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Theorem 7.11. If \( \omega \rightarrow {\left( \omega \right) }_{2}^{2} \) then \( \omega \rightarrow {\left( \omega \right) }_{m}^{n} \) for all \( n, m < \omega \) . | Proof. By proposition 7.9 it suffices to fix \( m \) and work by induction on \( n \) . So suppose \( \omega \rightarrow {\left( \omega \right) }_{m}^{n} \) and let \( \mathcal{P} = \left\{ {{P}_{1},..{P}_{m}}\right\} \) be a partition of \( {\left\lbrack \omega \right\rbrack }^{n + 1} \) . For each \( k \in \omega \) ... | Yes |
Theorem 7.13. If \( \kappa \) is weakly compact then \( \kappa \) is regular. | Proof. Suppose \( \kappa \) is weakly compact. Let \( \left\{ {{\kappa }_{\alpha } : \alpha < \operatorname{cf}\left( \kappa \right) }\right\} \) be an increasing sequence of cardinals cofinal in \( \kappa \) . For each ordinal \( \beta < \kappa \) we write \( f\left( \beta \right) = \) the least \( \alpha \) with \( \... | Yes |
Theorem 7.14. \( \forall \lambda {2}^{\lambda } \nrightarrow {\left( {\lambda }^{ + }\right) }_{2}^{2} \) . | Proof. First we prove a subclaim of independent interest:\n\nSubclaim 7.14.1. There are no increasing or decreasing chains of size \( {\lambda }^{ + } \) in \( {2}^{\lambda } \) under \( { \leq }_{L} \)\n\nProof. Suppose we have \( G = \left\{ {{g}_{\alpha } : \alpha < {\lambda }^{ + }}\right\} \subseteq {2}^{\lambda }... | Yes |
Corollary 7.15. A weakly compact cardinal is a regular strong limit cardinal. | Proof. We already know that weakly compact cardinals are regular. Suppose \( \kappa \) is not a strong limit. Then there is \( \lambda < \kappa \) with \( {2}^{\lambda } \geq \kappa \) . If \( \kappa \) were weakly compact, then by proposition \( {7.8\kappa } \rightarrow {\left( {\lambda }^{ + }\right) }_{2}^{2} \) hen... | Yes |
Theorem 7.26. Two elements in a tree are comparable iff they are compatible. | Proof. Note that \( p, q \) are comparable under \( \leq \) iff they are comparable under \( { \leq }^{ \leftarrow } \) . Note that in a partial order comparability implies compatibility. So assume \( p, q \) are compatible. There is \( r \) with \( p \leq r, q \leq r \) . If either \( p = r \) or \( q = r \) then \( p... | Yes |
Proposition 7.29. Singular cardinals do not have the tree property. | Proof. Let \( \kappa \) be singular, \( \lambda = \operatorname{cf}\left( \kappa \right) < \kappa \), and let \( \left\{ {{\kappa }_{\alpha } : \alpha < \lambda }\right\} \) be an increasing sequence whose union is \( \kappa \) . For \( \alpha < \lambda \) define \( {f}_{\alpha } : {\kappa }_{\alpha } \rightarrow \{ \a... | Yes |
Theorem 7.30. (König) \( \omega \) has the tree property. | Proof. Suppose \( T \) is an infinite tree with every level finite. By induction we construct an increasing sequence \( \left\{ {{t}_{n} : n < \omega }\right\} \) where each \( {t}_{n} \in T\left( n\right) \) and \( \forall n{t}_{n} \) has infinitely many successors.\n\nSuppose we have \( {t}_{0},\ldots {t}_{n} \) . Si... | Yes |
Theorem 7.41. (Cantor) Every nonempty countable dense linear order without endpoints is order-isomorphic to \( \mathbb{Q} \). | Proof. Let \( X \) be a countable linear order, \( X = \left\{ {{x}_{n} : n < \omega }\right\} \). Let \( \mathbb{Q} = \left\{ {{q}_{n} : n < \omega }\right\} \). We construct an order isomorphism \( \varphi : X \rightarrow \mathbb{Q} \) as follows:\n\nSuppose we have \( \varphi : {X}_{n} \rightarrow {Q}_{n} \) an orde... | Yes |
Corollary 7.42. Every countable dense linear ordering \( X \) is isomorphic to a subset of \( \mathbb{Q} \) . | Proof. Given a countable dense linear ordering, extend it (by countable recursion; see exercise 24) to a countable dense linear order without endpoints. Apply theorem 7.41. | No |
Proposition 7.45. Let \( T \) be a tree, \( B \) the set of branches of \( T \) . For each \( \alpha < {htT} \) let \( { \leq }_{\alpha } \) be a linear order on \( T\left( \alpha \right) \) . For \( b, c \in B \) we define \( b{ \leq }_{ * }c \) iff \( b = c \) or, for \( \alpha \) the least ordinal with \( b\left( \a... | The proof is left as an exercise. | No |
Theorem 7.46. A canonical linear order on the branches of a Suslin tree is a Suslin line. | Proof. Let \( T \) be a Suslin tree under \( { \leq }_{T} \), and let \( { \leq }_{ * } \) be a canonical linear order on the set of branches\n\n\( B \) .\n\nSubclaim 7.46.1. No countable set is dense in the linear order \( { \leq }_{ * } \).\n\nProof. Since every branch is countable, if \( A \subseteq B \) is countabl... | Yes |
Proposition 7.48. If there is a Suslin tree, then there is a Suslin splitting tree. | Proof. Let \( T \) be a Suslin tree. Let \( S = \{ t \in T \) : if \( t \uparrow \) is countable \( \} \), where \( t \uparrow = \{ s \in T : s > t\} \) . Let \( A \) be the set of minimal elements of \( S.A \) is an antichain, so it is countable. So \( T \smallsetminus \mathop{\bigcup }\limits_{{t \in A}}t \uparrow \)... | No |
Proposition 7.50. A dense linear order has a countable splitting tree of intervals. | Proof. Let \( X \) be a dense linear order. Let \( {x}_{0} < {x}_{1} < {x}_{2} \in X, T\left( 0\right) = \left\{ {\left( {{x}_{0},{x}_{1}}\right) ,\left( {{x}_{1},{x}_{2}}\right) }\right\} \) .\n\nGiven \( T\left( n\right) \) a collection of \( {2}^{n + 1} \) pairwise disjoint intervals, where \( T\left( n\right) \) re... | Yes |
Theorem 7.52. If there is a Suslin line there is a Suslin tree. | Proof. We may assume that our Suslin line \( X \) is dense.\n\nSubclaim 7.52.1. Every countable splitting tree of intervals \( T \) in \( X \) has an extension \( S \neq T \) which is also a countable splitting tree of intervals.\n\nProof. Let \( T \) be a countable splitting tree of intervals. Let \( A = \{ x \in X : ... | Yes |
Theorem 7.59. Every measurable cardinal is weakly compact. | Proof. Let \( \kappa \) be measurable, \( \mathcal{F} \) a \( \kappa \) -closed ultrafilter on \( \kappa \), and suppose \( {\left\lbrack \kappa \right\rbrack }^{2} = {P}_{0} \cup {P}_{1} \) where \( {P}_{0},{P}_{1} \) are disjoint. We imitate the proof of theorem 7.10:\n\nLet \( {\alpha }_{0} = 0 \) . There is a uniqu... | Yes |
Proposition 7.62. Suppose \( j : V \rightarrow M \) is a 1-1 map, where \( M \) is a transitive subclass of \( V, M \supset \) ON, and property (1) of theorem 7.61 holds. Then (c) If \( x = \left\{ {{y}_{\alpha } : \alpha < \lambda }\right\} \) then \( j\left( x\right) = \left\{ {{z}_{\alpha } : \alpha < j\left( \lambd... | For (c) Let \( x = \left\{ {{y}_{\alpha } : \alpha < \lambda }\right\} \) and let \( f : \lambda \rightarrow x \) with \( f\left( \alpha \right) = {y}_{\alpha } \) for all \( \alpha < \lambda \) . Then \( M \vDash j\left( f\right) : j\left( \lambda \right) \rightarrow j\left( x\right) \) and \( j\left( f\right) \) is o... | Yes |
Theorem 7.63. Suppose there is a 1-1 map \( j : V \rightarrow M \) as in theorem 7.61 where \( \kappa \) is the critical point (i.e., \( \kappa \) satisfies properties (2) and (3)). Then \( \kappa \) is measurable. | Proof. Define \( \mathcal{F} = \{ x \subseteq \kappa : \kappa \in j\left( x\right) \} \). By proposition 7.62, (i) \( \mathcal{F} \) is closed under finite intersection and superset. (ii) \( \forall x \subseteq \kappa \), either \( x \in \mathcal{F} \) or \( \kappa \smallsetminus x \in \mathcal{F} \). (iii) \( \varnoth... | Yes |
Theorem 7.64. (Scott) If \( V = L \) then there are no measurable cardinals. | Proof. . If there is a measurable cardinal then there is a smallest one, \( \kappa \) . Let \( j, M \) be as in theorem 7.61. Then \( M \vDash j\left( \kappa \right) \) is the smallest measurable cardinal. If \( V = L \) then \( V \subseteq M \subseteq V \), so \( M = V \) . Hence \( V \vDash {}^{\omega }\kappa < j\lef... | Yes |
Theorem 7.68. \( \omega < \mathfrak{p},\mathfrak{a},\mathfrak{b},\mathfrak{d} \) | Proof. For \( \omega < \mathfrak{p} \) : This follows from theorem 1.43.\n\nFor \( \omega < \mathfrak{a} \) : Let \( A \subset {\left\lbrack \omega \right\rbrack }^{\omega } \) be a countable almost disjoint family, \( A = \left\{ {{a}_{n} : n < \omega }\right\} \) . Let \( {b}_{0} = {a}_{0},{b}_{n + 1} = {a}_{n + 1} \... | No |
Theorem 7.70. (a) \( \mathfrak{b} \leq \mathfrak{d} \) | Proof. For (a): A dominating family is unbounded. | No |
Theorem 7.71. (Erdös-Rado). Assume CH. Then \( \omega \times {\omega }_{1} = H \cup K \) where \( H \cap K = \varnothing \) and there is no homogeneous subset of the form \( A \times B \) where \( A \in {\left\lbrack \omega \right\rbrack }^{\omega } \) and \( B \in {\left\lbrack {\omega }_{1}\right\rbrack }^{{\omega }_... | Proof. By \( \mathrm{{CH}} \), let \( \left\{ {{a}_{\alpha } : \alpha < {\omega }_{1}}\right\} \) enumerate \( {\left\lbrack \omega \right\rbrack }^{\omega } \) . Using a recursive construction of length \( {\omega }_{1} \), at stage \( \beta \) we decide which \( \left( {k,\beta }\right) \in H \) and which \( \left( {... | Yes |
Theorem 7.78. Assume \( M{A}_{{\omega }_{1}} \). Then \( {SH} \) holds. | Proof. First we need to show that if there's a Suslin tree, there's one in which every element has sucessors of arbitrarily high height.\n\nSubclaim 7.78.1. If \( T \) is a Suslin tree, there is a Suslin tree \( {T}^{ * } \subseteq T \) so that\n\n\[ \text{(*)}\forall t \in {T}^{ * }\forall \alpha < {\omega }_{1}\exist... | Yes |
Theorem 7.80. Assume \( M{A}_{{\omega }_{1}} \). Then \( {CH} \) fails. | Proof. Let \( F \in {\left\lbrack {2}^{\omega }\right\rbrack }^{{\omega }_{1}} \). We show that \( F \neq {2}^{\omega } \).\n\nDefine \( P = \mathop{\bigcup }\limits_{{n < \omega }}{2}^{n} \) where \( p \leq q \) iff \( p \supseteq q.{}^{111}P \) is countable, so ccc.\n\nFor \( f \in F \), define \( {D}_{f} = \{ p \in ... | Yes |
Theorem 7.83. Assume \( M{A}_{\sigma \text{-centered }} \) . If \( \kappa < \mathfrak{c} \) then \( {2}^{\kappa } = {2}^{\omega } \) . | Proof. First, note that if \( T = \mathop{\bigcup }\limits_{{n < \omega }}{2}^{n} \) is the binary tree of height \( \omega \), where the order is end-extension, then the set of branches of \( T \) is an almost disjoint family on \( T \) . Since \( T \) is countable, we can identify it with \( \omega \), so there is al... | Yes |
Corollary 7.84. Assume MA. Then \( {2}^{\omega } \) is regular. | Proof. If \( \kappa < {2}^{\omega } \) then \( {2}^{\kappa } = {2}^{\omega } \), so cf \( {2}^{\omega } = \) cf \( {2}^{\kappa } > \kappa \) . | Yes |
Theorem 7.85. (Baumgartner-Hajnal) Assume \( \mathfrak{p} = {\mathfrak{c}}^{.113} \) Then \( \omega \times {\omega }_{1} \rightarrow {\left( \omega \times {\omega }_{1}\right) }_{2}^{1} \), i.e., if \( \omega \times {\omega }_{1} = H \cup K \) where \( H \cap K = \varnothing \) then there are \( A \in {\left\lbrack \om... | Proof. Fix some \( \mathcal{F} \) a non-principal ultrafilter on \( \omega \) .\n\nFor each \( \alpha < {\omega }_{1} \) let \( {H}_{\alpha } = \{ n : \left( {n,\alpha }\right) \in H\} \) and \( {K}_{\alpha } = \{ n : \left( {n,\alpha }\right) \in K\} .\omega = {H}_{\alpha } \cup {K}_{\alpha } \), so for each \( \alpha... | Yes |
Theorem 7.89. Let \( \alpha \) be an orrdinal with uncountable cofinality. A continuous strictly increasing function from a club in \( \alpha \) to \( \alpha \) has a club of fixed points. | Proof. Let \( \alpha \) have uncounable cofinality, \( f : C \rightarrow \alpha \) where \( C \) is a club in \( \alpha \) and \( f \) is continuous and strictly increasing.\n\nIf \( A \) is a set of fixed points in \( C \) and \( \sup A < \alpha \) then by continuity \( \sup A \) is a fixed point, hence in \( C \) .\n... | Yes |
Theorem 7.91. (Fodor) Let \( \kappa \) be a regular uncountable cardinal. If \( f : S \rightarrow \kappa, S \) is stationary, and \( f\left( \alpha \right) < \alpha \) for all nonzero \( \alpha \in S \) (such a function \( f \) is called regressive), then there is a stationary \( R \subseteq S \) with \( f \) constant ... | Proof. Given \( f, S \) as in the hypothesis, \( \alpha < \kappa \), define \( {S}_{\alpha } = \{ \beta : f\left( \beta \right) = \alpha \} \) . If \( {S}_{\alpha } \) is not stationary, then there is a club \( {C}_{\alpha } \subseteq \kappa \) with \( {C}_{\alpha } \cap {S}_{\alpha } = \varnothing \) . Let \( C = \lef... | No |
Theorem 7.92. (Solovay) Every regular uncountable cardinal \( \kappa \) is the union of \( \kappa \) many disjoint stationary subsets of \( \kappa \) . | Proof. Let \( S = \{ \alpha < \kappa : \operatorname{cf}\left( \alpha \right) = \omega \} \), and for \( \alpha \in S \) let \( \left\{ {{\beta }_{n,\alpha } : n < \omega }\right\} \) be an increasing sequence converging to \( \alpha \) . For \( n < \omega ,\eta < \kappa \) let \( {S}_{n,\eta } = \left\{ {\alpha \in S ... | Yes |
Theorem 7.97. \( L \vDash \diamond \) . | The proof of theorem 7.97 requires careful inspection of how \( L \) is constructed, and is beyond the scope of this book. | No |
Example 1 (the complex affine line): \( {\mathbb{A}}_{\mathbb{C}}^{1} \mathrel{\text{:=}} \operatorname{Spec}\mathbb{C}\left\lbrack x\right\rbrack \) . Let’s find the prime ideals. As \( \mathbb{C}\left\lbrack x\right\rbrack \) is an integral domain,0 is prime. Also, \( \left( {x - a}\right) \) is prime, where \( a \in... | We now show that there are no other prime ideals. We use the fact that \( \mathbb{C}\left\lbrack \mathrm{x}\right\rbrack \) has a division algorithm, and is a unique factorization domain. Suppose \( \mathfrak{p} \) is a prime ideal. If \( \mathfrak{p} \neq \left( 0\right) \), then suppose \( f\left( x\right) \in \mathf... | Yes |
Show that for the last type of prime, of the form \( \left( {{x}^{2} + {ax} + b}\right) \), the quotient is always isomorphic to \( \mathbb{C} \) . | For example: \( \mathbb{R}\left\lbrack x\right\rbrack /\left( {x - 3}\right) \cong \mathbb{R} \) , \( \mathbb{R}\left\lbrack x\right\rbrack /\left( {{x}^{2} + 1}\right) \cong \mathbb{C}. \) | No |
Example 6 (the affine line over \( {\mathbb{F}}_{\mathrm{p}} \) ): \( {\mathbb{A}}_{{\mathbb{F}}_{\mathrm{p}}}^{1} = \operatorname{Spec}{\mathbb{F}}_{\mathrm{p}}\left\lbrack \mathrm{x}\right\rbrack \) . As in the previous examples, this has a division algorithm, so the prime ideals are of the form (0) or \( \left( {f\l... | Note that \( \operatorname{Spec}{\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) has \( p \) points corresponding to the elements of \( {\mathbb{F}}_{p} \), but also (infinitely) many more. This makes this space much richer than simply \( p \) points. For example, a polynomial \( f\left( x\right) \) is not determined by ... | No |
What are the prime ideals of \( \mathbb{C}\left\lbrack {x, y, z}\right\rbrack \) ? | Analogously to before, \( \left( {x - a, y - b, z - c}\right) \) is a prime ideal. This is a maximal ideal, with residue field \( \mathbb{C} \) ; we think of these as \ | No |
Corollary 2. If \( A \) is expressed as \( {a\alpha } \) and \( B \) as \( {b\beta } \), then \( \mathrm{S}{AB} = {ab}\cos {\alpha \beta } \) and \( \mathrm{V}{AB} = {ab}\sin {\alpha \beta } \cdot \overline{\alpha \beta } \), where \( \overline{\alpha \beta } \) denotes the direction which is normal to both \( \alpha \... | Example. Given \( A = r\overline{\phi //\theta } \) and \( B = {r}^{\prime }\overline{{\phi }^{\prime }//{\theta }^{\prime }} \) . Then\n\n\[ \mathrm{S}{AB} = r{r}^{\prime }\cos \overline{\phi //\theta }\overline{{\phi }^{\prime }//{\theta }^{\prime }} \]\n\n\[ = r{r}^{\prime }\left\{ {\cos \theta \cos {\theta }^{\prim... | Yes |
Think of a number, say 5 | Double it 10\nAdd 5 15\nAdd 12 27\nTake away 3 24\nHalve it 12\nTake away number first thought of -5\nThe answer will always be 7 | No |
Think of a number, say 8 | Square it 64\n\nSubtract the square of the number which is\n\n1 less than the number thought of - that\n\nis 7-whose square is 49-leaves 15\n\nAdd 1 16\n\nWhen this last number is told, halve it, and you will arrive at the original number-8 . | Yes |
Example 3. Think of a number, say 9 | Multiply by 3 27\nAdd 2 29\nMultiply by 3 87\nAdd 2 more than the number thought of (11) 98\nThe number of tens in the last answer gives the number thought of, viz., 9. | Yes |
Example 4. Think of a number, say\n\nMultiply by 3 2 1\n\n[If product be odd] add 1\n\nHalve it 11\n\nMultiply by 3 33\n\n[If product be odd] add 1\n\nHalve it 17\n\nAsk how many 9's are in the remainder, when, of course, the reply will be 1 .\n\nThe secret is to bear in mind whether the first sum be odd or even. If od... | In the above example, there being only one 9 in 17, this gives us 4 , which added to 3 produces the number thought of -7 . When even simply add 4 for every 9 in remainder. | No |
Theorem 29. If a triangle \( \Delta \) is decomposed by means of arbitrary straight lines into a finite number of triangles \( {\Delta }_{k} \), then the measure of area of \( \Delta \) is equal to the sum of the measures of area of the separate triangles \( {\Delta }_{k} \) . | Proof. From the distributive law of our algebra of segments, it follows immediately that the measure of area of an arbitrary triangle is equal to the sum of the measures of area of two such triangles as arise from any transversal decomposition of the given triangle. The repeated application of this proposition shows th... | Yes |
THEOREM 38. For an archimedean number system, the commutative law of multiplication is a necessary consequence of the remaining laws of operation; that is to say, if a number system possesses the properties \( 1 - {11},{13} - {17} \) given in \( §{13} \), it follows necessarily that this system satisfies also formula 1... | Proof. Let us observe first of all that, if \( a \) is an arbitrary number of the system, and, if\n\n\[ n = 1 + 1 + \cdots + 1 \]\n\nis a positive integral rational number, then for \( n \) and \( a \) the commutative law of multiplication always holds. In fact, we have\n\n\[ {an} = a\left( {1 + 1 + \cdots + 1}\right) ... | Yes |
Problem 1. To join two points with a straight line and to find the intersection of two straight lines, the lines not being parallel. | Axiom III renders possible the following construction: | No |
Problem 5. To draw a perpendicular to a given straight line. | We can solve problem 5 in the following manner. Let \( A \) be an arbitrary point of the given straight line. Then upon this straight line, lay off in both directions from \( A \) the two equal segments \( {AB} \) and \( {AC} \) . Determine, upon any two straight lines passing through the point \( A \), the points \( E... | Yes |
Theorem 42. Every totally positive number in \( k \) may be represented as the sum of four squares, whose bases are integral or fractional numbers of the field \( k \) . | The demonstration of this theorem presents serious difficulty. It depends essentially upon the theory of relatively quadratic number fields, which I have recently developed in several papers. \( {}^{17} \) We will here call attention only to that proposition in this theory which gives the condition that a ternary dioph... | No |
Theorem 43. Let, \( f\\left( x\\right) \) be an integral rational function of \( x \) whose coefficients are rational numbers and which never becomes negative for any real value of \( x \) . Then \( f\\left( x\\right) \) can always be represented as the quotient of two sums of squares of which the bases are all integra... | Proof. We will denote the degree of the function \( f\\left( x\\right) \) by \( m \), which, in any case, must evidently be even. When \( m = 0 \), that is to say, when \( f\\left( x\\right) \) is a rational number, the validity of theorem 43 follows immediately from Fermat's theorem concerning the representation of a ... | No |
Theorem 1. If two sets of rational numbers \( \left\lbrack r\right\rbrack \) and \( \left\lbrack s\right\rbrack \), having upper bounds, are such that no \( r \) is greater than every \( s \) and no \( s \) greater than every \( r \), then \( \bar{B}\left\lbrack r\right\rbrack \) and \( \bar{B}\left\lbrack s\right\rbra... | Proof. If \( \bar{B}\left\lbrack r\right\rbrack \) is rational, it is evident, and if \( \bar{B}\left\lbrack r\right\rbrack \) is irrational, it is a consequence of Axiom \( \mathrm{K} \) that\n\n\[ \bar{B}\left\lbrack r\right\rbrack > {s}^{\prime } \]\n\nwhere \( {s}^{\prime } \) is any rational number not an upper bo... | Yes |
Theorem 2. If \( a \) and \( b \) are any two distinct real numbers, then \( a < b \) or \( b < a \) ; if \( a < b \) , then not \( b < a \) ; if \( a < b \) and \( b < c \), then \( a < c \) . | Proof. Let \( a, b, c \) all be irrational and let \( \left\lbrack x\right\rbrack ,\left\lbrack y\right\rbrack ,\left\lbrack z\right\rbrack \) be sets of rational numbers determining \( a, b, c \) . In the two sets \( \left\lbrack x\right\rbrack \) and \( \left\lbrack y\right\rbrack \) there is either a number in one s... | Yes |
Theorem 3. If a and \( b \) are any two distinct numbers, then there exists a rational number \( c \) such that \( a < c \) and \( c < b \), or \( b < c \) and \( c < a \) . | Proof. Suppose \( a < b \) . When \( a \) and \( b \) are both rational \( \frac{b - a}{2} \) is a number of the required type. If \( a \) is rational and \( b \) irrational, then the theorem follows from the lemma and Corollary 2, page 4. If \( a \) and \( b \) are both irrational, it follows from Corollary 1, page 4.... | No |
Theorem 4. Every set of numbers \( \left\lbrack x\right\rbrack \) which has an upper bound, has a least upper bound. | Proof. Let \( \left\lbrack r\right\rbrack \) be the set of all rational numbers such that no number of the set \( \left\lbrack r\right\rbrack \) is greater than every number of the set \( \left\lbrack x\right\rbrack \) . Then \( \bar{B}\left\lbrack r\right\rbrack \) is an upper bound of \( \left\lbrack x\right\rbrack \... | Yes |
Theorem 5. Every set \( \left\lbrack x\right\rbrack \) of numbers which has a lower bound has a greatest lower bound. | Proof. The proof may be made by considering the least upper bound of the set \( \left\lbrack y\right\rbrack \) of all numbers, such that every number of \( \left\lbrack y\right\rbrack \) is less than every number of \( \left\lbrack x\right\rbrack \) . The details are left to the reader. | No |
Theorem 6. If all numbers are divided into two sets \( \left\lbrack x\right\rbrack \) and \( \left\lbrack y\right\rbrack \) such that \( x < y \) for every \( x \) and \( y \) of \( \left\lbrack x\right\rbrack \) and \( \left\lbrack y\right\rbrack \), then there is a greatest \( x \) or a least \( y \), but not both. | Proof. The proof is left to the reader. | No |
Theorem 7.\n\\[ \n e = \\bar{B}\\left\\lbrack {\\left( 1 + \\frac{1}{n}\\right) }^{n}\\right\\rbrack \n\\]\n\nwhere \\( \\left\\lbrack n\\right\\rbrack \\) is the set of all positive integers. | Proof. By the binomial theorem for positive integers\n\n\\[ \n{\\left( 1 + \\frac{1}{n}\\right) }^{n} = 1 + n\\left( \\frac{1}{n}\\right) + \\frac{n\\left( {n - 1}\\right) }{2!} \\cdot {\\left( \\frac{1}{n}\\right) }^{2} + \\ldots + {\\left( \\frac{1}{n}\\right) }^{n}. \n\\]\n\nHence\n\n\\[ \n{E}_{n} - {\\left( 1 + \\f... | Yes |
Theorem 11. If an interval \( \overrightarrow{ab} \) is covered by a set of segments \( \left\lbrack \sigma \right\rbrack \), then \( \overrightarrow{ab} \) may be divided into \( N \) equal intervals such that each interval is entirely within a \( \sigma \) . | Proof. By Theorem 10 \( \overrightarrow{ab} \) is covered by a finite set of \( \sigma \) ’s, \( {\sigma }_{1},{\sigma }_{2},\ldots ,{\sigma }_{n} \) . The end points of these \( \sigma \) ’s, together with \( a \) and \( b \), are a finite set of points. Let \( d \) be the smallest distance between any two distinct po... | Yes |
Theorem 14. If every point of the interior or boundary of a parallelogram \( P \) is interior to at least one parallelogram \( p \) of a set of parallelograms \( \left\lbrack p\right\rbrack \), then every point of \( P \) is interior to at least one parallelogram of a finite subset \( {p}_{1}\ldots {p}_{n} \) of \( \le... | Proof. Let \( x = 0, x = a > 0, y = 0, y = b > 0 \) determine the boundary of \( P \) . Let \( 0 \leqq {y}_{1} \leqq b \) . Upon the interval \( i \) of the line \( y = {y}_{1} \), cut off by \( P \), those parallelograms of \( \left\lbrack p\right\rbrack \) that include points of \( i \) as interior points determine a... | Yes |
Theorem 15. Every infinite bounded set \( \left\lbrack p\right\rbrack \) of points on a line has at least one limit point. | Proof. Since the set \( \left\lbrack p\right\rbrack \) is bounded, every one of its points lies on a certain interval \( \overrightarrow{ab} \) . If the set \( \left\lbrack p\right\rbrack \) has no limit point, then about every point of the interval \( \overrightarrow{ab} \) there is a segment \( \sigma \) which contai... | Yes |
Theorem 18. If \( \left\lbrack \left( {x, y}\right) \right\rbrack \) is any set of number pairs and if a is a limit point of the numbers \( \left\lbrack x\right\rbrack \), there is a value of \( b \), finite or \( + \infty \) or \( - \infty \), such that for every \( {V}^{ * }\left( a\right) \) and \( V\left( b\right) ... | Proof. Suppose there is no value \( b \) finite or \( + \infty \) or \( - \infty \) such as is required by the theorem. Since neither \( + \infty \) nor \( - \infty \) possesses the property required of \( b \), there is a \( \overline{{V}^{ * }}\left( a\right) \) and a \( V\left( \infty \right) \) and a \( V\left( {-\... | Yes |
Theorem 19. If on an interval \( \overrightarrow{a}\overrightarrow{b} \) a function has an upper bound \( M \), then it has a least upper bound \( \bar{B} \), and there is at least one value of \( x,{x}_{1} \) on \( \overset{\overleftrightarrow{} }{ab} \) such that the least upper bound of the function on every neighbo... | Proof. (1) The set of values of the function \( f\left( x\right) \) form a bounded set of numbers. By Theorem 4 the set has a least upper bound \( \bar{B} \).\n\n(2) Suppose there were no point \( {x}_{1} \) on \( \overrightarrow{ab} \) such that the least upper bound on every neighborhood of \( {x}_{1} \) contained in... | Yes |
Theorem 20. If \( y \) is a monotonic function of \( x \) on the interval \( \overrightarrow{ab} \), with bounds \( A \) and \( B \), then in turn \( x \) is a single-valued monotonic function of \( y \) on \( \overrightarrow{AB} \), whose upper and lower bounds are \( b \) and \( a \) . | Proof. It follows from the monotonic character of \( y \) as a function of \( x \) that for no two values of \( x \) does \( y \) have the same value. Hence for every value of \( y \) on \( \overrightarrow{AB} \) there exists one and only one value of \( x \). That is, \( x \) is a single-valued function of \( y{.}^{4}... | Yes |
Theorem 21. The function \( {a}^{x} \) for \( x \) on the set \( \left\lbrack \frac{m}{n}\right\rbrack \) is a monotonic increasing function if \( 1 < a \), and a monotonic decreasing function if \( 0 < a < 1 \) . | Proof. (a) For integral values of \( x \) the theorem is obvious.\n\n(b) If \( {x}_{1} = \frac{{m}_{1}}{{n}_{1}} \) and \( {x}_{2} = \frac{{m}_{2}}{{n}_{1}} \), where \( \frac{{m}_{2}}{{n}_{1}} > \frac{{m}_{1}}{{n}_{1}} \), then \( {a}^{{x}_{1}} < {a}^{{x}_{2}} \) if \( a > 1 \) and \( {a}^{{x}_{1}} > {a}^{{x}_{2}} \) ... | Yes |
Theorem 22. If \( x \) is any real number, and \( \left\lbrack \frac{m}{n}\right\rbrack \) the set of all rational numbers less than \( x \), and \( \left\lbrack \frac{p}{q}\right\rbrack \) the set of all rational numbers greater than \( x \), then\n\n\[ \bar{B}\left\lbrack {a}^{\frac{m}{n}}\right\rbrack = \underline{B... | Proof. We give the detailed proof only in the case \( a > 1 \), the other case being similar. By the lemma, since \( \underline{B}\left\lbrack {\frac{p}{q} - \frac{m}{n}}\right\rbrack \) is zero,\n\n\[ \underline{B}\left\lbrack {{a}^{\frac{p}{q}} - {a}^{\frac{m}{n}}}\right\rbrack = \underline{B}\left\lbrack {{a}^{\frac... | Yes |
Theorem 25. If \( f\left( x\right) \) is a non-oscillating function for a set of values \( \left\lbrack x\right\rbrack < a \), a being a limit point of \( \left\lbrack x\right\rbrack \), then as \( x \) approaches a from the left on the set \( \left\lbrack x\right\rbrack, f\left( x\right) \) approaches one and only one... | Proof. Consider an increasing non-oscillating function and let\n\n\[ b = \bar{B}f\left( x\right) \]\n\nfor \( x \) on \( \left\lbrack x\right\rbrack \) .\n\nIn view of the preceding theorem we need to prove only that no value \( {b}^{\prime } \neq b \) can be a value approached. Suppose \( {b}^{\prime } > b \) ; then s... | Yes |
Theorem 26. A necessary and sufficient condition \( {}^{1} \) that \( f\left( x\right) \) shall converge to a unique limit \( b \) as \( x \) approaches \( a \), i.e., that\n\n\[ \underset{x \doteq a}{L}f\left( x\right) = b \]\n\nis that for every \( V\left( b\right) \) there shall exist a \( {V}^{ * }\left( a\right) \... | Proof. (1) The condition is necessary. It is to be proved that if \( \underset{x \doteq a}{L}f\left( x\right) = b \), then for every \( V\left( b\right) \) there exists a \( {V}^{ * }\left( a\right) \) such that for every \( x \) in \( {V}^{ * }\left( a\right) \) the corresponding \( f\left( x\right) \) is in \( V\left... | Yes |
Theorem 27. A necessary and sufficient condition that \( f\left( x\right) \) shall converge to a finite limit as \( x \) approaches a is that for every \( \varepsilon > 0 \) there shall exist a \( {V}_{\varepsilon }^{ * }\left( a\right) \) such that if \( {x}_{1} \) and \( {x}_{2} \) are any two values of \( x \) in \(... | Proof. (1) The condition is necessary. If \( {Lf}\left( x\right) = b \) and \( b \) is finite, then by the preceding theorem for every \( \frac{\varepsilon }{2} > 0 \) there exists a \( {V}^{ * }\left( a\right) \) such that if \( {x}_{1} \) and \( {x}_{2} \) are in \( {V}^{ * }\left( a\right) \), then\n\n\[ \left| {f\l... | Yes |
Corollary 4. The expression\n\n\[ \underset{\begin{matrix} {x < a} \\ {x \doteq a} \end{matrix}}{L}f\left( x\right) = b \]\n\nis equivalent to\n\n\[ \underset{z \doteq + \infty }{L}f\left( {a + \frac{1}{z}}\right) = b \]\n\nwhere \( z = \frac{1}{x - a} \) . | The reader should verify these corollaries by writing down the necessary and sufficient condition for the existence of each limit. | No |
Corollary 5. If\n\n\\[ \n\\underset{x \\doteq a}{L}f\\left( x\\right) = b \n\\]\n\nthen\n\n\\[ \n\\underset{x \\doteq a}{L}\\left| {f\\left( x\\right) }\\right| = \\left| b\\right| \n\\] | Proof. By the necessary condition of Theorem 26 for every \\( \\varepsilon \\) there exists a \\( {V}_{\\varepsilon }^{ * }\\left( a\\right) \\) such that for every \\( {x}_{1} \\) of \\( {V}_{\\varepsilon }^{ * }\\left( a\\right) \\)\n\n\\[ \n\\left| {f\\left( {x}_{1}\\right) - b}\\right| < \\varepsilon \n\\]\n\nIf \\... | Yes |
Corollary 6. If a function \( f\left( x\right) \) is continuous at \( x = a \), then \( \left| {f\left( x\right) }\right| \) is continuous at \( x = a \) . | It should be noticed that\n\n\[\n\underset{x \doteq a}{L}\left| {f\left( x\right) }\right| = \left| b\right|\n\]\n\nis not equivalent to\n\n\[\n\underset{x \doteq a}{L}f\left( x\right) = b.\n\]\n\nSuppose \( f\left( x\right) = + 1 \) for all rational values of \( x \) and \( f\left( x\right) = - 1 \) for all irrational... | No |
If for every sequence of numbers \( \left\lbrack {x}_{n}\right\rbrack \) having \( a \) as a limit point, \[ \underset{\begin{matrix} {x \mid \left\lbrack {x}_{n}\right\rbrack } \\ {x \doteq a} \end{matrix}}{L}f\left( x\right) = b, \;\text{ then }\;\underset{x \doteq a}{L}f\left( x\right) = b. \] | Proof. In case two values \( b \) and \( {b}_{1} \) were approached by \( f\left( x\right) \) as \( x \) approaches \( a \), then, as in the first part of the proof of Theorem 26, two sequences could be chosen upon one of which \( f\left( x\right) \) approached \( b \) and upon the other of which \( f\left( x\right) \)... | No |
Theorem 29. If \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{b}_{k} \) is a convergent series all of whose terms are positive and \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k} \) is a series such that for every \( k,\left| {a}_{k}\right| \leqq {b}_{k} \), then\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k}... | Proof. By hypothesis\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{n}\left| {a}_{k}\right| \leqq \mathop{\sum }\limits_{{k = 0}}^{n}{b}_{k} \]\n\nHence\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{n}\left| {a}_{k}\right| \]\n\nis bounded, and being an increasing function of \( n \), the series is convergent according to Theorem 25. | Yes |
Theorem 30. If there exists a number, \( r,0 < r < 1 \), such that\n\n\[ \left| \frac{{a}_{n}}{{a}_{n - 1}}\right| < r \]\n\nfor every integral value of \( n \), then the series\n\n\[ {a}_{1} + {a}_{2} + \ldots + {a}_{n} + \ldots \]\n\n(1)\n\nis absolutely convergent. If \( \left| \frac{{a}_{n}}{{a}_{n - 1}}\right| \eq... | Proof. The series (1) may be written\n\n\[ {a}_{1} + {a}_{1}\frac{{a}_{2}}{{a}_{1}} + {a}_{1}\frac{{a}_{2}}{{a}_{1}} \cdot \frac{{a}_{3}}{{a}_{2}} + \ldots + {a}_{1}\frac{{a}_{2}}{{a}_{1}}\ldots \frac{{a}_{n}}{{a}_{n - 1}} \]\n\n(2)\n\n\( \left| \frac{{a}_{n}}{{a}_{n - 1}}\right| < r \), this is numerically less term b... | Yes |
Theorem 31. A necessary and sufficient condition that\n\n\[ \underset{x \doteq a}{L}f\\left( x\\right) = b \]\n\nis that for the function \( \\varepsilon \\left( x\\right) \) defined by the equation \( f\\left( x\\right) = b + \\varepsilon \\left( x\\right) \)\n\n\[ \underset{x \doteq a}{L}\\varepsilon \\left( x\\right... | Proof. Take \( \\varepsilon \\left( x\\right) = f\\left( x\\right) - b \) and apply Theorem 26. A special case of this theorem is: \( A \) necessary and sufficient condition for the convergence of a series to a finite value \( b \) is that for every \( \\varepsilon > 0 \) there exists an integer \( {N}_{\\varepsilon } ... | No |
Theorem 32. The sum, difference, or product of two infinitesimals is an infinitesimal. | Proof. Let the two infinitesimals be \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \). For every \( \varepsilon ,1 > \varepsilon > 0 \), there exists a \( {V}_{1}^{ * }\left( a\right) \) for every \( x \) of which\n\n\[ \left| {{f}_{1}\left( x\right) }\right| < \frac{\varepsilon }{2} \]\n\nand a \( {V}_{2}... | Yes |
Theorem 33. If \( f\left( x\right) \) is bounded on a certain \( \overline{{V}^{ * }}\left( a\right) \) and \( \varepsilon \left( x\right) \) is an infinitesimal as \( x \) approaches a, then \( \varepsilon \left( x\right) \cdot f\left( x\right) \) is also an infinitesimal as \( x \) approaches a. | Proof. By hypothesis there are two numbers \( m \) and \( M \), such that \( M > f\left( x\right) > m \) for every \( x \) on \( \overline{{V}^{ * }}\left( a\right) \) . Let \( k \) be the larger of \( \left| m\right| \) and \( \left| M\right| \) . Also by hypothesis there exists for every \( \varepsilon \) a \( {V}_{\... | Yes |
Theorem 35. If \( {f}_{2}\left( x\right) \) has a lower bound on some \( {V}^{ * }\left( a\right) \), and if\n\n\[ \underset{x \doteq 0}{L}{f}_{1}\left( x\right) = + \infty \]\n\nthen\n\n\[ \underset{x \doteq 0}{L}\left\{ {{f}_{2}\left( x\right) + {f}_{1}\left( x\right) }\right\} = + \infty . \] | Proof. Let \( M \) be the lower bound of \( {f}_{2}\left( x\right) \) . By hypothesis, for every number \( E \) there exists a \( {V}_{E}^{ * }\left( a\right) \) such that for \( x \) on \( {V}_{E}^{ * }\left( a\right) \)\n\n\[ {f}_{1}\left( x\right) > E - M. \]\n\nSince\n\n\[ {f}_{2}\left( x\right) > M \]\n\nthis give... | Yes |
Theorem 36. If \( \underset{x = a}{L}{f}_{1}\left( x\right) = + \infty \) or \( - \infty \), and if \( {f}_{2}\left( x\right) \) is such that for a \( \overline{{V}^{ * }}\left( a\right) ,{f}_{2}\left( x\right) \) has a lower bound greater than zero or an upper bound less than zero, then \( \underset{x \doteq a}{L}\lef... | Proof. Suppose \( {f}_{2}\left( x\right) \) has a lower bound greater than zero, say \( M \), and that \( \underset{x \doteq a}{L}{f}_{1}\left( x\right) = \) \( + \infty \) . Then for every \( E \) there exists a \( {V}_{E}^{ * }\left( a\right) \) within \( \overline{{V}^{ * }}\left( a\right) \) such that for every \( ... | Yes |
Theorem 37. If \( \underset{x \doteq a}{L}f\left( x\right) = + \infty \), then \( \underset{x \doteq a}{L}\frac{1}{f\left( x\right) } = 0 \), and there is a vicinity \( {V}^{ * }\left( a\right) \) upon which \( f\left( x\right) > 0 \) . Conversely, if \( \underset{x \doteq a}{L}f\left( x\right) = 0 \) and there is a \(... | Proof. If \( \underset{x \doteq a}{L}f\left( x\right) = + \infty \), then for every \( \varepsilon \) there exists a \( {V}_{\varepsilon }^{ * }\left( a\right) \) such that if \( x \) is in \( {V}_{\varepsilon }^{ * }\left( a\right) \), then \[ f\left( x\right) > \frac{1}{\varepsilon } \] and \[ \frac{1}{f\left( x\righ... | Yes |
Corollary 1. If \( {f}_{1}\left( x\right) \) has finite upper and lower bounds on some \( {V}^{ * }\left( a\right) \) and \( \underset{x \doteq a}{L}{f}_{2}\left( x\right) = + \infty \) or \( - \infty \), then | \[ \underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } = 0. \] | No |
Corollary 2. If \( {f}_{2}\left( x\right) \) is positive and \( {f}_{1}\left( x\right) \) has a positive lower bound on some \( {V}^{ * }\left( a\right) \) and \( \underset{x \doteq a}{L}{f}_{2}\left( x\right) = 0 \), then | \[ \underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } = + \infty . \] | Yes |
Theorem 38. (change of variable). If\n\n(1) \( \underset{x = a}{L}{f}_{1}\left( x\right) = {b}_{1} \) and \( \underset{x = {b}_{1}}{L}{f}_{2}\left( y\right) = {b}_{2} \) when \( y \) takes all valves of \( {f}_{1}\left( x\right) \) corresponding to values of \( x \) on some \( \overline{{V}^{ * }}\left( a\right) \), an... | Proof. ( \( \alpha \) ) Since \( \underset{y = {b}_{1}}{L}{f}_{2}\left( y\right) = {b}_{2} \), for every \( V\left( {b}_{2}\right) \) there exists a \( {V}^{ * }\left( {b}_{1}\right) \) such that if \( y \) is in \( {V}^{ * }\left( {b}_{1}\right) ,{f}_{2}\left( y\right) \) is in \( V\left( {b}_{2}\right) \) . Since \( ... | Yes |
Theorem 39. If \( \underset{x \doteq a}{L}{f}_{1}\left( x\right) = b \) and \( \underset{y \doteq b}{L}{f}_{2}\left( y\right) = {f}_{2}\left( b\right) \), where \( y \) takes all values taken by \( {f}_{1}\left( x\right) \) for \( x \) on some \( \overline{{V}^{ * }}\left( a\right) \), then | \[ \underset{x \doteq a}{L}{f}_{2}\left( {{f}_{1}\left( x\right) }\right) = {f}_{2}\left( b\right) \] Proof. The proof of the theorem is similar to that of Theorem 38. In this case the notation \( {f}_{2}\left( b\right) \) implies that \( b \) is a finite number. Thus for every \( {\varepsilon }_{1} \) there exists a \... | Yes |
Theorem 40. If \( f\left( x\right) \leqq b \) for all values of a set \( \left\lbrack x\right\rbrack \) on a certain \( {V}^{ * }\left( a\right) \), then every value approached by \( f\left( x\right) \) as \( x \) approaches \( a \) is less than or equal to \( b \) . Similarly if \( f\left( x\right) \geqq b \) for all ... | Proof. If \( f\left( x\right) \leqq b \) on \( {V}^{ * }\left( a\right) \), then if \( {b}^{\prime } \) is any value greater than \( b \), and \( V\left( {b}^{\prime }\right) \) any vicinity of \( {b}^{\prime } \) which does not include \( b \), there is no value of \( x \) on \( {V}^{ * }\left( a\right) \) for which \... | Yes |
Corollary 2. If \( {f}_{1}\left( x\right) \geqq {f}_{2}\left( x\right) \) in the neighborhood of \( x = a \), then\n\n\[ \underset{x \doteq a}{L}{f}_{1}\left( x\right) \geqq \underset{x \doteq a}{L}{f}_{2}\left( x\right) \]\n\nif both these limits exist. | Proof. Apply Corollary 1 to \( {f}_{1}\left( x\right) - {f}_{2}\left( x\right) \) . | Yes |
Corollary 4. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are both positive in the neighborhood of \( x = a \), and if \( {f}_{1}\left( x\right) \geqq {f}_{2}\left( x\right) \), then if \( \underset{x \doteq a}{L}{f}_{1}\left( x\right) = 0 \), it follows that | \[ \underset{x \doteq a}{L}{f}_{2}\left( x\right) = 0. \] | Yes |
Theorem 41. If \( \left\lbrack {x}^{\prime }\right\rbrack \) is a subset of \( \left\lbrack x\right\rbrack \), a being a limit point of \( \left\lbrack {x}^{\prime }\right\rbrack \), and if \( \underset{x = a}{L}f\left( x\right) \) exists, then \( {\iint }_{{x}^{\prime } \doteq a}f\left( {x}^{\prime }\right) \) exists ... | Proof. By hypothesis there exists for every \( V\left( b\right) \) a \( {V}^{ * }\left( a\right) \) such that for every \( x \) of the set \( \left\lbrack x\right\rbrack \) which is in \( {V}^{ * }\left( a\right), f\left( x\right) \) is in \( V\left( b\right) \) . Since \( \left\lbrack {x}^{\prime }\right\rbrack \) is ... | Yes |
Theorem 42. For every \( \varepsilon \) for which the set of values of \( {\delta }_{\varepsilon } \) has an upper bound there is a greatest \( {\delta }_{\varepsilon } \) . | Proof. Let \( \bar{B}\left\lbrack {\delta }_{\varepsilon }\right\rbrack \) be the least upper bound of the set of values of \( {\delta }_{\varepsilon } \), for a particular \( \varepsilon \) . If \( x \) is such that \( \left| {x - a}\right| < \bar{B}\left\lbrack {\delta }_{\varepsilon }\right\rbrack \), then there is ... | Yes |
Theorem 43. The limit of the least upper bound of a function \( f\left( x\right) \) on a variable segment \( \overline{ax}, a < x \), as the end point approaches \( a \), is the least upper bound of the values approached by the function as \( x \) approaches a from the right. | Proof. Let \( l \) be the least upper bound of the values approached by the function as \( x \) approaches \( a \) from the right, and let \( b\left( x\right) \) represent the upper bound of \( f\left( x\right) \) for all values of \( x \) on \( \bar{a}\bar{x} \) . Since \( \bar{B}f\left( x\right) \) on the segment \( ... | Yes |
Theorem 46. If \( f\left( x\right) \) is continuous at a point \( x = a \) and if \( f\left( a\right) \) is positive, then there is a neighborhood of \( x = a \) upon which the function is positive. | Proof. If there were values of \( x,\left\lbrack {x}^{\prime }\right\rbrack \) within every neighborhood of \( x = a \) for which the function is equal to or less than zero, then by Theorem 24 there would be a value approached by \( f\left( {x}^{\prime }\right) \) as \( {x}^{\prime } \) approaches \( a \) on the set \(... | Yes |
Theorem 47. If \( f\left( x\right) \) is continuous on a finite interval \( \overrightarrow{ab} \), then for every \( \varepsilon > 0,\overrightarrow{ab} \) can be divided into a finite number of equal intervals upon each of which the oscillation of \( f\left( x\right) \) is less than \( \varepsilon {.}^{1} \) | Proof. By Theorem 45 there is about every point of \( \overrightarrow{ab} \) a segment \( \sigma \) upon which the oscillation is less than \( \varepsilon \) . This set of segments \( \left\lbrack \sigma \right\rbrack \) covers \( {ab} \), and by Theorem \( {11ab} \) can be divided into a finite number of equal interva... | No |
Theorem 48. (Uniform continuity.) If a function is continuous on a finite interval \( \overrightarrow{a}\overrightarrow{b} \) , then for every \( \varepsilon > 0 \) there exists a \( {\delta }_{\varepsilon } > 0 \) such that for any two values of \( x,{x}_{1} \), and \( {x}_{2} \), on\n\n\( {ab} \) where \( \left| {{x}... | Proof. This theorem may be inferred in an obvious way from the preceding theorem, or it may be proved directly as follows:\n\nBy Theorem 27, for every \( \varepsilon \) there exists a neighborhood \( {V}_{\varepsilon }\left( {x}^{\prime }\right) \) of every \( {x}^{\prime } \) of \( \overrightarrow{ab} \) such that if ... | Yes |
Theorem 49. If a function is continuous on an interval \( \overrightarrow{a}\overrightarrow{b} \), it is bounded on that interval. | Proof. By Theorem 46 the interval \( \overrightarrow{ab} \) can be divided into a finite number of intervals, such that the oscillation on each interval is less than a given positive number \( \varepsilon \) . If the\n\nnumber of intervals is \( n \), then the oscillation on the interval \( {ab} \) is less than \( {n\v... | Yes |
Theorem 50. If a function \( f\left( x\right) \) is continuous on an interval \( \overrightarrow{a}\overrightarrow{b} \), then the function assumes as values its least upper and its greatest lower bound. | Proof. By the preceding theorem the function is bounded and hence the least upper and greatest lower bounds are finite. By Theorem 19 there is a point \( k \) on the interval \( \overrightarrow{ab} \) such that the least upper bound of the function on every neighborhood of \( x = k \) is the same as the least upper bou... | Yes |
Theorem 51. If a function is continuous on an interval \( \overrightarrow{a}\overrightarrow{b} \), then the function takes on all values between its least upper and its greatest lower bound. | Proof. If there is a value \( k \) between these bounds which is not assumed by a continuous function \( f\left( x\right) \), then by the corollary of the preceding theorem there is a value \( \Delta \) such that no values of \( f\left( x\right) \) are between \( k - \Delta \) and \( k + \Delta \) . With \( \varepsilon... | Yes |
Theorem 52. If \( y \) is a function, \( f\left( x\right) \), of \( x \), monotonic and continuous on an interval \( {ab} \), then \( x = {f}^{-1}\left( y\right) \) is a function of \( y \) which is monotonic and continuous on the interval \( f\left( a\right) f\left( b\right) \) . | Proof. By Theorem 20 the function \( {f}^{-1}\left( y\right) \) is monotonic and has as upper and lower bounds \( a \) and \( b \) . By Theorems 50 and 51 the function is defined for every value of \( y \) between and including \( f\left( a\right) \) and \( f\left( b\right) \) and for no other values. We prove the func... | Yes |
Theorem 53. If \( f\left( x\right) \) is single-valued and continuous with \( A, B \) as lower and upper bounds, on an interval \( \overrightarrow{ab} \) and has a single-valued inverse on the interval, \( \overrightarrow{AB} \) then \( f\left( x\right) \) is monotonic on \( \overrightarrow{a}\overrightarrow{b} \) . | Proof. If \( f\left( x\right) \) is not monotonic, then there must be three values of \( x \) , \[ {x}_{1} < {x}_{2} < {x}_{3} \] such that either \[ f\left( {x}_{1}\right) \leqq f\left( {x}_{2}\right) \geqq f\left( {x}_{3}\right) \] or \[ f\left( {x}_{1}\right) \geqq f\left( {x}_{2}\right) \leqq f\left( {x}_{3}\right)... | Yes |
Theorem 54. If the functions \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are continuous on the interval \( \overrightarrow{a}\overrightarrow{b} \), and if \( {f}_{1}\left( x\right) = {f}_{2}\left( x\right) \) on a set everywhere dense, then \( {f}_{1}\left( x\right) = {f}_{2}\left( x\right) \) on the ... | Proof. Let \( \left\lbrack {x}^{\prime }\right\rbrack \) be the set everywhere dense on \( \overrightarrow{ab} \) for which, by hypothesis, \( {f}_{1}\left( x\right) = {f}_{2}\left( x\right) \) . Let \( {x}^{\prime \prime } \) be any point of the interval not of the set \( \left\lbrack {x}^{\prime }\right\rbrack \) . B... | Yes |
Theorem 56. \[ \underset{n \doteq \infty }{L}{\left( 1 + \frac{x}{n}\right) }^{n} \] where \( \left\lbrack n\right\rbrack \) is the set of all positive integers, exists and is equal to \( e\left( x\right) \) for all values of \( x \) . | Proof. Let \[ {E}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{n}\frac{{x}^{k}}{k!} \] (where \( 0! = 1 \) ). Then, since \[ {\left( 1 + \frac{x}{n}\right) }^{n} = 1 + \frac{n!}{\left( {n - 1}\right) !} \cdot \frac{x}{n} + \frac{n!}{\left( {n - 2}\right) ! \cdot 2!}{\left( \frac{x}{n}\right) }^{2} + \ldots + \... | Yes |
Theorem 57. \[ \underset{z \doteq \infty }{L}\left( {1 + \frac{x}{z}}\right) \] where \( \left\lbrack z\right\rbrack \) is the set of all real numbers, exists and is equal to \( e\left( x\right) \) . | Proof. If \( z \) is any number greater than 1, let \( {n}_{z} \) be the integer such that \[ {n}_{z} \leqq z < {n}_{z} + 1 \] Hence, if \( x > 0 \), \[ 1 + \frac{x}{{n}_{z}} \geqq 1 + \frac{x}{z} > 1 + \frac{x}{{n}_{z} + 1} \] (1) Hence \[ {\left( 1 + \frac{x}{{n}_{z}}\right) }^{{n}_{z} + 1} \geqq {\left( 1 + \frac{x}... | Yes |
Theorem 58. The function \( e\left( x\right) \) is the same as \( {e}^{x} \) where\n\n\[ e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots \] | Proof. By the continuity of \( {z}^{x} \) as a function of \( z \) (see Corollary 2 of Theorem 39), it follows that, since\n\n\[ \underset{n \doteq \infty }{L}{\left( 1 + \frac{1}{n}\right) }^{n} = e \]\n\n\[ \underset{n \doteq \infty }{L}{\left( 1 + \frac{1}{n}\right) }^{nx} = {e}^{x} \]\n\nBut\n\n\[ {\left( 1 + \frac... | Yes |
Theorem 60. The function \( {cf}\left( x\right) \) is of the same order as \( f\left( x\right), c \) being any constant not zero. | Proof. By Theorem 34, \( \underset{x \doteq a}{L}\frac{{cf}\left( x\right) }{f\left( x\right) } = c \) . | Yes |
Theorem 61. If \( {f}_{1}\left( x\right) \) is of the same order as \( {f}_{2}\left( x\right) \), and \( {f}_{2}\left( x\right) \) is of the same order as \( {f}_{3}\left( x\right) \), then \( {f}_{1}\left( x\right) \) and \( {f}_{3}\left( x\right) \) are of the same order. | Proof. By hypothesis \( \underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } = {k}_{1} \) and \( \underset{x \doteq a}{L}\frac{{f}_{2}\left( x\right) }{{f}_{3}\left( x\right) } = {k}_{2} \) . By Theorem 34,\n\n\[ \underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } ... | Yes |
Theorem 62. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are infinitesimal (infinite) and neither is zero or changes sign on some \( {V}^{ * }\left( a\right) \), then \( {f}_{1}\left( x\right) \cdot {f}_{2}\left( x\right) \) is infinitesimal (infinite) of a higher order than either. | Proof.\n\[\n\underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) \cdot {f}_{2}\left( x\right) }{{f}_{2}\left( x\right) } = \underset{x \doteq a}{L}{f}_{1}\left( x\right) = 0.\left( {\pm \infty .}\right)\n\] | Yes |
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