Q
stringlengths 4
3.96k
| A
stringlengths 1
3k
| Result
stringclasses 4
values |
|---|---|---|
Theorem 55.7. If \( \Gamma \vDash \varphi \) then \( \Gamma \vdash \varphi \) .
|
Proof. We prove the contrapositive: Suppose \( \Gamma \nvdash \varphi \) . Then by Lemma 55.4, there is a prime set \( {\Gamma }^{ * } \supseteq \Gamma \) such that \( {\Gamma }^{ * } \nvdash \varphi \) . Consider the canonical model \( \mathfrak{M}\left( {\Gamma }^{ * }\right) \) for \( {\Gamma }^{ * } \) as defined in Definition 55.5. For any \( \psi \in \Gamma ,{\Gamma }^{ * } \vdash \psi \) . Note that \( {\Gamma }^{ * }\left( \Lambda \right) = {\Gamma }^{ * } \) . By the Truth Lemma (Lemma 55.6), we have \( \mathfrak{M}\left( {\Gamma }^{ * }\right) ,\Lambda \Vdash \psi \) for all \( \psi \in \Gamma \) and \( \mathfrak{M}\left( {\Gamma }^{ * }\right) ,\Lambda \nVdash \varphi \) . This shows that \( \Gamma \mathrel{\text{\vDash \not{} }} \varphi \) .
|
Yes
|
Lemma 56.6. If \( M\left\lbrack {{N}^{\varphi }/{x}^{\varphi }}\right\rbrack \) is a redex and \( M \neq x \), then one of the following cases holds:\n\n1. \( M \) is itself a redex, or\n\n2. \( M \) is of the form \( {\mathrm{p}}_{i}\left( x\right) \), and \( N \) is of the form \( \left\langle {{P}_{1},{P}_{2}}\right\rangle \)\n\n3. \( M \) is of the form case \( \left( {i,{x}_{1}.{P}_{1},{x}_{2}.{P}_{2}}\right) \), and \( N \) is of the form \( {\operatorname{in}}_{i}\left( Q\right) \)\n\n4. \( M \) is of the form \( {xQ} \), and \( N \) is of the form \( {\lambda x} \) . \( P \)\n\nIn the first case, \( \operatorname{cr}\left( {M\left\lbrack {N/x}\right\rbrack }\right) = \operatorname{cr}\left( M\right) \) ; in the other cases, \( \operatorname{cr}\left( {M\left\lbrack {N/x}\right\rbrack }\right) = \operatorname{len}\left( \varphi \right) \) ).
|
Proof. Proof by induction on \( M \) .\n\n1. If \( M \) is a single variable \( y \) and \( y ≢ x \), then \( y\left\lbrack {N/x}\right\rbrack \) is \( y \), hence not a redex.\n\n2. If \( M \) is of the form \( \left\langle {{N}_{1},{N}_{2}}\right\rangle \), or \( {\lambda x}.N \), or \( {\operatorname{in}}_{i}^{\varphi }\left( N\right) \), then \( M\left\lbrack {{N}^{\varphi }/{x}^{\varphi }}\right\rbrack \) is also of that form, and so is not a redex.\n\n3. If \( M \) is of the form \( {\mathrm{p}}_{i}\left( P\right) \), we consider two cases.\n\na) If \( P \) is of the form \( \left\langle {{P}_{1},{P}_{2}}\right\rangle \), then \( M \equiv {\mathrm{p}}_{i}\left( \left\langle {{P}_{1},{P}_{2}}\right\rangle \right) \) is a redex, and clearly\n\n\[ M\left\lbrack {N/x}\right\rbrack \equiv {\mathrm{p}}_{i}\left( \left\langle {{P}_{1}\left\lbrack {N/x}\right\rbrack ,{P}_{2}\left\lbrack {N/x}\right\rbrack }\right\rangle \right) \]\n\nis also a redex. The cut ranks are equal.\n\nb) If \( P \) is a single variable, it must be \( x \) to make the substitution a redex, and \( N \) must be of the form \( \left\langle {{P}_{1},{P}_{2}}\right\rangle \) . Now consider\n\n\[ M\left\lbrack {N/x}\right\rbrack \equiv {\mathrm{p}}_{i}\left( x\right) \left\lbrack {\left\langle {{P}_{1},{P}_{2}}\right\rangle /x}\right\rbrack \]\n\nwhich is \( {\mathrm{p}}_{i}\left( \left\langle {{P}_{1},{P}_{2}}\right\rangle \right) \) . Its cut rank is equal to \( \operatorname{cr}\left( x\right) \), which is \( \operatorname{len}\left( \varphi \right) \) .\n\nThe cases of case \( \left( {N,{x}_{1} \cdot {N}_{1},{x}_{2} \cdot {N}_{2}}\right) \) and \( {PQ} \) are similar.
|
Yes
|
Lemma 56.7. If \( M \) contracts to \( {M}^{\prime } \), and \( \operatorname{cr}\left( M\right) > \operatorname{cr}\left( N\right) \) for all proper redex sub-terms \( N \) of \( M \), then \( \operatorname{cr}\left( M\right) > \operatorname{mr}\left( {M}^{\prime }\right) \) .
|
Proof. Proof by cases.\n\n1. If \( M \) is of the form \( {\mathrm{p}}_{i}\left( \left\langle {{M}_{1},{M}_{2}}\right\rangle \right) \), then \( {M}^{\prime } \) is \( {M}_{i} \) ; since any sub-term of \( {M}_{i} \) is also proper sub-term of \( M \), the claim holds.\n\n2. If \( M \) is of the form \( \left( {\lambda {x}^{\varphi } \cdot N}\right) {Q}^{\varphi } \), then \( {M}^{\prime } \) is \( N\left\lbrack {{Q}^{\varphi }/{x}^{\varphi }}\right\rbrack \) . Consider a redex in \( {M}^{\prime } \) . Either there is corresponding redex in \( N \) with equal cut rank, which is less than \( \operatorname{cr}\left( M\right) \) by assumption, or the cut rank equals \( \operatorname{len}\left( \varphi \right) \) , which by definition is less than \( \operatorname{cr}\left( {\left( {\lambda {x}^{\varphi } \cdot N}\right) Q}\right) \).\n\n3. If \( M \) is of the form\n\n\[ \operatorname{case}\left( {{\operatorname{in}}_{i}\left( {N}^{{\varphi }_{i}}\right) ,{x}_{1}^{{\varphi }_{1}}.{N}_{1}^{\chi },{x}_{2}^{{\varphi }_{2}}.{N}_{2}^{\chi }}\right) ,\]\n\nthen \( {M}^{\prime } \equiv {N}_{i}\left\lbrack {N/{x}_{i}^{{\varphi }_{i}}}\right\rbrack \) . Consider a redex in \( {M}^{\prime } \) . Either there is corresponding redex in \( {N}_{i} \) with equal cut rank, which is less than \( \operatorname{cr}\left( M\right) \) by assumption; or the cut rank equals \( \operatorname{len}\left( {\varphi }_{i}\right) \), which by definition is less than \( \operatorname{cr}\left( {\operatorname{case}\left( {{\operatorname{in}}_{i}\left( {N}^{{\varphi }_{i}}\right) ,{x}_{1}^{{\varphi }_{1}} \cdot {N}_{1}^{\chi },{x}_{2}^{{\varphi }_{2}} \cdot {N}_{2}^{\chi }}\right) }\right) \) .
|
Yes
|
Theorem 56.8. All proof terms reduce to normal form; all derivations reduce to normal derivations.
|
Proof. The second follows from the first. We prove the first by complete induction on \( m = \operatorname{mr}\left( M\right) \), where \( M \) is a proof term.\n\n1. If \( m = 0, M \) is already normal.\n\n2. Otherwise, we proceed by induction on \( n \), the number of redexes in \( M \) with cut rank equal to \( m \) .\n\na) If \( n = 1 \), select any redex \( N \) such that \( m = \operatorname{cr}\left( N\right) > \operatorname{cr}\left( P\right) \) for any proper sub-term \( P \) which is also a redex of course. Such a redex must exist, since any term only has finitely many subterms.\n\nLet \( {N}^{\prime } \) denote the reductum of \( N \) . Now by the lemma \( \operatorname{mr}\left( {N}^{\prime }\right) < \) \( \mathrm{{mr}}\left( N\right) \), thus we can see that \( n \), the number of redexes with \( \operatorname{cr}\left( = \right) m \) is decreased. So \( m \) is decreased (by 1 or more), and we can apply the inductive hypothesis for \( m \) . 56.7. NORMALIZATION\n\nb) For the induction step, assume \( n > 1 \) . the process is similar, except that \( n \) is only decreased to a positive number and thus \( m \) does not change. We simply apply the induction hypothesis for \( n \) .
|
Yes
|
The sphere semantics invalidates the inference, i.e., we have \( p▱ \rightarrow r \mathrel{\text{\vDash \not{} }} \left( {p \land q}\right) ▱ \rightarrow r \) .
|
Consider the model \( \mathfrak{M} = \langle W, O, V\rangle \) where \( W = \left\{ {w,{w}_{1},{w}_{2}}\right\} ,{O}_{w} = \left\{ {\{ w\} ,\left\{ {w,{w}_{1}}\right\} ,\left\{ {w,{w}_{1},{w}_{2}}\right\} }\right\}, V\left( p\right) = \left\{ {{w}_{1},{w}_{2}}\right\}, V\left( q\right) = \left\{ {w}_{2}\right\} \), and \( V\left( r\right) = \left\{ {w}_{1}\right\} \). There is a \( p \) -admitting sphere \( S = \left\{ {w,{w}_{1}}\right\} \) and \( p \rightarrow r \) is true at all worlds in it, so \( \mathfrak{M}, w \Vdash p ▱ \rightarrow r \). There is also a \( \left( {p \land q}\right) \) -admitting sphere \( {S}^{\prime } = \left\{ {w,{w}_{1},{w}_{2}}\right\} \) but \( \mathfrak{M},{w}_{2} \nVdash \left( {p \land q}\right) \rightarrow r \), so \( \mathfrak{M}, w \nVdash \left( {p \land q}\right) ▱ \rightarrow r \) (see Figure 58.7).
|
Yes
|
Example 58.4. The sphere semantics invalidates the inference, i.e., we have \( p▱ \rightarrow q, q▱ \rightarrow r \mathrel{\text{\vDash \not{} }} p▱ \rightarrow r \) .
|
Consider the model \( \mathfrak{M} = \langle W, O, V\rangle \) where \( W = \) \( \left\{ {w,{w}_{1},{w}_{2}}\right\} ,{O}_{w} = \left\{ {\{ w\} ,\left\{ {w,{w}_{1}}\right\} ,\left\{ {w,{w}_{1},{w}_{2}}\right\} }\right\}, V\left( p\right) = \left\{ {w}_{2}\right\}, V\left( q\right) = \left\{ {{w}_{1},{w}_{2}}\right\} \) and \( V\left( r\right) = \left\{ {w}_{1}\right\} \) . There is a \( p \) -admitting sphere \( S = \left\{ {w,{w}_{1},{w}_{2}}\right\} \) and \( p \rightarrow q \) is true at all worlds in it, so \( \mathfrak{M}, w \Vdash p ▱ \rightarrow q \) . There is also a \( q \) -admitting sphere \( {S}^{\prime } = \left\{ {w,{w}_{1}}\right\} \) and \( \mathfrak{M} \nVdash q \rightarrow r \) is true at all worlds in it, so \( \mathfrak{M}, w \Vdash q▱ \rightarrow r \) . However, the \( p \) -admitting sphere \( \left\{ {w,{w}_{1},{w}_{2}}\right\} \) contains a world, namely \( {w}_{2} \), where \( \mathfrak{M},{w}_{2} \nVdash p \rightarrow r \) .
|
Yes
|
The sphere semantics invalidates contraposition, i.e., we have \( p▱ \rightarrow q \mathrel{\text{\vDash \not{} }} \neg q▱ \rightarrow \neg p \) .
|
Think of \( p \) as \
|
No
|
Theorem 59.1 (Russell’s Paradox). There is no set \( R = \{ x : x \notin x\} \)
|
Proof. For reductio, suppose that \( R = \{ x : x \notin x\} \) exists. Then \( R \in R \) iff \( R \notin R \), by Extensionality. Contradiction!
|
Yes
|
Lemma 59.2 (working in Grundgesetze). \( \forall F\forall a\left( {a \in {\epsilon xF}\left( x\right) \leftrightarrow {Fa}}\right) \)
|
Proof. Fix \( F \) and \( a \) . Now \( a \in {\epsilon xF}\left( x\right) \) iff \( \exists G\left( {{\epsilon xF}\left( x\right) = {\epsilon xG}\left( x\right) \land {Ga}}\right) \) (by the definition of membership) iff \( \exists G\left( {\forall x\left( {{Fx} \leftrightarrow {Gx}}\right) \land {Ga}}\right) \) (by Basic Law V) iff \( {Fa} \) (by elementary second-order logic).
|
Yes
|
Lemma 59.3 (working in Grundgesetze.). \( \forall F\exists s\forall a\left( {a \in s \leftrightarrow {Fa}}\right) \)
|
Proof. Fix \( F \) ; now Lemma 59.2 yields \( \forall a\left( {a \in {exF}\left( x\right) \leftrightarrow {Fa}}\right) \) ; so \( \exists s\forall a(a \in s \leftrightarrow \) \( {Fa} \) ) by existential generalisation. The result follows since \( F \) was arbitrary.
|
Yes
|
Theorem 60.1. There is no universal set, i.e., \( \{ x : x = x\} \) does not exist.
|
Proof. For reductio, suppose \( V \) is a universal set. Then by Separation, \( R = \) \( \{ x \in V : x \notin x\} = \{ x : x \notin x\} \) exists, contradicting Russell’s Paradox.
|
Yes
|
Proposition 60.3. \( A \smallsetminus B \) exists for any sets \( A \) and \( B \)
|
Proof. \( A \smallsetminus B = \{ x \in A : x \notin B\} \) exists by Separation.
|
Yes
|
Proposition 60.4. If \( A \neq \varnothing \), then \( \cap A = \{ x : \left( {\forall y \in A}\right) x \in y\} \) exists.
|
Proof. Let \( A \neq \varnothing \), so there is some \( c \in A \) . Then \( \cap A = \{ x : \left( {\forall y \in A}\right) x \in y\} = \) \( \{ x \in c : \left( {\forall y \in A}\right) x \in y\} \), which exists by Separation.
|
Yes
|
Proposition 60.5. For any sets \( a \) and \( b \), the following sets exist:\n\n1. \( \{ a\} \)\n\n2. \( a \cup b \)\n\n3. \( \langle a, b\rangle \)
|
Proof. (1). By Pairs, \( \{ a, a\} \) exists, which is \( \{ a\} \) by Extensionality.\n\n(2). By Pairs, \( \{ a, b\} \) exists. Now \( a \cup b = \bigcup \{ a, b\} \) exists by Union.\n\n(3). By (1), \( \{ a\} \) exists. By Pairs, \( \{ a, b\} \) exists. Now \( \{ \{ a\} ,\{ a, b\} \} = \langle a, b\rangle \) exists, by Pairs again.
|
Yes
|
Proposition 60.6. Given any sets \( A, B \), their Cartesian product \( A \times B \) exists.
|
Proof. The set \( \wp \left( {\wp \left( {A \cup B}\right) }\right) \) exists by Powersets and Proposition 60.5. So by Separation, this set exists:\n\n\[ \nC = \{ z \in \wp \left( {\wp \left( {A \cup B}\right) }\right) : \left( {\exists x \in A}\right) \left( {\exists y \in B}\right) z = \langle x, y\rangle \} .\n\]\n\nNow, for any \( x \in A \) and \( y \in B \), the set \( \langle x, y\rangle \) exists by Proposition 60.5. Moreover, since \( x, y \in A \cup B \), we have that \( \{ x\} ,\{ x, y\} \in \wp \left( {A \cup B}\right) \), and \( \langle x, y\rangle \in \wp \left( {\wp \left( {A \cup B}\right) }\right) \) . So \( A \times B = C \) .
|
Yes
|
Proposition 60.8. No natural number is Dedekind infinite.
|
Proof. The proof is by induction, i.e., Theorem 6.6. Clearly \( 0 = \varnothing \) is not Dedekind infinite. For the induction step, we will establish the contrapositive: if (absurdly) \( s\left( n\right) \) is Dedekind infinite, then \( n \) is Dedekind infinite.\n\nSo suppose that \( s\left( n\right) \) is Dedekind infinite, i.e., there is some injection \( f \) with \( \operatorname{ran}\left( f\right) \subsetneq \operatorname{dom}\left( f\right) = s\left( n\right) = n \cup \{ n\} \) . There are two cases to consider.\n\nCase 1: \( n \notin \operatorname{ran}\left( f\right) \) . Let \( g = {\left. f\right| }_{n} \) . Now \( \operatorname{ran}\left( f\right) \subseteq n \), and \( f\left( n\right) \in n \), so that \( \operatorname{ran}\left( g\right) = \operatorname{ran}\left( f\right) \smallsetminus \{ f\left( n\right) \} \subsetneq n = \operatorname{dom}\left( g\right) \) . Hence \( n \) is Dedekind infinite.\n\nCase 2: \( n \in \operatorname{ran}\left( f\right) \) . Fix \( m \in \operatorname{dom}\left( f\right) \smallsetminus \operatorname{ran}\left( f\right) \), and define a function \( {f}^{ * } \):\n\n\[ \n{f}^{ * }\left( x\right) = \left\{ \begin{array}{ll} f\left( x\right) & \text{ if }x \neq {f}^{-1}\left( n\right) \\ m & \text{ if }x = {f}^{-1}\left( n\right) \end{array}\right.\n\]\n\nSo \( {f}^{ * } \) and \( f \) agree everywhere, except that \( {f}^{ * }\left( {{f}^{-1}\left( n\right) }\right) = m \neq n = f\left( {{f}^{-1}\left( n\right) }\right) \) . Since \( f \) is an injection, \( n \notin \operatorname{ran}\left( {f}^{ * }\right) \) ; and \( \operatorname{ran}\left( {f}^{ * }\right) \subsetneq \operatorname{dom}\left( {f}^{ * }\right) = s\left( n\right) \) . Now \( n \) is Dedekind infinite, as in Case 1.
|
Yes
|
Proposition 61.2. If \( < \) well-orders \( A \), then every non-empty subset of \( A \) has a \( < \) - least member, and \( < \) is irreflexive, asymmetric and transitive.
|
Proof. If \( X \) is a non-empty subset of \( A \), it has a \( < \) -minimal element \( m \), i.e., \( \left( {\forall z \in X}\right) z \nless m \) . Since \( < \) is connected, \( \left( {\forall z \in X}\right) m \leq z \) . So \( m \) is \( < \) -least.\n\nFor irreflexivity, fix \( a \in A \) ; since \( \{ a\} \) has a <-least element, \( a \nless a \) . For transitivity, if \( a < b < c \), then since \( \{ a, b, c\} \) has a <-least element, \( a < c \) . Asymmetry follows from irreflexivity and transitivity
|
Yes
|
Proposition 61.3. If \( < \) well-orders \( A \), then for any formula \( \varphi \left( x\right) : {}^{1} \)\n\n\[ \text{if}\left( {\forall a \in A}\right) \left( {\left( {\forall b < a}\right) \varphi \left( b\right) \rightarrow \varphi \left( a\right) }\right) \text{, then}\left( {\forall a \in A}\right) \varphi \left( a\right) \text{.} \]
|
Proof. We will prove the contrapositive. Suppose \( \neg \left( {\forall a \in A}\right) \varphi \left( a\right) \), i.e., that \( X = \{ x \in A : \neg \varphi \left( x\right) \} \neq \varnothing \) . Then \( X \) has an \( < \) -minimal element, \( a \) . So \( \left( {\forall b < a}\right) \varphi \left( b\right) \) but \( \neg \varphi \left( a\right) \) .
|
Yes
|
Lemma 61.5. Compositions of isomorphisms are isomorphisms, i.e.: if \( f : A \rightarrow B \) and \( g : B \rightarrow C \) are isomorphisms, then \( \left( {g \circ f}\right) : A \rightarrow C \) are isomorphisms. (It follows that \( X \cong Y \) is an equivalence relation.)
|
Proof. Left as an exercise.
|
No
|
Proposition 61.6. If \( \langle A, < \rangle \) and \( \langle B, \lessdot \rangle \) are isomorphic well-orderings, then the isomorphism between them is unique.
|
Proof. Let \( f \) and \( g \) be isomorphisms \( A \rightarrow B \) . Fix \( a \in A \), and suppose that \( \left( {\forall b < a}\right) f\left( b\right) = g\left( b\right) \), and fix \( x \in B \) . If \( x < f\left( a\right) \), then \( {f}^{-1}\left( x\right) < a \), so \( g\left( {{f}^{-1}\left( x\right) < g\left( a\right) }\right) \), invoking the fact that \( f \) and \( g \) are isomorphisms. But since \( {f}^{-1}\left( x\right) < a \), by our supposition \( x = \) \( f\left( {{f}^{-1}\left( x\right) }\right) = g\left( {{f}^{-1}\left( x\right) }\right) \) . So \( x < g\left( a\right) \) . Similarly, if \( x < g\left( a\right) \) then \( x < f\left( a\right) \) . Generalising, \( \left( {\forall x \in B}\right) \left( {x < f\left( a\right) \leftrightarrow x \lessdot g\left( a\right) }\right) \) . It follows that \( f\left( a\right) = g\left( a\right) \) by Proposition 2.28. So \( \left( {\forall a \in A}\right) f\left( a\right) = g\left( a\right) \) by Proposition 61.3.
|
Yes
|
Lemma 61.8. If \( \langle A, < \rangle \) is a well-ordering with \( a \in A \), then \( \langle A, a\rangle \ncong \left\langle {{A}_{a},{ < }_{a}}\right\rangle \)
|
Proof. For reductio, suppose \( f : A \rightarrow {A}_{a} \) is an isomorphism. Since \( f \) is a bijection and \( {A}_{a} \subsetneq A \), let \( b \in A \) be the \( < \) -least element of \( A \) such that \( b \neq f\left( b\right) \) . We’ll show that \( \left( {\forall x \in A}\right) \left( {x < b \leftrightarrow x < f\left( b\right) }\right) \), from which it will follow by Proposition 2.28 that \( b = f\left( b\right) \), completing the reductio.\n\nSuppose \( x < b \) . So \( x = f\left( x\right) \), by the choice of \( b \) . And \( f\left( x\right) < f\left( b\right) \), as \( f \) is an isomorphism. So \( x < f\left( b\right) \).\n\nSuppose \( x < f\left( b\right) \) . So \( {f}^{-1}\left( x\right) < b \), since \( f \) is an isomorphism, and so \( {f}^{-1}\left( x\right) = x \) by the choice of \( b \) . So \( x < b \).
|
Yes
|
Lemma 61.9. Let \( \langle A, < \rangle \) and \( \langle B, \lessdot \rangle \) be well-orderings. If \( f : A \rightarrow B \) is an isomorphism and \( a \in A \), then \( f{ \upharpoonright }_{{A}_{a}} : {A}_{a} \rightarrow {B}_{f\left( a\right) } \) is an isomorphism.
|
Proof. Since \( f \) is an isomorphism:\n\n\[ f\left\lbrack {A}_{a}\right\rbrack = f\left\lbrack {\{ x \in A : x < a\} }\right\rbrack \]\n\n\[ = f\left\lbrack \left\{ {{f}^{-1}\left( y\right) \in A : {f}^{-1}\left( y\right) < a}\right\} \right\rbrack \]\n\n\[ = \{ y \in B : y < f\left( a\right) \} \]\n\n\[ = {B}_{f\left( a\right) } \]\n\nAnd \( f{ \upharpoonright }_{{A}_{a}} \) preserves order because \( f \) does.
|
Yes
|
Lemma 61.10. Let \( \langle A, < \rangle \) and \( \langle B, \lessdot \rangle \) be well-orderings. If \( \left\langle {{A}_{{a}_{1}},{ < }_{{a}_{1}}}\right\rangle \cong \left\langle {{B}_{{b}_{1}},{ \lessdot }_{{b}_{1}}}\right\rangle \) and \( \left\langle {{A}_{{a}_{2}},{ < }_{{a}_{2}}}\right\rangle \cong \left\langle {{B}_{{b}_{2}},{ < }_{{b}_{2}}}\right\rangle \), then \( {a}_{1} < {a}_{2} \) iff \( {b}_{1} < {b}_{2} \)
|
Proof. We will prove left to right; the other direction is similar. Suppose both \( \left\langle {{A}_{{a}_{1}},{ < }_{{a}_{1}}}\right\rangle \cong \left\langle {{B}_{{b}_{1}},{ < }_{{b}_{1}}}\right\rangle \) and \( \left\langle {{A}_{{a}_{2}},{ < }_{{a}_{2}}}\right\rangle \cong \left\langle {{B}_{{b}_{2}},{ < }_{{b}_{2}}}\right\rangle \), with \( f : {A}_{{a}_{2}} \rightarrow {B}_{{b}_{2}} \) our isomorphism. Let \( {a}_{1} < {a}_{2} \) ; then \( \left\langle {{A}_{{a}_{1}},{ < }_{{a}_{1}}}\right\rangle \cong \left\langle {{B}_{f\left( {a}_{1}\right) },{ < }_{f\left( {a}_{1}\right) }}\right\rangle \) by Lemma 61.9. So \( \left\langle {{B}_{{b}_{1}},{ \lessdot }_{{b}_{1}}}\right\rangle \cong \left\langle {{B}_{f\left( {a}_{1}\right) },{ \lessdot }_{f\left( {a}_{1}\right) }}\right\rangle \), and so \( {b}_{1} = f\left( {a}_{1}\right) \) by Lemma 61.8. Now \( {b}_{1} < {b}_{2} \) as \( {f}^{\prime }\mathrm{s} \) domain is \( {B}_{{b}_{2}} \) .
|
Yes
|
Theorem 61.11. Given any two well-orderings, one is isomorphic to an initial segment (not necessarily proper) of the other.
|
Proof. Let \( \langle A, < \rangle \) and \( \langle B, \lessdot \rangle \) be well-orderings. Using Separation, let\n\n\[ f = \left\{ {\langle a, b\rangle \in A \times B : \left\langle {{A}_{a},{ < }_{a}}\right\rangle \cong \left\langle {{B}_{b},{ \lessdot }_{b}}\right\rangle }\right\} .\n\]\n\nBy Lemma 61.10, \( {a}_{1} < {a}_{2} \) iff \( {b}_{1} < {b}_{2} \) for all \( \left\langle {{a}_{1},{b}_{1}}\right\rangle ,\left\langle {{a}_{2},{b}_{2}}\right\rangle \in f \) . So \( f : \operatorname{dom}\left( f\right) \rightarrow \) \( \operatorname{ran}\left( f\right) \) is an isomorphism.\n\nIf \( {a}_{2} \in \operatorname{dom}\left( f\right) \) and \( {a}_{1} < {a}_{2} \), then \( {a}_{1} \in \operatorname{dom}\left( f\right) \) by Lemma 61.9; so \( \operatorname{dom}\left( f\right) \) is an initial segment of \( A \) . Similarly, \( \operatorname{ran}\left( f\right) \) is an initial segment of \( B \) . For reductio, suppose both are proper initial segments. Then let \( a \) be the \( < \) -least element of \( A \smallsetminus \operatorname{dom}\left( f\right) \), so that \( \operatorname{dom}\left( f\right) = {A}_{a} \), and let \( b \) be the \( < \) -least element of \( B \smallsetminus \operatorname{ran}\left( f\right) \), so that \( \operatorname{ran}\left( f\right) = {B}_{b} \) . So \( f : {A}_{a} \rightarrow {B}_{b} \) is an isomorphism, and hence \( f\left( a\right) = b \), a contradiction.
|
Yes
|
Lemma 61.13. Every element of an ordinal is an ordinal.
|
Proof. Let \( \alpha \) be an ordinal with \( b \in \alpha \) . Since \( \alpha \) is transitive, \( b \subseteq \alpha \) . So \( \in \) well-orders \( b \) as \( \in \) well-orders \( \alpha \) . For transitivity, suppose \( x \in c \in b \) . So \( c \in \alpha \) as \( b \subseteq \alpha \) . Again, as \( \alpha \) is transitive, \( c \subseteq \alpha \), so that \( x \in \alpha \) . So \( x, c, b \in \alpha \) . But \( \in \) well-orders \( \alpha \), so that \( \in \) is a transitive relation on \( \alpha \) by Proposition 61.2. So since \( x \in c \in b \), we have \( x \in b \) . Generalising, \( c \subseteq b \)
|
Yes
|
Corollary 61.14. \( \alpha = \{ \beta \in \alpha : \beta \) is an ordinal \( \} \), for any ordinal \( \alpha \)
|
Proof. Immediate from Lemma 61.13.
|
No
|
Theorem 61.15 (Transfinite Induction). For any formula \( \varphi \left( x\right) { : }^{2} \)\n\n\[ \n\text{if}\exists {\alpha \varphi }\left( \alpha \right) \text{, then}\exists \alpha \left( {\varphi \left( \alpha \right) \land \left( {\forall \beta \in \alpha }\right) \neg \varphi \left( \beta \right) }\right)\n\]\n\nwhere the displayed quantifiers are implicitly restricted to ordinals.
|
Proof. Suppose \( \varphi \left( \alpha \right) \), for some ordinal \( \alpha \) . If \( \left( {\forall \beta \in \alpha }\right) \neg \varphi \left( \beta \right) \), then we are done. Otherwise, as \( \alpha \) is an ordinal, it has some \( \in \) -least element which is \( \varphi \), and this is an ordinal by Lemma 61.13.
|
No
|
Theorem 61.16 (Trichotomy). \( \alpha \in \beta \vee \alpha = \beta \vee \beta \in \alpha \), for any ordinals \( \alpha \) and \( \beta \) .
|
Proof. The proof is by double induction, i.e., using Theorem 61.15 twice. Say that \( x \) is comparable with \( y \) iff \( x \in y \vee x = y \vee y \in x \) . For induction, suppose that every ordinal in \( \alpha \) is comparable with every ordinal. For further induction, suppose that \( \alpha \) is comparable with every ordinal in \( \beta \) . We will show that \( \alpha \) is comparable with \( \beta \) . By induction on \( \beta \), it will follow that \( \alpha \) is comparable with every ordinal; and so by induction on \( \alpha \), every ordinal is comparable with every ordinal, as required. It suffices to assume that \( \alpha \notin \beta \) and \( \beta \notin \alpha \), and show that \( \alpha = \beta \) . To show that \( \alpha \subseteq \beta \), fix \( \gamma \in \alpha \) ; this is an ordinal by Lemma 61.13. So by the first induction hypothesis, \( \gamma \) is comparable with \( \beta \) . But if either \( \gamma = \beta \) or \( \beta \in \gamma \) then \( \beta \in \alpha \) (invoking the fact that \( \alpha \) is transitive if necessary), contrary to our assumption; so \( \gamma \in \beta \) . Generalising, \( \alpha \subseteq \beta \) . Exactly similar reasoning, using the second induction hypothesis, shows that \( \beta \subseteq \alpha \) . So \( \alpha = \beta \) .
|
Yes
|
Corollary 61.17. If \( \exists {\alpha \varphi }\left( \alpha \right) \), then \( \exists \alpha \left( {\varphi \left( \alpha \right) \land \forall \beta \left( {\varphi \left( \beta \right) \rightarrow \alpha \leq \beta }\right) }\right) \) . Moreover, for any ordinals \( \alpha ,\beta ,\gamma \), both \( \alpha \notin \alpha \) and \( \alpha \in \beta \in \gamma \rightarrow \alpha \in \gamma \) .
|
Proof. Just like Proposition 61.2.
|
No
|
Corollary 61.18. \( A \) is an ordinal iff \( A \) is a transitive set of ordinals.
|
Proof. Left-to-right. By Lemma 61.13. Right-to-left. If \( A \) is a transitive set of ordinals, then \( \in \) well-orders \( A \) by Theorem 61.15 and Theorem 61.16.
|
Yes
|
Theorem 61.19 (Burali-Forti Paradox). There is no set of all the ordinals
|
Proof. For reductio, suppose \( O \) is the set of all ordinals. If \( \alpha \in \beta \in O \), then \( \alpha \) is an ordinal, by Lemma 61.13, so \( \alpha \in O \) . So \( O \) is transitive, and hence \( O \) is an ordinal by Corollary 61.18. Hence \( O \in O \), contradicting Corollary 61.17.
|
Yes
|
Proposition 61.20. Any strictly descending sequence of ordinals is finite.
|
Proof. Any infinite strictly descending sequence of ordinals \( \ldots \in {\alpha }_{3} \in {\alpha }_{2} \in \) \( {\alpha }_{1} \in {\alpha }_{0} \) has no \( \in \) -minimal member, contradicting Theorem 61.15.
|
Yes
|
Proposition 61.21. \( \alpha \subseteq \beta \vee \beta \subseteq \alpha \), for any ordinals \( \alpha ,\beta \) .
|
Proof. If \( \alpha \in \beta \), then \( \alpha \subseteq \beta \) as \( \beta \) is transitive. Similarly, if \( \beta \in \alpha \), then \( \beta \subseteq \alpha \) . And if \( \alpha = \beta \), then \( \alpha \subseteq \beta \) and \( \beta \subseteq \alpha \) . So by Theorem 61.16 we are done.
|
Yes
|
Proposition 61.22. \( \alpha = \beta \) iff \( \alpha \cong \beta \), for any ordinals \( \alpha ,\beta \) .
|
Proof. The ordinals are well-orders; so this is immediate from Trichotomy (Theorem 61.16) and Lemma 61.8.
|
No
|
Corollary 61.23. For any term \( \tau \left( x\right) ,{}^{6} \) and any set \( A \), this set exists:\n\n\[ \{ \tau \left( x\right) : x \in A\} = \{ y : \left( {\exists x \in A}\right) y = \tau \left( x\right) \} . \]
|
Proof. Since \( \tau \) is a term, \( \forall x\exists !{y\tau }\left( x\right) = y \) . A fortiori, \( \left( {\forall x \in A}\right) \exists !{y\tau }\left( x\right) = y \) . So \( \{ y : \left( {\exists x \in A}\right) \tau \left( x\right) = y\} \) exists by Replacement.
|
Yes
|
Theorem 61.25. Every well-ordering is isomorphic to a unique ordinal.
|
Proof. Let \( \langle A, < \rangle \) be a well-order. By Proposition 61.22, it is isomorphic to at most one ordinal. So, for reductio, suppose \( \langle A, < \rangle \) is not isomorphic to any ordinal. We will first \
|
No
|
Corollary 61.27. Where \( \langle A, < \rangle \) and \( \langle B, \lessdot \rangle \) are well-orderings:\n\n\[ \operatorname{ord}\left( {A, < }\right) = \operatorname{ord}\left( {B, < }\right) \text{iff}\langle A, < \rangle \cong \langle B, < \rangle \]\n\n\[ \operatorname{ord}\left( {A, < }\right) \in \operatorname{ord}\left( {B, \lessdot }\right) \text{iff}\langle A, < \rangle \cong \left\langle {{B}_{b},{ \lessdot }_{b}}\right\rangle \text{for some}b \in B \]
|
Proof. The identity holds as isomorphism is an equivalence relation. To prove the second claim, let \( \operatorname{ord}\left( {A, < }\right) = \alpha \) and \( \operatorname{ord}\left( {B, < }\right) = \beta \), and let \( f : \beta \rightarrow \langle B, < \rangle \) be our isomorphism. Then:\n\n\[ \alpha \in \beta \text{iff}{\left. f\right| }_{\alpha } : \alpha \rightarrow {B}_{f\left( \alpha \right) }\text{is an isomorphism} \]\n\n\[ \text{iff}\langle A, < \rangle \cong \left\langle {{B}_{f\left( \alpha \right) },{ < }_{f\left( \alpha \right) }}\right\rangle \]\n\n\[ \text{iff}\langle A, < \rangle \cong \left\langle {{B}_{b},{ < }_{b}}\right\rangle \text{for some}b \in B \]\n\nby Proposition 61.6, Lemma 61.9, and Corollary 61.14.
|
Yes
|
Proposition 61.29. For any ordinal \( \alpha \) :\n\n1. \( \alpha \in {\alpha }^{ + } \) ;\n\n2. \( {\alpha }^{ + } \) is an ordinal;\n\n3. there is no ordinal \( \beta \) such that \( \alpha \in \beta \in {\alpha }^{ + } \) .
|
Proof. Trivially, \( \alpha \in \alpha \cup \{ \alpha \} = {\alpha }^{ + } \) . Equally, \( {\alpha }^{ + } \) is a transitive set of ordinals, and hence an ordinal by Corollary 61.18. And it is impossible that \( \alpha \in \beta \in {\alpha }^{ + } \) , since then either \( \beta \in \alpha \) or \( \beta = \alpha \), contradicting Corollary 61.17.
|
Yes
|
Theorem 61.30 (Simple Transfinite Induction). Let \( \varphi \left( x\right) \) be a formula such that: \( {}^{8} \n\n1. \( \varphi \left( \varnothing \right) \) ; and\n\n2. for any ordinal \( \alpha \), if \( \varphi \left( \alpha \right) \) then \( \varphi \left( {\alpha }^{ + }\right) \) ; and\n\n3. if \( \alpha \) is a limit ordinal and \( \left( {\forall \beta \in \alpha }\right) \varphi \left( \beta \right) \), then \( \varphi \left( \alpha \right) \) .\n\nThen \( \forall {\alpha \varphi }\left( \alpha \right) \) .
|
Proof. We prove the contrapositive. So, suppose there is some ordinal which is \( \neg \varphi \) ; let \( \gamma \) be the least such ordinal. Then either \( \gamma = \varnothing \), or \( \gamma = {\alpha }^{ + } \) for some \( \alpha \) such that \( \varphi \left( \alpha \right) \) ; or \( \gamma \) is a limit ordinal and \( \left( {\forall \beta \in \gamma }\right) \varphi \left( \beta \right) \) .
|
No
|
Proposition 61.32. If \( X \) is a set of ordinals, \( \operatorname{lsub}\left( X\right) \) is the least ordinal greater than every ordinal in \( X \).
|
Proof. Let \( Y = \left\{ {{\alpha }^{ + } : \alpha \in X}\right\} \), so that \( \operatorname{Isub}\left( X\right) = \bigcup Y \) . Since ordinals are transitive and every member of an ordinal is an ordinal, \( \operatorname{lsub}\left( X\right) \) is a transitive set of ordinals, and so is an ordinal by Corollary 61.18.\n\nIf \( \alpha \in X \), then \( {\alpha }^{ + } \in Y \), so \( {\alpha }^{ + } \subseteq \cup Y = \operatorname{lsub}\left( X\right) \), and hence \( \alpha \in \operatorname{lsub}\left( X\right) \) . So \( \operatorname{lsub}\left( X\right) \) is strictly greater than every ordinal in \( X \).\n\nConversely, if \( \alpha \in \operatorname{lsub}\left( X\right) \), then \( \alpha \in {\beta }^{ + } \in Y \) for some \( \beta \in X \), so that \( \alpha \leq \beta \in X \) . So \( \operatorname{lsub}\left( X\right) \) is the least strict upper bound on \( X \) .
|
Yes
|
Theorem 62.2 (Bounded Recursion). For any term \( \tau \left( x\right) \) and any ordinal \( \alpha ,{}^{2} \) there is a unique function \( f \) with domain \( \alpha \) such that \( \left( {\forall \beta \in \alpha }\right) f\left( \beta \right) = \tau \left( {f{ \upharpoonright }_{\beta }}\right) \)
|
Proof. We will show that, for any \( \delta \leq \alpha \), there is a unique \( {g}_{\delta } \) with domain \( \delta \) such that \( \left( {\forall \beta \in \delta }\right) g\left( \beta \right) = \tau \left( {\left. g\right| }_{\beta }\right) \). We first establish uniqueness. Given \( {g}_{{\delta }_{1}} \) and \( {g}_{{\delta }_{2}} \), a transfinite induction on their arguments shows that \( g\left( \beta \right) = h\left( \beta \right) \) for any \( \beta \in \operatorname{dom}\left( g\right) \cap \operatorname{dom}\left( h\right) = \) \( {\delta }_{1} \cap {\delta }_{2} = \min \left( {{\delta }_{1},{\delta }_{2}}\right) \). So our functions are unique (if they exist), and agree on all values. To establish existence, we now use a simple transfinite induction (Theorem 61.30) on ordinals \( \delta \leq \alpha \). Let \( {g}_{\varnothing } = \varnothing \) ; this trivially behaves correctly. Given \( {g}_{\delta } \), let \( {g}_{{\delta }^{ + }} = {g}_{\delta } \cup \left\{ \left\langle {\delta ,\tau \left( {g}_{\delta }\right) }\right\rangle \right\} \). This behaves correctly as \( {\left. {g}_{{\delta }^{ + }}\right| }_{\delta } = {g}_{\delta } \). Given \( {g}_{\gamma } \) for all \( \gamma \leq \delta \) with \( \delta \) a limit ordinal, let \( {g}_{\delta } = \mathop{\bigcup }\limits_{{\gamma \in \delta }}{g}_{\gamma } \). This is a function, since our various \( {g}_{\beta } \) ’s agree on all values. And if \( \beta \in \delta \) then \( {g}_{\delta }\left( \beta \right) = \) \( {g}_{{\beta }^{ + }}\left( \beta \right) = \tau \left( {{g}_{{\beta }^{ + }}{ \upharpoonright }_{\beta }}\right) = \tau \left( {{g}_{\delta }{ \upharpoonright }_{\beta }}\right) . This completes the proof by transfinite induction. Now just let \( f = {g}_{\alpha } \).
|
Yes
|
Theorem 62.3 (General Recursion). For any term \( \tau \left( x\right) \) we can explicitly define a term \( \sigma \left( x\right) ,{}^{3} \) such that \( \sigma \left( \alpha \right) = \tau \left( {\sigma { \upharpoonright }_{\alpha }}\right) \) for any ordinal \( \alpha \) .
|
Proof. For each \( \alpha \), by Theorem 62.2 are unique \( {\alpha }^{ + } \) -approximations, \( {f}_{{\alpha }^{ + }} \), and:\n\n\[ \n{f}_{{\alpha }^{ + }}\left( \alpha \right) = \tau \left( {{f}_{{\alpha }^{ + }}{ \upharpoonright }_{\alpha }}\right) = \tau \left( \left\{ {\left\langle {\gamma ,{f}_{{\alpha }^{ + }}\left( \gamma \right) }\right\rangle : \gamma \in \alpha }\right\} \right) .\n\]\n\nSo define \( \sigma \left( \alpha \right) \) as \( {f}_{{\alpha }^{ + }}\left( \alpha \right) \) . Repeating the induction of Theorem 62.2, but without the upper bound, this is well-defined.
|
Yes
|
Theorem 62.4 (Simple Recursion). For any terms \( {\tau }_{1}\left( x\right) \) and \( {\tau }_{2}\left( x\right) \) and any set \( A \), we can explicitly define a term \( \sigma \left( x\right) \) such that: \( {}^{4} \)\n\n\[ \sigma \left( \varnothing \right) = A \]\n\n\[ \sigma \left( {\alpha }^{ + }\right) = {\tau }_{1}\left( {\sigma \left( \alpha \right) }\right) \] for any ordinal \( \alpha \)\n\n\[ \sigma \left( \alpha \right) = {\tau }_{2}\left( {\operatorname{ran}\left( {\left. \sigma \right| }_{\alpha }\right) }\right) \] when \( \alpha \) is a limit ordinal
|
Proof. We start by defining a term, \( \xi \left( x\right) \), as follows:\n\n\[ \xi \left( x\right) = \left\{ \begin{matrix} A & \text{ if }x\text{ is not a function whose domain is an ordinal; } \\ & \text{ otherwise: } \\ {\tau }_{1}\left( {x\left( \alpha \right) }\right) & \text{ if dom }\left( x\right) = {\alpha }^{ + } \\ {\tau }_{2}\left( {\operatorname{ran}\left( x\right) }\right) & \text{ if dom }\left( x\right) \text{ is a limit ordinal } \end{matrix}\right. \]\n\nBy Theorem 62.3, there is a term \( \sigma \left( x\right) \) such that \( \sigma \left( \alpha \right) = \xi \left( {\left. \sigma \right| }_{\alpha }\right) \) for every ordinal \( \alpha \) ; moreover, \( {\left. \sigma \right| }_{\alpha } \) is a function with domain \( \alpha \) . We show that \( \sigma \) has the required properties, by simple transfinite induction (Theorem 61.30).\n\nFirst, \( \sigma \left( \varnothing \right) = \xi \left( \varnothing \right) = A \).\n\nNext, \( \sigma \left( {\alpha }^{ + }\right) = \xi \left( {\left. \sigma \right| }_{{\alpha }^{ + }}\right) = {\tau }_{1}\left( {{\left. \sigma \right| }_{{\alpha }^{ + }}\left( \alpha \right) }\right) = {\tau }_{1}\left( {\sigma \left( \alpha \right) }\right) \).\n\nFinally, if \( \alpha \) is a limit ordinal, \( \sigma \left( \alpha \right) = \xi \left( {\sigma { \upharpoonright }_{\alpha }}\right) = {\tau }_{2}\left( {\operatorname{ran}\left( {\sigma { \upharpoonright }_{\alpha }}\right) }\right) \).
|
Yes
|
Lemma 62.5. For each ordinal \( \alpha \) :\n\n1. Each \( {V}_{\alpha } \) is a transitive set.\n\n2. Each \( {V}_{\alpha } \) is a sublative set, \( {}^{5} \) i.e., \( \forall A\left( {\exists B\left( {A \subseteq B \in {V}_{\alpha }}\right) \rightarrow A \in {V}_{\alpha }}\right) \) .\n\n3. If \( \gamma \in \alpha \), then \( {V}_{\gamma } \in {V}_{\alpha } \) (and hence also \( {V}_{\gamma } \subseteq {V}_{\alpha } \) by (1))
|
Proof. We prove this by a (simultaneous) transfinite induction. For induction, suppose that (1)-(3) holds for each ordinal \( \beta < \alpha \) .\n\nThe case of \( \alpha = \varnothing \) is trivial.\n\nSuppose \( \alpha = {\beta }^{ + } \) . To show (3), if \( \gamma \in \alpha \) then \( {V}_{\gamma } \subseteq {V}_{\beta } \) by hypothesis, so \( {V}_{\gamma } \in \wp \left( {V}_{\beta }\right) = {V}_{\alpha } \) . To show (2), suppose \( A \subseteq B \in {V}_{\alpha } \) i.e., \( A \subseteq B \subseteq {V}_{\beta } \) ; then\n\n--- \n\n\( {}^{4} \) The terms may have parameters.\n\n\( {}^{5} \) There’s no standard terminology for \
|
No
|
Lemma 62.6. For each ordinal \( \alpha ,{V}_{\alpha } \notin {V}_{\alpha } \) .
|
Proof. By transfinite induction. Evidently \( {V}_{\varnothing } \notin {V}_{\varnothing } \) .\n\nIf \( {V}_{{\alpha }^{ + }} \in {V}_{{\alpha }^{ + }} = \wp \left( {V}_{\alpha }\right) \), then \( {V}_{{\alpha }^{ + }} \subseteq {V}_{\alpha } \) ; and since \( {V}_{\alpha } \in {V}_{{\alpha }^{ + }} \) by Lemma 62.5, we have \( {V}_{\alpha } \in {V}_{\alpha } \) . Conversely: if \( {V}_{\alpha } \notin {V}_{\alpha } \) then \( {V}_{{\alpha }^{ + }} \notin {V}_{{\alpha }^{ + }} \)\n\nIf \( \alpha \) is a limit and \( {V}_{\alpha } \in {V}_{\alpha } = \mathop{\bigcup }\limits_{{\beta \in \alpha }}{V}_{\beta } \), then \( {V}_{\alpha } \in {V}_{\beta } \) for some \( \beta \in \alpha \) ; but then also \( {V}_{\beta } \in {V}_{\alpha } \) so that \( {V}_{\beta } \in {V}_{\beta } \) by Lemma 62.5 (twice). Conversely, if \( {V}_{\beta } \notin {V}_{\beta } \) for all \( \beta \in \alpha \), then \( {V}_{\alpha } \notin {V}_{\alpha } \).
|
Yes
|
For any ordinals \( \alpha ,\beta : \alpha \in \beta \) iff \( {V}_{\alpha } \in {V}_{\beta } \)
|
Proof. Lemma 62.5 gives one direction. Conversely, suppose \( {V}_{\alpha } \in {V}_{\beta } \) . Then \( \alpha \neq \beta \) by Lemma 62.6; and \( \beta \notin \alpha \), for otherwise we would have \( {V}_{\beta } \in {V}_{\alpha } \) and hence \( {V}_{\beta } \in {V}_{\beta } \) by Lemma 62.5 (twice), contradicting Lemma 62.6. So \( \alpha \in \beta \) by Trichotomy.
|
Yes
|
Proposition 62.9. \( A \subseteq \operatorname{trcl}\left( A\right) \) and \( \operatorname{trcl}\left( A\right) \) is a transitive set.
|
Proof. Evidently \( A = {\operatorname{cl}}_{0}\left( A\right) \subseteq \operatorname{trcl}\left( A\right) \) . And if \( x \in b \in \operatorname{trcl}\left( A\right) \), then \( b \in \) \( {\operatorname{cl}}_{n}\left( A\right) \) for some \( n \), so \( x \in {\operatorname{cl}}_{n + 1}\left( A\right) \subseteq \operatorname{trcl}\left( A\right) \) .
|
Yes
|
Lemma 62.10. If \( A \) is a transitive set, then there is some \( \alpha \) such that \( A \subseteq {V}_{\alpha } \) .
|
Proof. Recalling the definition of \
|
No
|
Theorem 62.11. Regularity holds.
|
Proof. Fix \( A \) ; now \( A \subseteq \operatorname{trcl}\left( A\right) \) by Proposition 62.9, which is transitive. So there is some \( \alpha \) such that \( A \subseteq \operatorname{trcl}\left( A\right) \subseteq {V}_{\alpha } \) by Lemma 62.10
|
Yes
|
Proposition 62.14. For any ordinal \( \alpha ,{V}_{\alpha } = \{ x : \operatorname{rank}\left( x\right) \in \alpha \} \) .
|
Proof. If \( \operatorname{rank}\left( x\right) \in \alpha \) then \( x \subseteq {V}_{\operatorname{rank}\left( x\right) } \in {V}_{\alpha } \), so \( x \in {V}_{\alpha } \) as \( {V}_{\alpha } \) is sublative (invoking Lemma 62.5 multiple times). Conversely, by definition of \
|
No
|
Proposition 62.15. If \( B \in A \), then \( \operatorname{rank}\left( B\right) \in \operatorname{rank}\left( A\right) \) .
|
Proof. \( A \subseteq {V}_{\operatorname{rank}\left( A\right) } = \{ x : \operatorname{rank}\left( x\right) \in \operatorname{rank}\left( A\right) \} \) by Proposition 62.14.
|
No
|
Theorem 62.16 ( \( \\in \) -Induction Scheme). For any formula \( \\varphi : {}^{7} \n\n\[ \n\\forall A\\left( {\\left( {\\forall x \\in A}\\right) \\varphi \\left( x\\right) \\rightarrow \\varphi \\left( A\\right) }\\right) \\rightarrow \\forall {A\\varphi }\\left( A\\right) .\n\]
|
Proof. We will prove the contrapositive. So, suppose \( \\neg \\forall {A\\varphi }\\left( A\\right) \) . Since every set has a rank, Transfinite Induction (Theorem 61.15) tells us that there is a non- \( \\varphi \) of least possible rank. That is: there is some \( A \) such that \( \\neg \\varphi \\left( A\\right) \) and \( \\forall x\\left( {\\operatorname{rank}\\left( x\\right) \\in \\operatorname{rank}\\left( A\\right) \\rightarrow \\varphi \\left( x\\right) }\\right) \) . Now if \( x \\in A \) then \( \\operatorname{rank}\\left( x\\right) \\in \\operatorname{rank}\\left( A\\right) \), by Proposition 62.15. So \( \\left( {\\forall x \\in A}\\right) \\varphi \\left( x\\right) \\land \\neg \\varphi \\left( A\\right) \), falsifying the antecedent.
|
Yes
|
Proposition 62.17. \( \operatorname{rank}\left( A\right) = {\operatorname{lsub}}_{x \in A}\operatorname{rank}\left( x\right) \) .
|
Proof. Let \( \alpha = {\operatorname{lsub}}_{x \in A}\operatorname{rank}\left( x\right) \) . By Proposition 62.15, \( \alpha \leq \operatorname{rank}\left( A\right) \) . But if \( x \in A \) then \( \operatorname{rank}\left( x\right) \in \alpha \), so that \( x \in {V}_{\alpha } \), and hence \( A \subseteq {V}_{\alpha } \), i.e., \( \operatorname{rank}\left( A\right) \leq \alpha \) . Hence \( \operatorname{rank}\left( A\right) = \alpha \) .
|
Yes
|
For any ordinal \( \alpha ,\operatorname{rank}\left( \alpha \right) = \alpha \) .
|
Proof. Suppose for transfinite induction that \( \operatorname{rank}\left( \beta \right) = \beta \) for all \( \beta \in \alpha \) . Now \( \operatorname{rank}\left( \alpha \right) = {\operatorname{lsub}}_{\beta \in \alpha }\operatorname{rank}\left( \beta \right) = {\operatorname{lsub}}_{\beta \in \alpha }\beta = \alpha \) by Proposition 62.17.
|
Yes
|
Proposition 62.19 (working in \( {\mathrm{{ZF}}}^{ - } + \) Regularity). Foundation holds.
|
Proof. Fix \( A \neq \varnothing \), and some \( B \in A \) of least possible rank. If \( c \in B \) then \( \operatorname{rank}\left( c\right) \in \operatorname{rank}\left( B\right) \) by Proposition 62.15, so that \( c \notin A \) by choice of \( B \) .
|
Yes
|
Lemma 63.2. For each \( 1 \leq i \leq k \), let \( {\varphi }_{i}\left( {{\bar{v}}_{i}, x}\right) \) be a formula. \( {}^{5} \) Then for each \( \alpha \) there is some \( \beta > \alpha \) such that, for any \( {\bar{a}}_{1},\ldots ,{\bar{a}}_{k} \in {V}_{\beta } \) and each \( 1 \leq i \leq k \):\n\n\[ \exists x{\varphi }_{i}\left( {{\bar{a}}_{i}, x}\right) \rightarrow \left( {\exists x \in {V}_{\beta }}\right) {\varphi }_{i}\left( {{\bar{a}}_{i}, x}\right) \]
|
Proof. We define a term \( \mu \) as follows: \( \mu \left( {{\bar{a}}_{1},\ldots ,{\bar{a}}_{k}}\right) \) is the least stage, \( V \), which satisfies all of the following conditionals, for \( 1 \leq i \leq k \):\n\n\[ \left. {\exists x{\varphi }_{i}\left( {{\bar{a}}_{i}, x}\right) \rightarrow \left( {\exists x \in V}\right) {\varphi }_{i}\left( {{\bar{a}}_{i}, x}\right) }\right) \]\n\nUsing Replacement and our recursion theorem, define:\n\n\[ {S}_{0} = {V}_{\alpha + 1} \]\n\n\[ {S}_{m + 1} = {S}_{m} \cup \bigcup \left\{ {\mu \left( {{\bar{a}}_{1},\ldots ,{\bar{a}}_{k}}\right) : {\bar{a}}_{1},\ldots ,{\bar{a}}_{k} \in {S}_{m}}\right\} \]\n\n\[ S = \mathop{\bigcup }\limits_{{m < \omega }}{S}_{m} \]\n\nEach \( {S}_{m} \), and hence \( S \) itself, is a stage after \( {V}_{\alpha } \) . Now fix \( {\bar{a}}_{1},\ldots ,{\bar{a}}_{k} \in S \) ; so there is some \( m < \omega \) such that \( {\bar{a}}_{1},\ldots ,{\bar{a}}_{k} \in {S}_{m} \) . Fix some \( 1 \leq i \leq k \), and suppose that \( \exists x{\varphi }_{i}\left( {{\bar{a}}_{i}, x}\right) \) . So \( \left( {\exists x \in \mu \left( {{\bar{a}}_{1},\ldots ,{\bar{a}}_{k}}\right) }\right) {\varphi }_{i}\left( {{\bar{a}}_{i}, x}\right) \) by construction, so \( \left( {\exists x \in {S}_{m + 1}}\right) {\varphi }_{i}\left( {{\bar{a}}_{i}, x}\right) \) and hence \( \left( {\exists x \in S}\right) {\varphi }_{i}\left( {{\bar{a}}_{i}, x}\right) \) . So \( S \) is our \( {V}_{\beta } \) .
|
Yes
|
Lemma 63.3 (in \( \mathbf{Z} + \) Weak-Reflection.). For any formulas \( \psi ,\chi \), there is a transitive set \( S \) such that 0 and 1 (and any parameters to the formulas) are elements of \( S \) , and \( \left( {\forall \bar{x} \in S}\right) \left( {\left( {\psi \leftrightarrow {\psi }^{S}}\right) \land \left( {\chi \leftrightarrow {\chi }^{S}}\right) }\right) \) .
|
Proof. Let \( \varphi \) be the formula \( \left( {z = 0 \land \psi }\right) \vee \left( {z = 1 \land \chi }\right) \). Here we use an abbreviation; we should spell out \
|
No
|
Theorem 63.4 (in \( \mathrm{Z} \) + Weak-Reflection). For any formula \( \varphi \left( {v, w}\right) ,{}^{7} \) and any \( A \) , if \( \left( {\forall x \in A}\right) \exists !{y\varphi }\left( {x, y}\right) \), then \( \{ y : \left( {\exists x \in A}\right) \varphi \left( {x, y}\right) \) exists.
|
Proof. Fix \( A \) such that \( \left( {\forall x \in A}\right) \exists !{y\varphi }\left( {x, y}\right) \), and define some formulas:\n\n\[ \psi \text{is}\left( {\varphi \left( {x, z}\right) \land A = A}\right) \]\n\n\[ \chi \text{is}\exists {y\varphi }\left( {x, y}\right) \]\n\nUsing Lemma 63.3, since \( A \) is a parameter to \( \psi \), there is a transitive \( S \) such that \( 0,1, A \in S \) (along with any other parameters), and such that:\n\n\[ \left( {\forall x, z \in S}\right) \left( {\left( {\psi \leftrightarrow {\psi }^{S}}\right) \land \left( {\chi \leftrightarrow {\chi }^{S}}\right) }\right) \]\n\nSo in particular:\n\n\[ \left( {\forall x, z \in S}\right) \left( {\varphi \left( {x, z}\right) \leftrightarrow {\varphi }^{S}\left( {x, z}\right) }\right) \]\n\n\[ \left( {\forall x \in S}\right) \left( {\exists {y\varphi }\left( {x, y}\right) \leftrightarrow \left( {\exists y \in S}\right) {\varphi }^{S}\left( {x, y}\right) }\right) \]\n\nCombining these, and observing that \( A \subseteq S \) since \( A \in S \) and \( S \) is transitive:\n\n\[ \left( {\forall x \in A}\right) \left( {\exists {y\varphi }\left( {x, y}\right) \leftrightarrow \left( {\exists y \in S}\right) \varphi \left( {x, y}\right) }\right) \]\n\nNow \( \left( {\forall x \in A}\right) \left( {\exists !y \in S}\right) \varphi \left( {x, y}\right) \), because \( \left( {\forall x \in A}\right) \exists !{y\varphi }\left( {x, y}\right) \) . Now Separation yields \( \{ y \in S : \left( {\exists x \in A}\right) \varphi \left( {x, y}\right) \} = \{ y : \left( {\exists x \in A}\right) \varphi \left( {x, y}\right) \} \) .
|
Yes
|
Lemma 64.4. \( \langle \alpha \sqcup \beta , \vartriangleleft \rangle \) is a well-order, for any ordinals \( \alpha \) and \( \beta \) .
|
Proof. Obviously \( \vartriangleleft \) is connected on \( \alpha \sqcup \beta \) . To show it is well-founded, fix a non-empty \( X \subseteq \alpha \sqcup \beta \), and let\n\n\[ \n{X}_{0} = \{ \langle a, b\rangle \in X : \left( {\forall \langle x, y\rangle \in X}\right) b \leq y\} .\n\]\n\nNow choose the element of \( {X}_{0} \) with smallest first coordinate.
|
No
|
Proposition 64.5. \( \alpha + 1 = {\alpha }^{ + } \), for any ordinal \( \alpha \) .
|
Proof. Consider the isomorphism \( f \) from \( {\alpha }^{ + } = \alpha \cup \{ \alpha \} \) to \( \alpha \sqcup 1 = \left( {\alpha \times \{ 0\} }\right) \sqcup \) \( \left( {\{ 0\} \times \{ 1\} }\right) \) given by \( f\left( \gamma \right) = \langle \gamma ,0\rangle \) for \( \gamma \in \alpha \), and \( f\left( \alpha \right) = \langle 0,1\rangle \) .
|
Yes
|
Lemma 64.6. For any ordinals \( \alpha ,\beta \), we have:\n\n\[ \alpha + 0 = \alpha \]\n\n\[ \alpha + \left( {\beta + 1}\right) = \left( {\alpha + \beta }\right) + 1 \]\n\n\[ \alpha + \beta = \mathop{\operatorname{lsub}}\limits_{{\delta < \beta }}\left( {\alpha + \delta }\right) \] if \( \beta \) is a limit ordinal
|
Proof. We check case-by-case; first:\n\n\[ \alpha + 0 = \operatorname{ord}\left( {\left( {\alpha \times \{ 0\} }\right) \sqcup \left( {0\times \{ 1\} }\right) , \vartriangleleft }\right) \]\n\n\[ = \operatorname{ord}\left( {\left( {\alpha \times \{ 0\} }\right) \times \{ 0\} , \vartriangleleft }\right) \]\n\n\[ = \alpha \]\n\n\[ \alpha + \left( {\beta + 1}\right) = \operatorname{ord}\left( {\left( {\alpha \times \{ 0\} }\right) \cup \left( {{\beta }^{ + }\times \{ 1\} }\right) , \vartriangleleft }\right) \]\n\n\[ = \operatorname{ord}\left( {\left( {\alpha \times \{ 0\} }\right) \cup \left( {\beta \times \{ 1\} }\right) , \vartriangleleft }\right) + 1 \]\n\n\[ = \left( {\alpha + \beta }\right) + 1 \]\n\nNow let \( \beta \neq \varnothing \) be a limit. If \( \delta < \beta \) then also \( \delta + 1 < \beta \), so \( \alpha + \delta \) is a proper initial segment of \( \alpha + \beta \) . So \( \alpha + \beta \) is a strict upper bound on \( X = \{ \alpha + \delta : \delta < \beta \} \) . Moreover, if \( \alpha \leq \gamma < \alpha + \beta \), then clearly \( \gamma = \alpha + \delta \) for some \( \delta < \beta \) . So \( \alpha + \beta = {\operatorname{lsub}}_{\delta < \beta }\left( {\alpha + \delta }\right) .\)
|
Yes
|
Lemma 64.7. If \( \alpha ,\beta ,\gamma \) are ordinals, then:\n\n1. if \( \beta < \gamma \), then \( \alpha + \beta < \alpha + \gamma \)\n\n2. if \( \alpha + \beta = \alpha + \gamma \), then \( \beta = \gamma \)\n\n3. \( \alpha + \left( {\beta + \gamma }\right) = \left( {\alpha + \beta }\right) + \gamma \), i.e., addition is associative\n\n4. If \( \alpha \leq \beta \), then \( \alpha + \gamma \leq \beta + \gamma \)
|
Proof. We prove (3), leaving the rest as an exercise. The proof is by Simple Transfinite Induction on \( \gamma \), using Lemma 64.6. When \( \gamma = 0 \) :\n\n\[ \left( {\alpha + \beta }\right) + 0 = \alpha + \beta = \alpha + \left( {\beta + 0}\right) \]\n\nWhen \( \gamma = \delta + 1 \), suppose for induction that \( \left( {\alpha + \beta }\right) + \delta = \alpha + \left( {\beta + \delta }\right) \) ; now:\n\n\[ \left( {\alpha + \beta }\right) + \left( {\delta + 1}\right) = \left( {\left( {\alpha + \beta }\right) + \delta }\right) + 1 \]\n\n\[ = \left( {\alpha + \left( {\beta + \delta }\right) }\right) + 1 \]\n\n\[ = \alpha + \left( {\left( {\beta + \delta }\right) + 1}\right) \]\n\n\[ = \alpha + \left( {\beta + \left( {\delta + 1}\right) }\right) \]\n\nWhen \( \gamma \) is a limit ordinal, suppose for induction that if \( \delta \in \gamma \) then \( \left( {\alpha + \beta }\right) + \delta = \alpha + \left( {\beta + \delta }\right) \) ; now:\n\n\[ \left( {\alpha + \beta }\right) + \gamma = \mathop{\operatorname{lub}}\limits_{{\delta < \gamma }}\left( {\left( {\alpha + \beta }\right) + \delta }\right) \]\n\n\[ = \mathop{\operatorname{lsub}}\limits_{{\delta < \gamma }}\left( {\alpha + \left( {\beta + \delta }\right) }\right) \]\n\n\[ = \alpha + \mathop{\operatorname{lsub}}\limits_{{\delta < \gamma }}\left( {\beta + \delta }\right) \]\n\n\[ = \alpha + \left( {\beta + \gamma }\right) \]
|
No
|
Proposition 64.8. Ordinal addition is not commutative; \( 1 + \omega = \omega < \omega + 1 \) .
|
Proof. Note that \( 1 + \omega = {\operatorname{lsub}}_{n < \omega }\left( {1 + n}\right) = \omega \in \omega \cup \{ \omega \} = {\omega }^{ + } = \omega + 1 \) .
|
Yes
|
Lemma 64.9. If \( \operatorname{rank}\left( A\right) = \alpha \) and \( \operatorname{rank}\left( B\right) = \beta \), then:\n\n1. \( \operatorname{rank}\left( {\wp \left( A\right) }\right) = \alpha + 1 \)
|
Proof. Throughout, we invoke Proposition 62.17 repeatedly.\n\n(1). If \( x \subseteq A \) then \( \operatorname{rank}\left( x\right) \leq \operatorname{rank}\left( A\right) \) . So \( \operatorname{rank}\left( {\wp \left( A\right) }\right) \leq \alpha + 1 \) . Since \( A \in \wp \left( A\right) \) in particular, \( \operatorname{rank}\left( {\wp \left( A\right) }\right) = \alpha + 1 \) .
|
Yes
|
Lemma 64.10. For any ordinal \( \alpha \), the following are equivalent:\n\n1. \( \alpha \notin \omega \), i.e., \( \alpha \) is not a natural number\n\n2. \( \omega \leq \alpha \)\n\n3. \( 1 + \alpha = \alpha \)\n\n4. \( \alpha \approx \alpha + 1 \), i.e., \( \alpha \) and \( \alpha + 1 \) are equinumerous\n\n5. \( \alpha \) is Dedekind infinite
|
Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) . By Trichotomy.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) . Fix \( \alpha \geq \omega \) . By Transfinite Induction, there is some least ordinal \( \gamma \) (possibly 0 ) such that there is a limit ordinal \( \beta \) with \( \alpha = \beta + \gamma \) . Now:\n\n\[ 1 + \alpha = 1 + \left( {\beta + \gamma }\right) = \left( {1 + \beta }\right) + \gamma = \mathop{\operatorname{lub}}\limits_{{1 + \delta }}\left( {\delta < \beta }\right) + \gamma = \beta + \gamma = \alpha . \]\n\n\( \left( 3\right) \Rightarrow \left( 4\right) \) . There is clearly a bijection \( f : \left( {\alpha \sqcup 1}\right) \rightarrow \left( {1 \sqcup \alpha }\right) \) . If \( 1 + \alpha = \alpha \), there is an isomorphism \( g : \left( {1 \sqcup \alpha }\right) \rightarrow \alpha \) . Now consider \( g \circ f \).\n\n\( \left( 4\right) \Rightarrow \left( 5\right) \) . If \( \alpha \approx \alpha + 1 \), there is a bijection \( f : \left( {\alpha \sqcup 1}\right) \rightarrow \alpha \) . Define \( g\left( \gamma \right) = \) \( f\left( {\gamma ,0}\right) \) for each \( \gamma < \alpha \) ; this injection witnesses that \( \alpha \) is Dedekind infinite, since \( f\left( {0,1}\right) \in \alpha \smallsetminus \operatorname{ran}\left( g\right) \).\n\n\( \left( 5\right) \Rightarrow \left( 1\right) \) . This is Proposition 60.8 .
|
Yes
|
Lemma 64.12. \( \langle \alpha \times \beta , \vartriangleleft \rangle \) is a well-order, for any ordinals \( \alpha \) and \( \beta \) .
|
Proof. Exactly as for Lemma 64.4.
|
No
|
Lemma 64.13. For any ordinals \( \alpha ,\beta \) :\n\n\[ \alpha \cdot 0 = 0 \]\n\n\[ \alpha \cdot \left( {\beta + 1}\right) = \left( {\alpha \cdot \beta }\right) + \alpha \]\n\n\[ \alpha \cdot \beta = \mathop{\operatorname{lub}}\limits_{{\delta < \beta }}\left( {\alpha \cdot \delta }\right) \;\text{when}\beta \text{is a limit ordinal.} \]
|
Proof. Left as an exercise.
|
No
|
Lemma 64.14. If \( \alpha ,\beta ,\gamma \) are ordinals, then:\n\n1. if \( \alpha \neq 0 \) and \( \beta < \gamma \), then \( \alpha \cdot \beta < \alpha \cdot \gamma \) ;\n\n2. if \( \alpha \neq 0 \) and \( \alpha \cdot \beta = \alpha \cdot \gamma \), then \( \beta = \gamma \) ;\n\n3. \( \alpha \cdot \left( {\beta \cdot \gamma }\right) = \left( {\alpha \cdot \beta }\right) \cdot \gamma \) ;\n\n4. If \( \alpha \leq \beta \), then \( \alpha \cdot \gamma \leq \beta \cdot \gamma \) ;\n\n5. \( \alpha \cdot \left( {\beta + \gamma }\right) = \left( {\alpha \cdot \beta }\right) + \left( {\alpha \cdot \gamma }\right) .
|
Proof. Left as an exercise.
|
No
|
Proposition 64.15. Ordinal multiplication is not commutative: \( 2 \cdot \omega = \omega < \omega \cdot 2 \)
|
Proof. \( 2 \cdot \omega = {\operatorname{lsub}}_{n < \omega }\left( {2 \cdot n}\right) = \omega \in {\operatorname{lsub}}_{n < \omega }\left( {\omega + n}\right) = \omega + \omega = \omega \cdot 2 \) .
|
Yes
|
Lemma 65.2. For every set \( A \) :\n\n1. \( \left| A\right| \) exists and is unique;\n\n2. \( \left| A\right| \approx A \) ;\n\n3. \( \left| A\right| \) is a cardinal, i.e., \( \left| A\right| = \parallel A\parallel \) ;
|
Proof. Fix \( A \) . By Well-Ordering, there is a well-ordering \( \langle A, R\rangle \) . By Theorem 61.25, \( \langle A, R\rangle \) is isomorphic to a unique ordinal, \( \beta \) . So \( A \approx \beta \) . By Transfinite Induction, there is a uniquely least ordinal, \( \gamma \), such that \( A \approx \gamma \) . So \( \left| A\right| = \gamma \), establishing (1) and (2). To establish (3), note that if \( \delta \in \gamma \) then \( \delta \prec A \), by our choice of \( \gamma \), so that also \( \delta \prec \gamma \) since equinumerosity is an equivalence relation (Proposition 4.20). So \( \gamma = \left| \gamma \right| \) .
|
Yes
|
Lemma 65.3. For any sets \( A \) and \( B \) :\n\n\[ A \approx B\text{ iff }\left| A\right| = \left| B\right| \]\n\n\[ A \preccurlyeq B\text{iff}\left| A\right| \leq \left| B\right| \]\n\n\[ A \prec B\text{ iff }\left| A\right| < \left| B\right| \]
|
Proof. We will prove the left-to-right direction of the second claim (the other cases are similar, and left as an exercise). So, consider the following diagram:\n\n\n\nThe double-headed arrows indicate bijections, whose existence is guaranteed by Lemma 65.2. In assuming that \( A \preccurlyeq B \), there is some an injection to \( A \rightarrow B \) . Now, chasing the arrows around from \( \left| A\right| \) to \( A \) to \( B \) to \( \left| B\right| \), we obtain an injection \( \left| A\right| \rightarrow \left| B\right| \) (the dashed arrow).
|
No
|
Proposition 65.5. Let \( n, m \in \omega \) . Then \( n = m \) iff \( n \approx m \) .
|
Proof. Left-to-right is trivial. To prove right-to-left, suppose \( n \approx m \) although \( n \neq m \) . By Trichotomy, either \( n \in m \) or \( m \in n \) ; suppose \( n \in m \) without loss of generality. Then \( n \subsetneq m \) and there is a bijection \( f : m \rightarrow n \), so that \( m \) is Dedekind infinite, contradicting Proposition 60.8.
|
Yes
|
Corollary 65.6. If \( n \in \omega \), then \( n \) is a cardinal.
|
Proof. Immediate.
|
No
|
Theorem 65.7. For any set \( A \), the following are equivalent:\n\n1. \( \left| A\right| \notin \omega \), i.e., \( A \) is not a natural number;\n\n2. \( \omega \leq \left| A\right| \) ;\n\n3. A is Dedekind infinite.
|
Proof. From Lemma 64.10, Lemma 65.3, and Corollary 65.6.
|
No
|
Corollary 65.9. \( \omega \) is the least infinite cardinal.
|
Proof. \( \omega \) is a cardinal, since \( \omega \) is Dedekind infinite and if \( \omega \approx n \) for any \( n \in \omega \) then \( n \) would be Dedekind infinite, contradicting Proposition 60.8. Now \( \omega \) is the least infinite cardinal by definition.
|
No
|
Corollary 65.10. Every infinite cardinal is a limit ordinal.
|
Proof. Let \( \alpha \) be an infinite successor ordinal, so \( \alpha = \beta + 1 \) for some \( \beta \) . By Proposition 65.5, \( \beta \) is also infinite, so \( \beta \approx \beta + 1 \) by Lemma 64.10. Now \( \left| \beta \right| = \) \( \left| {\beta + 1}\right| = \left| \alpha \right| \) by Lemma 65.3, so that \( \alpha \neq \left| \alpha \right| \) .
|
Yes
|
Proposition 65.11. \( A \) is enumerable iff \( \left| A\right| \leq \omega \), and \( A \) is non-enumerable iff \( \omega < \) \( \left| A\right| \) .
|
Proof. By Trichotomy, the two claims are equivalent, so it suffices to prove that \( A \) is enumerable iff \( \left| A\right| \leq \omega \) . For right-to-left: if \( \left| A\right| \leq \omega \), then \( A \preccurlyeq \omega \) by Lemma 65.3 and Corollary 65.9. For left-to-right: suppose \( A \) is enumerable; then by Definition 4.27 there are three possible cases:\n\n1. if \( A = \varnothing \), then \( \left| A\right| = 0 \in \omega \), by Corollary 65.6 and Lemma 65.3.\n\n2. if \( n \approx A \), then \( \left| A\right| = n \in \omega \), by Corollary 65.6 and Lemma 65.3.\n\n3. if \( \omega \approx A \), then \( \left| A\right| = \omega \), by Corollary 65.9.\n\nSo in all cases, \( \left| A\right| \leq \omega \) .
|
Yes
|
Corollary 65.12. \( \omega \) is the only enumerable infinite cardinal.
|
Proof. Let \( \mathfrak{a} \) be a enumerable infinite cardinal. Since \( \mathfrak{a} \) is infinite, \( \omega \leq \mathfrak{a} \) . Since \( \mathfrak{a} \) is a enumerable cardinal, \( \mathfrak{a} = \left| \mathfrak{a}\right| \leq \omega \) . So \( \mathfrak{a} = \omega \) by Trichotomy.
|
Yes
|
Proposition 65.13. If every member of \( X \) is a cardinal, then \( \bigcup X \) is a cardinal.
|
Proof. It is easy to check that \( \bigcup X \) is an ordinal. Let \( \alpha \in \bigcup X \) be an ordinal; then \( \alpha \in \mathfrak{b} \in X \) for some cardinal \( \mathfrak{b} \). Since \( \mathfrak{b} \) is a cardinal, \( \alpha \prec \mathfrak{b} \). Since \( \mathfrak{b} \subseteq \bigcup X \), we have \( \mathfrak{b} \preccurlyeq \bigcup X \), and so \( \alpha ≉ \bigcup X \). Generalising, \( \bigcup X \) is a cardinal.
|
Yes
|
Theorem 65.14. There is no largest cardinal.
|
Proof. For any cardinal \( \mathfrak{a} \), Cantor’s Theorem (Theorem 4.24) and Lemma 65.2 entail that \( \mathfrak{a} < \left| {\wp \left( \mathfrak{a}\right) }\right| \) .
|
Yes
|
Theorem 65.15. The set of all cardinals does not exist.
|
Proof. For reductio, suppose \( C = \{ \mathfrak{a} : \mathfrak{a} \) is a cardinal \( \} \) . Now \( \cup C \) is a cardinal by Proposition 65.13, so by Theorem 65.14 there is a cardinal \( \mathfrak{b} > \bigcup C \) . By definition \( \mathfrak{b} \in C \), so \( \mathfrak{b} \subseteq \cup C \), so that \( \mathfrak{b} \leq \cup C \), a contradiction.
|
Yes
|
Proposition 66.2. \( \oplus \) and \( \otimes \) are commutative and associative.
|
Proof. For commutativity, by Lemma 65.3 it suffices to observe that \( \left( {\mathfrak{a} \sqcup \mathfrak{b}}\right) \approx \) \( \left( {\mathfrak{b} \sqcup \mathfrak{a}}\right) \) and \( \left( {\mathfrak{a} \times \mathfrak{b}}\right) \approx \left( {\mathfrak{b} \times \mathfrak{a}}\right) \) . We leave associativity as an exercise.
|
No
|
Lemma 66.4. \( \\left| {\\wp \\left( A\\right) }\\right| = {2}^{\\left| A\\right| } \), for any \( A \) .
|
Proof. For each subset \( B \\subseteq A \), let \( {\\chi }_{B} \\in {}^{A}2 \) be given by:\n\n\[ \n{\\chi }_{B}\\left( x\\right) \\mathrel{\\text{:=}} \\left\\{ \\begin{array}{ll} 1 & \\text{ if }x \\in B \\\\ 0 & \\text{ otherwise. } \\end{array}\\right.\n\]\n\nNow let \( f\\left( B\\right) = {\\chi }_{B} \) ; this defines a bijection \( f : \\wp \\left( A\\right) \\rightarrow {}^{A}2 \) . So \( \\wp \\left( A\\right) \\approx {}^{A}2 \) . Hence \( \\wp \\left( A\\right) \\approx {}^{\\left| A\\right| }2 \), so that \( \\left| {\\wp \\left( A\\right) }\\right| = \\left| {{}^{\\left| A\\right| }2}\\right| = {2}^{\\left| A\\right| } \).
|
Yes
|
Corollary 66.5. \( \mathfrak{a} < {2}^{\mathfrak{a}} \) for any cardinal \( \mathfrak{a} \) .
|
Proof. From Cantor's Theorem (Theorem 4.24) and Lemma 66.4.
|
No
|
Theorem 66.6. \( \\left| \\mathbb{R}\\right| = {2}^{\\omega } \)
|
Proof skeleton. There are plenty of ways to prove this. The most straightforward is to argue that \( \\wp \\left( \\omega \\right) \\preccurlyeq \\mathbb{R} \) and \( \\mathbb{R} \\preccurlyeq \\wp \\left( \\omega \\right) \), and then use Schröder-Bernstein to infer that \( \\mathbb{R} \\approx \\wp \\left( \\omega \\right) \), and Lemma 66.4 to infer that \( \\left| \\mathbb{R}\\right| = {2}^{\\omega } \). We leave it as an (illuminating) exercise for the reader to define injections \( f : \\wp \\left( \\omega \\right) \\rightarrow \\mathbb{R} \) and \( g : \\mathbb{R} \\rightarrow \\wp \\left( \\omega \\right) \).
|
No
|
Lemma 66.8. \( \langle \alpha \times \alpha , \vartriangleleft \rangle \) is a well-order, for any ordinal \( \alpha \) .
|
Proof. Evidently \( \vartriangleleft \) is connected on \( \alpha \times \alpha \) . For suppose that neither \( \left\langle {{\alpha }_{1},{\alpha }_{2}}\right\rangle \) nor \( \left\langle {{\beta }_{1},{\beta }_{2}}\right\rangle \) is \( \vartriangleleft \) -less than the other. Then \( \max \left( {{\alpha }_{1},{\alpha }_{2}}\right) = \max \left( {{\beta }_{1},{\beta }_{2}}\right) \) and \( {\alpha }_{1} = {\beta }_{1} \) and \( {\alpha }_{2} = {\beta }_{2} \), so that \( \left\langle {{\alpha }_{1},{\alpha }_{2}}\right\rangle = \left\langle {{\beta }_{1},{\beta }_{2}}\right\rangle \) .\n\nTo show well-ordering, let \( X \subseteq \alpha \times \alpha \) be non-empty. Since \( \alpha \) is an ordinal, some \( \delta \) is the least member of \( \left\{ {\max \left( {{\gamma }_{1},{\gamma }_{2}}\right) : \left\langle {{\gamma }_{1},{\gamma }_{2}}\right\rangle \in X}\right\} \) . Now discard all pairs from \( \left\{ {\left\langle {{\gamma }_{1},{\gamma }_{2}}\right\rangle \in X : \max \left( {{\gamma }_{1},{\gamma }_{2}}\right) = \delta }\right\} \) except those with least first coordinate; from among these, the pair with least second coordinate is the \( \vartriangleleft \) -least element of \( X \) .
|
Yes
|
Lemma 66.10. \( \\alpha \\approx \\alpha \\times \\alpha \), for any infinite ordinal \( \\alpha \)
|
Proof. For reductio, let \( \\alpha \) be the least infinite ordinal for which this is false. Proposition 4.12 shows that \( \\omega \\approx \\omega \\times \\omega \), so \( \\omega \\in \\alpha \). Moreover, \( \\alpha \) is a cardinal: suppose otherwise, for reductio; then \( \\left| \\alpha \\right| \\in \\alpha \), so that \( \\left| \\alpha \\right| \\approx \\left| \\alpha \\right| \\times \\left| \\alpha \\right| \), by hypothesis; and \( \\left| \\alpha \\right| \\approx \\alpha \) by definition; so that \( \\alpha \\approx \\alpha \\times \\alpha \) by Proposition 66.9.\n\nNow, for each \( \\left\\langle {{\\gamma }_{1},{\\gamma }_{2}}\\right\\rangle \\in \\alpha \\times \\alpha \), consider the segment:\n\n\[ \n\\operatorname{Seg}\\left( {{\\gamma }_{1},{\\gamma }_{2}}\\right) = \\left\\{ {\\left\\langle {{\\delta }_{1},{\\delta }_{2}}\\right\\rangle \\in \\alpha \\times \\alpha : \\left\\langle {{\\delta }_{1},{\\delta }_{2}}\\right\\rangle \\vartriangleleft \\left\\langle {{\\gamma }_{1},{\\gamma }_{2}}\\right\\rangle }\\right\\} \n\]\n\nLet \( \\gamma = \\max \\left( {{\\gamma }_{1},{\\gamma }_{2}}\\right) \). When \( \\gamma \) is infinite, observe:\n\n\( \\operatorname{Seg}\\left( {{\\gamma }_{1},{\\gamma }_{2}}\\right) \\precsim \\left( {\\left( {\\gamma + 1}\\right) \\cdot \\left( {\\gamma + 1}\\right) }\\right) \), by the first clause defining \( \\vartriangleleft \)\n\n\( \\approx \\left( {\\gamma \\cdot \\gamma }\\right) \), by Lemma 64.10 and Proposition 66.9\n\n\( \\approx \\gamma \), by the induction hypothesis\n\n\( \\prec \\alpha \), since \( \\alpha \) is a cardinal\n\nSo \( \\operatorname{ord}\\left( {\\alpha \\times \\alpha , \\vartriangleleft }\\right) \\leq \\alpha \), and hence \( \\alpha \\times \\alpha \\preccurlyeq \\alpha \). Since of course \( \\alpha \\preccurlyeq \\alpha \\times \\alpha \), the result follows by Schröder-Bernstein.
|
Yes
|
Theorem 66.11. If \( \mathfrak{a},\mathfrak{b} \) are infinite cardinals, \( \mathfrak{a} \otimes \mathfrak{b} = \mathfrak{a} \oplus \mathfrak{b} = \max \left( {\mathfrak{a},\mathfrak{b}}\right) \) .
|
Proof. Without loss of generality, suppose \( \mathfrak{a} = \max \left( {\mathfrak{a},\mathfrak{b}}\right) \) . Then invoking Lemma 66.10, \( \mathfrak{a} \otimes \mathfrak{a} = \mathfrak{a} \leq \mathfrak{a} \oplus \mathfrak{b} \leq \mathfrak{a} \oplus \mathfrak{a} \leq \mathfrak{a} \otimes \mathfrak{a} \) .
|
Yes
|
Proposition 66.12. Let \( \\mathfrak{a} \) be an infinite cardinal. For each ordinal \( \\beta \\in \\mathfrak{a} \), let \( {X}_{\\beta } \) be a set with \( \\left| {X}_{\\beta }\\right| \\leq \\mathfrak{a} \). Then \( \\left| {\\mathop{\\bigcup }\\limits_{{\\beta \\in \\mathfrak{a}}}{X}_{\\beta }}\\right| \\leq \\mathfrak{a} \).
|
Proof. For each \( \\beta \\in \\mathfrak{a} \), fix an injection \( {f}_{\\beta } : {X}_{\\beta } \\rightarrow \\mathfrak{a} \). Define an injection \( g : \\mathop{\\bigcup }\\limits_{{\\beta \\in \\mathfrak{a}}}{X}_{\\beta } \\rightarrow \) \( \\mathfrak{a} \\times \\mathfrak{a} \) by \( g\\left( v\\right) = \\left\\langle {\\beta ,{f}_{\\beta }\\left( v\\right) }\\right\\rangle \), where \( v \\in {X}_{\\beta } \) and \( v \\notin {X}_{\\gamma } \) for any \( \\gamma \\in \\beta \). Now \( \\mathop{\\bigcup }\\limits_{{\\beta \\in \\mathfrak{a}}}{X}_{\\beta } \\preccurlyeq \\mathfrak{a} \\times \\mathfrak{a} \\approx \\mathfrak{a} \) by Theorem 66.11.
|
Yes
|
Proposition 66.13. \( {\mathfrak{a}}^{\mathfrak{b} \oplus \mathfrak{c}} = {\mathfrak{a}}^{\mathfrak{b}} \otimes {\mathfrak{a}}^{\mathfrak{c}} \) and \( {\left( {\mathfrak{a}}^{\mathfrak{b}}\right) }^{\mathfrak{c}} = {\mathfrak{a}}^{\mathfrak{b} \otimes \mathfrak{c}} \), for any cardinals \( \mathfrak{a},\mathfrak{b},\mathfrak{c} \) .
|
Proof. For the first claim, consider a function \( f : \left( {\mathfrak{b} \sqcup \mathfrak{c}}\right) \rightarrow \mathfrak{a} \) . Now \
|
No
|
Proposition 66.14. If \( 2 \leq \mathfrak{a} \leq \mathfrak{b} \) and \( \mathfrak{b} \) is infinite, then \( {\mathfrak{a}}^{\mathfrak{b}} = {2}^{\mathfrak{b}} \)
|
Proof.\n\n\[ \n{2}^{\mathfrak{b}} \leq {\mathfrak{a}}^{\mathfrak{b}},\text{ as }2 \leq \mathfrak{a} \n\] \n\n\[ \n\leq {\left( {2}^{\mathfrak{a}}\right) }^{\mathfrak{b}}\text{, by Lemma 66.4} \n\] \n\n\[ \n= {2}^{\mathfrak{a} \otimes \mathfrak{b}}\text{, by Proposition 66.13} \n\] \n\n\[ \n= {2}^{b}\text{, by Theorem 66.11} \n\] \n\nWe should not really expect to be able to simplify this any further, since \( \mathfrak{b} < {2}^{\mathfrak{b}} \) by Lemma 66.4. However, this does not tell us what to say about \( {\mathfrak{a}}^{\mathfrak{b}} \) when \( \mathfrak{b} < \mathfrak{a} \) . Of course, if \( \mathfrak{b} \) is finite, we know what to do.
|
No
|
Proposition 66.15. If \( \mathfrak{a} \) is infinite and \( n \in \omega \) then \( {\mathfrak{a}}^{n} = \mathfrak{a} \)
|
Proof. \( {\mathfrak{a}}^{n} = \mathfrak{a} \otimes \mathfrak{a} \otimes \ldots \otimes \mathfrak{a} = \mathfrak{a} \), by \( n - 1 \) applications of Theorem 66.11.
|
No
|
Proposition 66.16. If \( 2 \leq \mathfrak{b} < \mathfrak{a} \leq {2}^{\mathfrak{b}} \) and \( \mathfrak{b} \) is infinite, then \( {\mathfrak{a}}^{\mathfrak{b}} = {2}^{\mathfrak{b}} \)
|
Proof. \( {2}^{\mathfrak{b}} \leq {\mathfrak{a}}^{\mathfrak{b}} \leq {\left( {2}^{\mathfrak{b}}\right) }^{\mathfrak{b}} = {2}^{\mathfrak{b} \otimes \mathfrak{b}} = {2}^{\mathfrak{b}} \), reasoning as in Proposition 66.14.
|
Yes
|
Proposition 66.18. Both \( {\aleph }_{\alpha } \) and \( {\beth }_{\alpha } \) are cardinals, for every ordinal \( \alpha \) .
|
Proof. Both results hold by a simple transfinite induction. \( {\aleph }_{0} = {\beth }_{0} = \omega \) is a cardinal by Corollary 65.9. Assuming \( {\aleph }_{\alpha } \) and \( {\beth }_{\alpha } \) are both cardinals, \( {\aleph }_{\alpha + 1} \) and \( {\beth }_{\alpha + 1} \) are explicitly defined as cardinals. And the union of a set of cardinals is a cardinal, by Proposition 65.13.
|
Yes
|
Proposition 66.19. If \( \mathfrak{a} \) is an infinite cardinal, then \( \mathfrak{a} = {\aleph }_{\gamma } \) for some \( \gamma \) .
|
Proof. By transfinite induction on cardinals. For induction, suppose that if \( \mathfrak{b} < \mathfrak{a} \) then \( \mathfrak{b} = {\aleph }_{{\gamma }_{\mathfrak{b}}} \) . If \( \mathfrak{a} = {\mathfrak{b}}^{ \oplus } \) for some \( \mathfrak{b} \), then \( \mathfrak{a} = {\aleph }_{{\gamma }_{\mathfrak{b}}}^{ \oplus } = {\aleph }_{{\gamma }_{\mathfrak{b}} + 1} \) . If \( \mathfrak{a} \) is not the successor of any cardinal, then since cardinals are ordinals \( \mathfrak{a} = \mathop{\bigcup }\limits_{{\mathfrak{b} < \mathfrak{a}}}\mathfrak{b} = \) \( \mathop{\bigcup }\limits_{{\mathfrak{b} < \mathfrak{a}}}{\aleph }_{{\gamma }_{\mathfrak{b}}} \), so \( \mathfrak{a} = {\aleph }_{\gamma } \) where \( \gamma = \mathop{\bigcup }\limits_{{\mathfrak{b} < \mathfrak{a}}}{\gamma }_{\mathfrak{b}} \) .
|
Yes
|
Proposition 66.20. There is an \( \aleph \) -fixed-point.
|
Proof. Using recursion, define:\n\n\[{\kappa }_{0} = 0\]\n\n\[{\kappa }_{n + 1} = {\aleph }_{{\kappa }_{n}}\]\n\n\[\kappa = \mathop{\bigcup }\limits_{{n < \omega }}{\kappa }_{n}\]\n\nNow \( \kappa \) is a cardinal by Proposition 65.13. But now:\n\n\[\kappa = \mathop{\bigcup }\limits_{{n < \omega }}{\kappa }_{n + 1} = \mathop{\bigcup }\limits_{{n < \omega }}{\aleph }_{{\kappa }_{n}} = \mathop{\bigcup }\limits_{{\alpha < \kappa }}{\aleph }_{\alpha } = {\aleph }_{\kappa }\]
|
Yes
|
Proposition 66.21. There is a \( \beth \) -fixed-point, i.e., \( {a\kappa } \) such that \( \kappa = {\beth }_{\kappa } \)
|
Proof. As in Proposition 66.20, using \
|
No
|
Proposition 66.22. \( \left| {V}_{\omega + \alpha }\right| = {\beth }_{\alpha } \). If \( \omega \cdot \omega \leq \alpha \), then \( \left| {V}_{\alpha }\right| = {\beth }_{\alpha } \).
|
Proof. The first claim holds by a simple transfinite induction. The second claim follows, since if \( \omega \cdot \omega \leq \alpha \) then \( \omega + \alpha = \alpha \). To establish this, we use facts about ordinal arithmetic from chapter 64. First note that \( \omega \cdot \omega = \omega \cdot \left( {1 + \omega }\right) = \left( {\omega \cdot 1}\right) + \left( {\omega \cdot \omega }\right) = \omega + \left( {\omega \cdot \omega }\right) \). Now if \( \omega \cdot \omega \leq \alpha \), i.e., \( \alpha = \left( {\omega \cdot \omega }\right) + \beta \) for some \( \beta \), then \( \omega + \alpha = \omega + \left( {\left( {\omega \cdot \omega }\right) + \beta }\right) = \left( {\omega + \left( {\omega \cdot \omega }\right) }\right) + \beta = \left( {\omega \cdot \omega }\right) + \beta = \alpha \).
|
Yes
|
Corollary 66.23. There is a \( \kappa \) such that \( \left| {V}_{\kappa }\right| = \kappa \) .
|
Proof. Let \( \kappa \) be a \( \beth \) -fixed point, as given by Proposition 66.21. Clearly \( \omega \cdot \omega < \) \( \kappa \) . So \( \left| {V}_{\kappa }\right| = {\beth }_{\kappa } = \kappa \) by Proposition 66.22.
|
Yes
|
Lemma 67.3 (in \( \mathbf{ZF} \)). For any set \( A \), there is an ordinal \( \alpha \) such that \( \alpha \npreceq A \)
|
Proof. If \( B \subseteq A \) and \( R \subseteq {B}^{2} \), then \( \langle B, R\rangle \subseteq {V}_{\operatorname{rank}\left( A\right) + 4} \) by Lemma 64.9. So, using Separation, consider:\n\n\[ C = \left\{ {\langle B, R\rangle \in {V}_{\operatorname{rank}\left( A\right) + 5} : B \subseteq A}\right. \text{and}\langle B, R\rangle \text{is a well-ordering}\} \]\n\nUsing Replacement and Theorem 61.25, form the set:\n\n\[ \alpha = \{ \operatorname{ord}\left( {B, R}\right) : \langle B, R\rangle \in C\} . \]\n\nBy Corollary \( {61.18},\alpha \) is an ordinal, since it is a transitive set of ordinals. After all, if \( \gamma \in \beta \in \alpha \), then \( \beta = \operatorname{ord}\left( {B, R}\right) \) for some \( B \subseteq R \), whereupon \( \gamma = \) \( \operatorname{ord}\left( {{B}_{b},{R}_{b}}\right) \) for some \( b \in B \) by Lemma 61.9, so that \( \gamma \in \alpha \) .\n\nFor reductio, suppose there is an injection \( f : \alpha \rightarrow A \) . Then, where:\n\n\[ B = \operatorname{ran}\left( f\right) \]\n\n\[ R = \{ \langle f\left( \alpha \right), f\left( \beta \right) \rangle \in A \times A : \alpha \in \beta \} . \]\n\nClearly \( \alpha = \operatorname{ord}\left( {B, R}\right) \) and \( \langle B, R\rangle \in C \) . So \( \alpha \in \alpha \), which is a contradiction.
|
Yes
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.