, b3 ). Part (b) is again sufficiently straightforward that nothing more need be -l said. Arithmetic of Order Types 225 (a 2, b2>(R * S)(a 3, b 3> ~ ______ ~A~ ______ ~, b2 Sb 3 b2 = b 3 & a 2 Ra 3 b l Sb 2 b l Sb 3 b l Sb 3 b l = b2 & a l Ra 2 b l Sb 3 b l = b 3 & a l Ra 3 Fig. 47. Transitivity of R * S. Examples The product 0·2 is "0, 2 times." The set w x 2 under Hebrew lexicographic ordering is: <0,0) (1,0), (2,0),..., <0, 1), (1,1), (2, 1),.... On the other hand 2· 0 lexicogra phic order is: is "2, 0 times." The set 2 x w under Hebrew <0,0) (1,0), <0,1), (1,1), <0,2), (1,2),.... This ordering, unlike the other, is isomorphic to the natural numbers, i.e., 2. 0 = 0. In particular, 2. 0 =f. 0. 2. The next theorem gives some of the general laws that addition and multiplication obey. We have already seen that neither operation obeys the commutative law in general; for example, and Furthermore the right distributive law fails; for example, (0 + f). 2 =f. 0' 2 + f· 2 (Exercise 15). Theorem 81 The following identities hold for any order types. (a) Associative laws (p + 0') + r = p + (0' + r), (p. 0'). r = p. (0'. r). (b) Left distributive law p. (0' + r) = (p. 0') + (p. r). 226 8. Ordinals and Order Types (c) Identity elements p + 0 = 0 + p = p, p' 1 = 1· p = p, p |
' 0 = o· p = O. Proof Both (p + 0') + rand p + (0' + r) produce the "first p then 0' and then r" ordering. Let <A, R), <B, S), and <C, T) be of type p, 0', and r, respectively, with A, B, and C disjoint. Then both (R EB S) EB T and R EB (S EB T) are easily shown to be CuB x C. Both (p' 0'). rand p. (0'. r) produce Hebrew lexicographic order on A x B x C. (R * S) * T is an ordering on (A x B) x C, and R * (S * T) is an ordering on A x (B x C), but we can establish an isomorphism. In detail, the condition for <ai' bi, Ci ) to be less than <a 2, b2, c 2 ) under (R * S) * T is, when expanded, ci TC 2 or (c i = c2 & bi Sb 2 ) or (c i = c2 & bi = b2 & a i Ra 2 ). The condition for <ai' <bi' Ci» to be less than <a 2, <b2, c2» under R * (S * T) is, when expanded, exactly the same. In the left distributive law, both R * (S EB T) and (R * S) EB (R * T) are orderings on the set (A x B) u (A x C). And in both cases the condition for <ai' x) to be less than <a 2, y) turns out to be (x E B & y E C) or xSy or xTy or (x = y & a i Ra 2 ). Part (c) is straightforward. -l Exercises 11. Show that for any order types p and 0' there exist structures <A, R) and <B, S) of types p and 0', respectively, such that A n B = 0. 12. Prove that for any linearly ordered structures, it<A, R) + it<B, S) = it«{O} x A) u ({I} x B), <L)' where < L is lexicographic ordering. 13. Supply proofs for part ( |
b) of Lemma 8G and part (b) of Lemma 8H. 14. Assume that p. 0' = O. Show that either p = 0 or 0' = O. 15. Show that (0 + 1). 2 is not the same as (0' 2) + (1' 2). 16. Supply a proof for part (c) of Theorem 81. Ordinal Arithmetic 227 17. A partial ordering R is said to be dense iff whenever xRz, then xRy and yRz for some y. For example, the usual ordering <Q on the set iQl of rational numbers is dense. Assume that <A, R) is a linearly ordered structure with A countable and R dense. Show that <A, R) is isomorphic to <B, <(2) for some subset B of iQl. [Suggestion: Suppose that A = {ao, ai'... }. Define f(aJ by recursion on i.J 18. Assume that <A, R) and <B, S) are both linearly ordered structures with dense orderings. Assume that A and B are countable and nonempty. Assume that neither ordering has a first or last element. Show that <A, R) ~ <B, S). [Suppose that A = {a o, ai'... } and B = {bo, bi,... }. At stage 2n be sure that an is paired with some suitable bj and at stage 2n + 1 be sure that bn is paired with some suitable ai.J 19. Use the preceding exercise to show that '1 + '1 = '1. '1 = '1. ORDINAL ARITHMETIC There are two available methods for defining addition and multiplication on the ordinals. One method uses transfinite recursion on the ordinals, extending the definitions by recursion used for the finite ordinals in Cha pter 4. The second method uses our more recent work with order types, and defines IX + p to be the ordinal y such that Fi + II = y (under addition of order types). Since both methods have their advantages, we will show that they produce exactly the same operations. We will then be able to draw on either method as the occasion demands. We will say that an order type p is well ordered iff the structures of type p are well ordered |
. Theorem 8J The sum and the product of well-ordered types is again well ordered. Proof Assume that <A, R) and <B, S) are well-ordered structures with -A and B disjoint. What is to be proved is that <A u B, REB S) and <A x B, R * S) are well-ordered structures. If C is a nonempty subset of A u B, then either C n A =I- 0 (in which case its R-Ieast element is (R EB S)-least in C) or else C ~ B (in which case its S-least element is (R EB S)-least. Similarly if D is a nonempty subset of A x B, we first take the S-least bo in ran D. Let ao be the R-Ieast member of {a I <a, bo) ED}. Then <ao' bo) is the (R * S)-least element of D. -l The above theorem lets us transfer addition and multiplication from well ordered types directly to the ordinals. For ordinals IX and p, we have the well- 228 8. Ordinals and Order Types ordered types Fi. + 11 and Fi. • 11. The ordinals of these types are defined to be IX + P and IX • p. Definition Let IX and p be ordinal numbers. Define the sum IX + p to be the unique ordinal y such that Fi. + 11 = y. Define the product IX' P to be the unique ordinal b such that Fi.·11 = b. 'To be more specific, observe that Fi. + 11 is the order type of ({O} x IX) u ({I} x P) in lexicographic order < L (compare Exercise 12). Consequently IX + p is the ordinal number of the structure «{O} x IX) u ({I} x P), <L)' This is the ordinal number that measures the "first IX and then p" ordering. Similarly Fi. ·11 is the order type of IX x P with Hebrew lexicographic order <H' Consequently IX • P is the ordinal number of the structure <IX x p, < H)' This is the ordinal number that measures the "IX, P times" ordering. We use the same symbols + and, for both order type arithmetic and ordinal arithmetic, but we will try |
always to be clear about which is intended. Example To calculate the sum 1 + W, we shift to the order types r + w. From a previous example r + w = w, so going back to ordinals we have 1 + W = w. Similarly from the known equations w + r = W"" and :2. w = w, we obtain the corresponding equations for ordinals: W + 1 = W + and 2·w=w. Example We claim that IX + 1 = IX+ for any ordinal IX. The sum Fi. + r is the order type of IX u {s} (where sf/: IX) under the ordering that makes s the largest element. This is the same as the order type of IX U {IX} under the epsilon ordering, which is just i? So we have Fi. + r = i?, whereupon IX + 1 = IX+. Note that the definition of ordinal addition and multiplication yields the equations and To verify an equation, e.g., w. 2 = w + w, we can use the following strategy. If suffices to show that the order types w-:-L and W + ware the same, since the assignment of order types to ordinals is one-to-one. By the above equations, this reduces to verifying that w' :2 = w + w. And this can be done by selecting representative structures for each side of the equation and showing them to be isomorphic. In recent notation, the fact that <w x 2, <H) ~ «{O} x w) u ({I} x w), <L) Ordinal Arithmetic 229 does the job. (Alternatively, we can appeal to the next theorem to obtain w·2 = w + w.) The laws previously established for all order types must in particular be valid for ordinals: Theorem 8K For any ordinal numbers: (IX + P) + y = IX + (P + y), (IX. P). y = IX. (P. y), IX' (P + y) = (IX' P) + (IX' y), IX+O=O+IX=IX, IX' 1 = l'lX = IX, IX' 0 = O'IX = O. Proof We know from Theorem 81 that addition of order types is associative, so (iX + J3) + y = iX + (fJ + y). Hence the |
ordinals (IX + P) + Y and IX + (P + y) have the same order type, and thus are equal. The same argument is applicable to the other parts of the theorem. -l Our work with order types also provides us with counterexamples to the commutative laws and the right distributive laws: 1 + w =I- oj + 1, 2· w =I- W· 2, (w + 1)' 2 =I- (w' 2) + (1. 2). The next theorem gives the equations that characterize addition and multiplication by recursion. Recall that for a set S of ordinals, its supremum sup S is just US. Theorem 8L For any ordinal numbers IX and P and any limit ordinal A _ the following equations hold. (AI) (A2) (A3) (M1) (M2) (M3) IX+O=IX, IX + p+ = (IX + Pt, IX + A = SUp{1X + PI PEA}, IX ·0= 0, IX. P + = IX. P + IX, IX' A = SUp{1X. PI PEA}. 230 8. Ordinals and Order Types Proof (AI) and (Ml) are contained in the previous theorem. To prove (A2), recall that p+ = f3 + 1. Hence 1X+f3+=1X+(f3+1) = (IX + f3) + 1 = (IX + f3t. Similarly for (M2) we have IX' p+ = IX • (f3 + 1) = (IX' f3) + (IX' 1) = (IX' f3) + IX. It remains to prove (A3) and (M3). For that proof we need a lemma on chains of well orderings. Say that a structure <B, < a) is an end extension of a structure <A, < A) iff A £ B, < A £ < a' and every element of B - A is larger than anything in A: a E A & bE B - A = a < ab. Lemma 8M Assume that re is a set of well-ordered structures. Further assume that if <A, < A) and <B, < a) are structures in re, then one is an end extension of the other. Let <W |
, < w) be the union of all these structures in the sense that W = U{A I <A, < A) E re for some < A}' < w = U{ < A I <A, < A) Ere for some A}. Then < w, < w) is a well-ordered structure whose ordinal number is the supremum of the ordinals of the members of re. Proof of the Lemma For each structure <A, < A) in re, we have the usual isomorphism E A onto its ordinal number, ordered by epsilon. If <B, < a) is an end extension of <A, < a)' then E a is an extension of E A' i.e., E 04= E a ~ A. (This is easy to see, but formally we verify E A(a) = E a(a) by induction on a in A.) Thus the set of all possible E A'S {E A I <A, < A) E re for some < A} is a chain of one-to-one functions. So its union (call it E) is a one-to-one fun~tion. The domain of E is the union of all possible A's; that is, dom E = W. The range of E is the union (supremum) of the ordinals of the structures in re; call this ordinal O. Thus E maps W one-to-one onto O. Ordinal Arithmetic 231 Furthermore it preserves order: x<wY ¢> X<AY ¢> E A(X) E E A(Y) ¢> E(x) E E(y). for some A in rc for some A in rc Thus E is an isomorphism of <W, <w) onto <0, Eo)' Hence <W, <w) is well ordered, and its ordinal is O. -l Now return to the proof of Theorem 8L. Since A is a limit ordinal, we have A = UA. The sum IX + A is the ordinal number of the following set (under lexicographic order): ({O} x IX) u ({I} x A) = ({O} x IX) u ({I} x U{P I PEA}) = ({O} x IX) u U{{l} x PIP E A} = U{({O} x |
IX) u ({I} x P) I PEA} = U{Ap I PEA}, where Ap = ({O} x IX) u ({I} x /1). The ordinal of Ap (under lexicographic order) is IX + p. The ordinal of U{Ap I PEA} is provided by the lemma. Once we verify that the lemma is applicable, it will tell us that the ordinal of U{Ap I PEA} is SUp{1X + PIP E A}, which is what we need for (A3). For any P and y in A, either P § y or y § p. If it is the former, then Ay (under lexicographic order) is clearly an end extension of Ap. Thus the lemma is applicable. The proof of (M3) is similar. The product IX' A is the ordinal number of IX x A under < H' Hebrew lexicographic order: IX x A = IX x U{P I PEA} =U{lXxPIPEA} and the ordinal of IX x P is IX' p. Again the lemma is applicable, because whenever P § y, then IX x y is an end extension of IX x P (under <H)' The lemma tells us that the ordinal of U{IX x PIP E A} is SUp{IX' PIP E A}, which completes the proof of (M3). -l For finite ordinals (the natural numbers) we have now defined addition and multiplication three times: first in Chapter 4 (by recursion), then in Chapter 6 (as finite cardinals), and now again (by use of order types). All three agree on the natural numbers; Theorem 8L shows that the recent definition is in agreement with Chapter 4. Theorem 8L also indicates how addition and multiplication could have been (equivalently) defined by transfinite recursion on the ordinals. The six equations in the theorem describe how, for fixed IX, to form IX + P from earlier values IX + y for y less than p. 232 8. Ordinals and Order Types We will give the details of this approach of exponentiation, but they are easily modified to cover addition and multiplication. Consider then, a fixed nonzero ordinal IX. We propose to define IXP in such a way as to satisfy the following three equations: (El) (E2) (E3 |
) IXA = SUp{IXP I PEA} for a limit ordinal A. Now we activate the transfinite recursion machinery. The procedure is similar to that used for the beth operation. Let y(f, y) be the formula: Either or (i) (ii) or or (iii) (iv) f is a function with domain 0 and y = 1, f is a function whose domain is a successor ordinal P+ and y = f(P). IX, f is a function whose domain is a limit ordinal A and y = U ranf, none of the above and y = 0. Then transfinite recursion replies with a formula rp that will define the exponentiation operation. We select the symbol "aE" and define aEp = the unique y such that rp(P, y) for every ordinal p. Then we are assured that "Y(aE ~ p, aEp)." For P = 0, this fact becomes aEO = 1. For a successor ordinal P+ in place of P it becomes and for a limit ordinal A in place of P it becomes aEp+ = aEp' IX aEA = sUP{aEy lYE A}. These three displayed equations are (El)-(E3), except for notational differences. We henceforth dispense with any special symbol for exponenti ation, and instead use the traditional placement of letters: IXP = aEp. There is a special problem in defining Op. If we were to follow blindly (El)-(E3), we would have Ow = 1. This is undesirable, which is our reason for having specified in the foregoing that IX is a fixed nonzero ordinal. We can simply define oP directly: 00 = 1 and oP = 0 for P =f. O. Ordinal Arithmetic 233 Example We have 2w = sup{2n I nEw} = w. Thus ordinal exponentiation is very different from cardinal exponentiation, since 2" =I- K for cardinals. Ordinal addition and multiplication are also very different from cardinal addition and multiplication. Please do not confuse them. Theorem 6J tells us that the operations agree on finite numbers; that theorem does not extend to infinite numbers. We can now apply our remarks on ordinal operations to derive informa tion concerning ordinal arithmetic. Consider afixed ordinal number a. Then the operation of a-addition is the |
operation assigning to each ordinal p the sum a + p. (In our earlier notation, tp = a + p, where a is fixed.) Similarly, the operation of a-multiplication assigns to p the product a' p, and a exponentiation assigns to p the power aP• Theorem 8N addition is normal. (a) For any ordinal number a, the operation of a- (b) If I § a, then the operation of a-multiplication is normal. If 2 § a, then the operation of a-exponentiation is normal. (c) Proof Continuity is immediate from (A3) and (M3) of Theorem 8L and from (E3). For monotonicity, we use Theorem 8C, which tells us that it suffices to show that: a+pEa+p+, a,pEa·p+ aP E aP+ if 1 § a, if 2 § a. The first of these is immediate from (A2) of Theorem 8L: a + p E (a + f3) + = a + p+, whence a-addition is normal. For multiplication we have l§a = OEa and so by the monotonicity of (a. p)-addition, a. pEa. p + a. This together with (M2) gives a. pEa' p+. Exponentiation is similar; 2 §a = 1 E a, whence by the mono tonicity of aP-multiplication and (E3) aP E aP • a = aP + • This completes the proof. -l 234 8. Ordinals and Order Types Example Assume that A. is a limit ordinal. By Exercise 4, t A (for a normal operation) is also a limit ordinal. So by the preceding theorem, IX + A. is a limit ordinal for any ordinal IX. If 1 § IX, then both IX' A. and A.. IX are limit ordinals. (In the second case we use the fact that A.. p+ = A.. P + A..) Similarly if 2 § IX, then both IXA and A.a are limit ordinals. Corollary 8P (a) The following order-preserving laws hold. IX + P E IX + y. p E Y ¢> If 1 § IX, then If 2 |
§ IX, then P E Y ¢> IX' P E IX • y. P E Y ¢> IXP E IXY• (b) The following left cancellation laws hold: IX + P = IX + y = P = y. If 1 § IX, then If 2 § IX, then IX • P = IX • Y = P = y. P = IX IX Y = P = y. Proof These are consequences of monotonicity. Any monotone opera tion assigning tp to P has the properties PEY ¢> tpEty' tp = ty = P = y, by Exercise 3. -l Part (b) of this theorem gives only left cancellation laws. Right cancellation laws fail in general. For example, 2 + W= 3 +w=w, 2· W = 3· W =W, 2w = 3w = w, but we cannot cancel the w's to get 2 = 3. There is a weakened version of part (a) that holds: Theorem 8Q The following weak order-preserving laws hold for any ordinal numbers: (a) P § Y = P + IX § Y + IX. (b) P § Y = P. IX § Y. IX. (c) P§y=pa§t. Ordinal Arithmetic 235 Proof Each part can be proved by transfinite induction on IX. But parts (a) and (b) also can be proved using concepts from order types. Assume that p § y; then also p £ y. Thus ({O} x P) u ({I} x IX) £ ({O} x y) u ({I} x IX) and p x IX £ Y X IX. Furthermore in each case the relevant ordering (lexicographic and Hebrew lexicographic, respectively) on the subset is the restriction of the ordering on the larger set. Hence when we take the ordinal numbers of the sets, we get p + IX § Y + IX and p. IX § Y. IX (compare Exercise 17 of Chapter 7). We will prove (c) by transfinite induction on IX. SO suppose that p § y and that po § l whenever bE IX; we must show that pa § ya. If IX = 0, then pa = ya = 1. Next take the case where IX is a successor ordinal b +. Then pa = po. p § l· P § l· y = ya. since po |
§ l by Corollary 8P Finally take the case where IX is a limit ordinal. We may suppose that neither p nor y is 0, since those cases are clear. Since po § l for b in IX, we have whence pa § ya. -l H is not possible to replace "§" by "E" in the above theorem, since 2 + wf/: 3 + w, 2 'wf/:3 'w, 2w f/: 3w, despite the fact that 2 E 3. Subtraction Theorem If IX § P (for ordinal numbers IX and P), then there exists a unique ordinal number b (their "difference") such that IX + b = p. Proof The uniqueness of b is immediate from the left cancellation law (Corollary 8P). We will indicate two proofs of existence. Since IX-addition is a normal operation and p is at least as large as IX + 0, we know from Theorem 80 that there is a largest b for which IX + b § p. But equality must hold, for if IX + b E p, then contradicting the maximality of b. lX+b+=(IX+b)+§P 236 8. Ordinals and Order Types The other proof starts from the fact that P - IX (i.e., {x E PI x f/: IX}) is well ordered by epsilon. Let 15 be its ordinal. Then IX + 15 is the ordinal of ({O} x IX) u ({I} x (P - IX)) under lexicographic order, which is isomorphic to P under the epsilon ordering. -l For example, if IX = 3 and P = w, then the "difference" is w, since 3 + W = w. If IX E p, we cannot in general find a 15 for which 15 + IX = P; for example, there is no 15 for which 15 + 3 = w. (Why?) Division Theorem Let IX and 15 be ordinal numbers with 15 (the divisor) nonzero. Then there is a unique pair of ordinal numbers P and Y (the quotient and the remainder) such that and Proof First we will prove existence. Since b-multiplication is a normal operation, there is a largest P for which 15. P § IX. Then by the subtraction theorem there is some y for which 15. P + y = IX. We |
must show that y E b. If to the contrary 15 § y, then b· p+ = b· P + 15 §b' p.:r y = IX, contradicting the maximality of p. Having established existence, we now turn to uniqueness. Assume that IX = 15. Pi + Yl = 15. P2 + Y2' where Yl and Y2 are less than b. To show that Pi = P2' it suffices to eliminate the alternatives. Suppose, contrary to our hopes, that Pi E P2 • Then Pt § P2 and so 15. (pt) § 15. P2' Since Yl E 15, we have IX = 15. Pi + Yl E 15. Pi + 15 = 15. (pt) § b· P2 § 15. P2 + Y2 = IX. But this is impossible, so Pi f/: P2· By symmetry, P2 f/: Pl' Hence Pi = P2; call it simply p. We now have IX = 15. P + Yl = 15. P + Y2' whence by left cancellation (Corollary 8P), Yl = Y2' -l Digression Ordinal addition and multiplication are often introduced by transfinite recursion instead of by order types. In that case, the foregoing theorems can be used to establish the connection with order types, without need of Theorem 8L or Lemma 8M. The statements to be proved are those that we took in our development as definitions: The order type of an ordinal sum (or product) is the sum (or product) of the order types. That is, the equations and Ordinal Arithmetic 237 are, in this alternative development, to be proved. For addition, it suffices ({O} x IX) u ({I} x P) with lexicographic to give an isomorphism from ordering onto IX + P with epsilon ordering. The isomorphismfis defined by the equations f«O, y») = y f(1,b»)=IX+b for y in IX, 15 in p. for Then it can be verified that ranf = IX + P and that f preserves order. For multiplication, the isomorphism g from IX x p with Hebrew lexicographic order onto IX. P with epsilon ordering is defined by the equation g«y, b») = IX ·15 + y for y in IX |
and 15 in p. Again it can be verified that ran g = IX. P and that g preserves order. Logarithm Theorem Assume that IX and p are ordinal numbers with IX =f. 0 and p (the base) greater than 1. Then there are unique ordinal numbers y, 15, and p (the logarithm, the coefficient, and the remainder) such that IX = f3Y. 15 + P & 0 =f. 15 E P & p E f3Y. Proof Since p-exponentiation is a normal operation, there is (by Theorem 80) a largest y such that f3Y § IX. Apply the division theorem to IX...;- f3Y to obtain 15 and p such that and P E f3Y. Note that 15 =f. 0 since p E pY § IX. We must show that 15 E p. If to the contrary p § 15, then f3Y+ = f3Y. p § f3Y. 15 § f3Y. 15 + p = IX, contradicting the maximality of y. Thus we have the existence of y, 15, and p meeting the prescribed conditions. To show uniqueness, consider any representation IX = f3Y. 15 + P for which 0 =f. 15 E P and pEP. We first claim that y must be exactly the one we used in the preceding paragraph. We have f3Y§IX=f3Y·b+p since 1 E 15 Ef3Y·b+f3Y =f3Y·b+ § pY. P = pY+. since p E f3Y since 15 E P Thus f3Y § IX E f3Y+. This double inequality uniquely determines y; it must be the largest ordinal for which f3Y § IX. Once y is fixed, the division theorem tells us that 15 and p are unique. -l 238 8. Ordinals and Order Types Observe that in the logarithm theorem, we always have P E fJY § 0(. An interesting special case is where the base f3 is w. Then given any nonzero ordinal 0(, we can write 0( = w Y1 • n l + Pi' where n l is a nonzero natural number and Pi E w Y1 § 0(. If Pi is nonzero, then we can repeat: Pi = w Y2 • n 2 + P |
2' where n 2 is a nonzero natural number and P2 E w Y2 § Pi E w Y1 • Hence both P2 E Pi and Y2 E Yl' Continuing in this way, we construct progressively smaller ordinals Pi' P2'.... This descending chain cannot be infinite, so Pk = 0 for some (finite!) k. We then have 0( represented as an "w polynomial" 0( = w Y1 • n l +... + wYk • n where n l,..., nk are nonzero natural numbers and Yk E Yk-l E'" E Yl' This polynomial representation is called the Cantor normal form of 0(. You are asked in Exercise 26 to show that it is unique., k Theorem 8R For any ordinal numbers, Proof We use transfinite induction on y. In the limit ordinal case we will use the normality of O(-exponentiation. But normality is true only for 0( greater than 1. So a separate proof is needed for the cases 0( = 0 and 0( = 1. We leave this separate proof, which does not require induction, to you (Exercise 27). Henceforth we assume that 0( is at least 2. Suppose, as the inductive hypothesis, that O(P +fJ = O(P. 0(0 for all bless than Y; we must prove that O(P+Y = O(P.O(Y. Case I Y = O. Then on the left side we have O(P+o = O(P. On the right side we have O(P. 0(0 = O(P. 1 = O(P, so this case is done. Case II Y = b + for some b. Then O(P+Y = O(p+o+ = O((p+o)+ = O(p+o'O( = O(P '0(0'0( = O(P '0(0+ = O(P. O(Y. by (A2) by (E2) by the inductive hypothesis by (E2) (The associative law was also used here.) Ordinal Arithmetic 239 Case I I I y is a limit ordinal. Then rxP + Y = rxsup{P +" I " E Y} = sup{rxP+Ii I bEY} = S |
Up{rxP. rxO I bEY} by (A3) by Theorem 8E by the inductive hypothesis. On the other side, rxP. rx Y = rxP. sup{rxo I bEY} = sup{rxP. rxO I bEY} by Theorem 8E again, this time applied to rxP-multiplication. Thus the two sides agree. -l Theorem 8S For any ordinal numbers, (rxP)Y = rxP. Y. Proof We again use transfinite induction on y. And again we leave to you the case rx = 0 or rx = 1 (Exercise 27). Henceforth we assume that rx is at least 2. Suppose, as the inductive hypothesis, that (rxP)O = rxP' ° for all b less than Y; we must prove that (rxP)Y = rxP' Y. Case I Y = O. Obviously both sides are equal to 1. Case II Y = b +. Then we calculate (rxP)Y = (rxP)o+ = (rxP)O. rxP = rxP. 0 • rxP = rxP 'O+P =rxP'o+ = rx P. Y by the inductive hypothesis by the previous theorem as desired. Case III Y is a limit ordinal. Then rxP. Y = rxsup{P'" I" E Y) = sup{rxP' o I bEY} = sup{ (rxP)O I bEY} = (rxP)Y by Theorem 8E by the inductive hypothesis as desired. -l 240 8. Ordinals and Order Types Example As with finite numbers, aPY always means a<py). The other grouping, (aP)Y, equals aP' Y by Theorem 8S. Now suppose that we start with wand apply exponentiation over and over again. Let eo = sup w, w,w,.... WOO W } { Then by Theorem 8E weo = sup{WW, w W OO, ••• } = eo. The equation eo = weo gives the Cantor normal form representation for eo. More generally, the epsilon numbers are the ordinals e for which e = we. The smallest epsilon number is eo. It is a count |
able ordinal, being the countable union of countable sets. By the Veblen fixed-point theorem, the class of epsilon numbers is unbounded. Exercises 20. Show that every ordinal number is expressible in the form A. + n where nEW and A. is either zero or a limit ordinal. Show further that the representation in this form is unique. In Exercise 4 of Chapter 7, find the ordinal number of <P, R). 21. 22. Prove parts (a) and (b) of Theorem 8Q by transfinite induction on a. 23. (a) Show that w + w 2 = w 2 (b) Show that whenever w 2 ~ p, then w + p = p. • 24. Assume that w ~ a and 11. the preceding exercise.) 25. Consider any fixed ordinals a and O. Show that 2 ~ p. Show that a + p = p. (This generalizes a + 0 = a u {a + bib EO}. 26. Prove that the representation of an ordinal number in Cantor normal form is unique. 27. Supply proofs for Theorem 8R and Theorem 8S when a is 0 or 1. 28. Show that for any given ordinal number, there is a larger epsilon number. 29. Show that the class of epsilon numbers is closed, i.e., if S is a non empty set of epsilon numbers, then sup S is an epsilon number. CHAPTER 9 SPECIAL TOPICS In this final chapter, we present three topics that stand somewhat apart from our previous topics, yet are too interesting to omit from the book. The three sections are essentially independent. WELL-FOUNDED RELATIONS Some of the important properties of well orderings (such as transfinite induction and recursion) depend more on the" well" than on the" ordering." In this section, we extend those properties to a larger class of relations. Recall that for a partial ordering R and a set, D, an element m of D was said to be a minimal element if there was no x in D with xRm. This terminology can actually be applied to any set R: Definition An element m of a set D is said to be an R-minimal element of D iff there is no x in D for which xRm. Definition A relation R is said to be well founded iff every non |
empty set D contains an R-minimal element. 241 242 9. Special Topics If the set D in this definition is not a subset of fld R, then it is certain fld R). Thus the to contain an R-minimal element (namely any m in D only sets D that matter here are the subsets of fld R. Examples The finite relation R = {<I, 2), <2,6), (3, 6)} is well founded. Its field is {l, 2, 3, 6}, and all fifteen nonempty subsets of the field have R-minimal elements. The empty relation is also well founded, but for uninteresting reasons. Example Consider any set S and the membership relation Es = {<x,y) E S x S I x E y} on S. Then we can show from the regularity axiom that Es is well founded. For consider any nonempty set D. By regularity there some m in D with m n D = 0, i.e., there is no x in D with x E m. So certainly there is no x in D with x Es m. Hence m is Es-minimal in D. In fact the regularity axiom can be summarized by the statement: The membership relation is well founded. The following theorem extends Theorem 7B, which asserted that a linear ordering was a well ordering iff it had no descending chains. Theorem 9A A relation R is well founded iff there is no function f with domain w such thatf(n+)Rf(n) for each n. Proof The proof is exactly the same as for Theorem 7B. If R is not well founded, then there exists a nonempty set A without an R-minimal, element, i.e., (Vx E A)(3y E A) yRx. So we can apply Exercise 20 of Chapter 6 -l to obtain a descending chain. If R is a binary relation on A (i.e., R s A x A) that is well founded, then we can say that R is a well-founded relation on A or that <A, R) is a well-founded structure. Transfinite Induction Principle Assume that R is a well-founded relation on A. Assume that B is a subset of A with the special property that for every t in A, {x E A I xRt} s B = t E B |
. Then B coincides with A. This theorem is the direct analogue of the transfinite induction principle for well-ordered relations (in Chapter 7). It asserts that any R-inductive subset of A must actually be A itself. In fact the earlier theorem is a special case of the above theorem, wherein R linearly orders A. Well-Founded Relations 243 Proof The proof is the same as before. If B is a proper subset of A, the minimality, then A - B has an R-minimal element m. By {x E A I xRm} ~ B. But then by the special property of B (the property of being R-inductive), m E B after all. -l Next we want to describe how any relation R (well founded or not) can be extended to R" the smallest transitive relation extending R. Theorem 9B Let R be a relation. Then there exists a unique relation R' such that: (a) R' is a transitive relation and R ~ R'. (b) If Q is any transitive relation that includes R, then R' ~ Q. Proof There are two equivalent ways to obtain R'. The brute force method is to define R* "from above" by R* = n{Q I R ~ Q ~ fld R x fld R & Q is a transitive relation}. The collection of all such Q's is nonempty, since fld R x fld R is a member. Hence it is permissible to take the intersection. Then R ~ R* (because R ~ Q for each of those Q's). R* is a transitive relation (recall Exercise 34 of Chapter 3). And R* is as small as possible, being the intersection of all candidates. Hence we may take R' = R*. Uniqueness is immediate; by (b) any two contenders must be subsets of each other. Although the theorem is now proved, we will ignore that fact and give the construction of R' "from below." Define Rn by recursion for n in w by the equations and and then define R* = UnewRn. For example, we can describe Rn for small n as follows: Ro = R = {<x, y) I xRy}, R1 = R 0 R = {<x, y) I xRtRy for some t}, R2 = R 0 R 0 R = {<x, y) I xRt1 |
Rt2Ry for some t1 and t 2}. In general, and R* = {<x, y) I xRnY for some n}. Clearly R ~ R*. To show that R* is a transitive relation, suppose that xR*yR*z. Then xRmyRnz for some m and n. We claim that xRm+n+ 1z. 244 9. Special Topics This is clear from the above characterization of Rn; the "three dots" technique can, as usual, be replaced by an induction on n. Hence xR* z. To see that R* also satisfies the minimality clause (b), consider any transitive relation Q that includes R. To show that R* ~ Q, it suffices to show that Rn ~ Q for each n. We do this by induction on n. It holds for n = 0 by assumption. If Rk ~ Q and xRk+ lZ, then we have xRkyRz for some y, whence xQyQz and (by transitivity) xQz. So Rk + 1 ~ Q and the induction is complete. We are now entitled to conclude that Rt = R* = R*. -l Example Assume that S is a transitive set. Define the binary relation ES = {<x, y) E S X S I x E y} on S. Then for x and y in S, X E~y ¢> XEs'" EsY ¢> X E'" E y, where the intermediate points are automatically in S (by transitivity). Hence we have X E~ Y ¢> X E TC y. Recall from Exercise 7 of Chapter 7 that TC y, the transitive closure of y, is the smallest transitive set that includes y. Its members are roughly the members of members of... of members of y. A little more precisely, TC.... Theorem 9C If R is a well-founded relation, then its transitive extension Rt is also well founded. Proof We will give a proof that uses the axiom of choice. Exercise 1 requests an alternative proof that does not require choice. Suppose that, contrary to our expectations, Rt is not well founded. Then by Theorem 9A there is a descending chain f. That is, f is a function with domain wand f(n+)RY(n) for each n. The idea is to fill in this descending |
chain to get a descending chain for R, and thereby to contradict the assumption that R is well founded. Sincef(n+)RY(n), we know that f(n+)RxlR"· RxkRf(n) for some Xl'..., x k • By interpolating these intermediate points between f(n+) and f(n) (and doing this for each n), we get a descending chain g such that g(m+)Rg(m) for each m. -l Well-Founded Relations 245 Corollary 9D If R is a well-founded relation on A, then R' is a partial ordering on A. Proof Certainly R' £ A x A and R' is a transitive relation. So we need show only that R' is irreflexive. But any well-founded relation is irreflexive; we cannot have XR'X lest {x} fail to have a R'-minimal element. -l Because of this corollary, transitive well-founded relations are sometimes called partial well orderings. Transfinite Recursion Theorem Schema For any formula y(x, y, z) the following is a theorem: Assume that R is a well-founded relation on a set A. Assume that for any f and t there is a unique z such that y(f, t, z). Then there exists a unique function F with domain A such that y(F ~ {x E A I xRt}, t, F(t)) for all t in A. Here y defines a function-class G, and the equation F(t) = G(F ~ {x E A I xRt}, t) holds for all t in A. A comparison of the above theorem schema with its predecessor in Chapter 7 will show that we have added an additional variable to y (and to G). This is because knowing what F ~ {x E A I xRt} is does not tell us what t is. The proof of the above theorem schema is much like the proofs of recursion theorems given in Chapters 4 and 7. Proof As before, we form F by taking the union of many approximating functions. For any x in A, define seg x to be {t I tRx}. For the purposes of this proof, call a function v acceptable iff dom v £ A and whenever x E dom v, then seg x £ |
dom v and y(v ~ seg x, x, v(x). 1. First we claim that any two acceptable functions Vi and v2 agree at all points (if any) belonging to both domains. Should this fail, there is an R-minimal element x of dom Vi n dom v2 for which Vi (x) =f. v2 (x). By the minimality, Vi ~ seg x = v2 ~ seg x. But then acceptability together with our assumption on y tells us that v1(x) = v2(x) after all. We can now use a replacement axiom to form the set % of all acceptable functions. Take rp(u, v) to be the formula: u £ A & v is an acceptable function with domain u. It follows from the preceding paragraph that 246 9. Special Topics Hence by replacement there is a set % such that v E % I define F to be the set U%. Thus ({:r) ¢> <x, y) E F ¢> v(x) = y (3u E f!jJ A) rp(u, v). for some acceptable v. Letting % be the set of all acceptable functions, we can explicitly Observe that F is a function, because any two acceptable functions agree wherever both are defined. 2. Next we claim that the function F is acceptable. Consider any x in dom F. Then there exists some acceptable v with x E dom v. Hence x E A and seg x ~ dom v ~ dom F. We must check that y(F ~ seg x, x, F(x)) holds. We have y(v ~ seg x, x, v(x)) v ~ seg x = F ~ seg x v(x) = F(x) by acceptability, by ({:r) and (1), by ({:r), from which we can conclude that y(F ~ seg x, x, F(x)). 3. We now claim that dom F is all of A. If this fails, then there is an R-minimal element t of A - dom F. By minimality, seg t s dom F. Take the unique y such that y(F ~ seg t, t, y) and let v = F u {<t, y)}. We must verify that v is acceptable. Clearly v is a function and dom v |
~ A. in dom v. One possibility is Consider any x that x E dom F. Then seg x ~ dom F ~ dom v and from the equations v ~ seg x = F ~ seg x v(x) = F(x) and we conclude that y(v ~ seg x, x, v(x)). The other possibility is that x = t. Then seg t ~ dom F ~ dom v and from the equations v ~ seg t = F ~ seg t v(t) = y and we conclude that y(v ~ seg t, t, v(t)). Hence v is acceptable, and so t E dom F after all. Consequently, dom F = A. 4. F is unique, by an inductive argument like those used before. -l We now proceed to show how transfinite recursion can be applied to the membership relation to produce the rank of a set (and thereby generate all of the ordinal numbers). Recall that in Chapter 7 we applied transfinite recursion to a well-ordered structure <A, <) to obtain a function E with domain A such that E(a) = {E(x)lx E a} for each a E A. It then turned out that ran E was (by definition) the ordinal number of <A, <). Well-Founded Relations 247 Now we intend to perform the analogous construction using, in place of a well ordering, the membership relation on some set. After all, the regularity axiom assures us that the membership relation is always well founded. So we can apply transfinite recursion. What will we get? The answer is that if we do it right, we will get the rank of the set! To be more specific, let S be the set < 1, 0), or {{{0}}, {0, {0}}} to use its full name. We can illustrate the formation of this set by Fig. 48. The sets appearing below S in this figure are the sets in TC S, the transitive closure of S. {{{0}}. {0. {0}}} {{0]] [2] /~ ~/ {0. {0}} [2] {0:~ Fig. 48. The membership relation on TC{(l, D)}. o [0] Let M be the membership relation on TC S: M = {<x, y) I x EyE TC S}. The pairs |
in this relation are illustrated by straight lines in Fig. 48. The relation M is sure to be well founded; in the present example it is also finite. We have seen that Mt is a well-founded partial ordering on TC S; in fact it is the partial ordering corresponding to Fig. 48. Since Mt is well founded, we can apply transfinite recursion to obtain the unique function E on TC S for which E(a) = {E(x) I xMta} for every a in TC S. (In the transfinite recursion theorem schema, take y(f, t, z) to be the formula z = ran f. Then E(a) = ran(E ~ {x E TC S I xMta}) = E[{x I xMta}] = {E(x) I xMta} 248 9. Special Topics as predicted.) The values of E are shown by the bracketed numerals in Figure 48. Observe that ran E = E[TC S] = 3, which is indeed rank S. And furthermore E(a) = rank a for every a in TC S. Was it just a lucky coincidence that for S = (1, 0) we reproduced the rank function? (Would we have gone through the example if it were?) The following theorem says that the outcome is inevitable. Theorem 9E Let S be any set and let M = {(x, y) I x EyE TC S}. Let E be the unique function with domain TC S such that E(a) = {E(x)lxMta} for each a in TC S. Then ran E = rank S. Furthermore E(a) = rank a for each a in TC S. Proof This theorem is a consequence of Theorem 7V, which character ized rank a as the least ordinal strictly greater than rank x for every x in a. The heart of the proof consists of showing that rank a = {rank x I xMta} for all a in TC S. To prove the" 2" half of (1:r), we use Theorem 7V(a): xMta => x E... E a => => rank x E... E rank a rank x E rank a. The "s" half of (1:r) is proved by transfinite induction on a over the well-founded structure (TC S, Mt). Suppose then that (1:r) holds of b |
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