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Theorem 7.5 (Tarski). Every filter can be extended to an ultrafilter. | Proof. Let \( {F}_{0} \) be a filter on \( S \) . Let \( P \) be the set of all filters \( F \) on \( S \) such that \( F \supset {F}_{0} \) and consider the partially ordered set \( \left( {P, \subset }\right) \) . If \( \mathcal{C} \) is a chain in \( P \), then by Lemma 7.2(ii), \( \bigcup \mathcal{C} \) is a filter... | Yes |
Theorem 7.6 (Pospíšil). For every infinite cardinal \( \kappa \), there exist \( {2}^{{2}^{\kappa }} \) uniform ultrafilters on \( \kappa \) . | We prove first the following lemma. Let us call a family \( \mathcal{A} \) of subsets of \( \kappa \) independent if for any distinct sets \( {X}_{1},\ldots ,{X}_{n},{Y}_{1},\ldots ,{Y}_{m} \) in \( \mathcal{A} \), the intersection\n\n(7.4)\n\n\[ \n{X}_{1} \cap \ldots \cap {X}_{n} \cap \left( {\kappa - {Y}_{1}}\right) ... | No |
Lemma 7.7. There exists an independent family of subsets of \( \kappa \) of cardinality \( {2}^{\kappa } \) . | Proof. Let us consider the set \( P \) of all pairs \( \left( {F,\mathcal{F}}\right) \) where \( F \) is a finite subset of \( \kappa \) and \( \mathcal{F} \) is a finite set of finite subsets of \( \kappa \) . Since \( \left| P\right| = \kappa \), it suffices to find an independent family \( \mathcal{A} \) of subsets ... | Yes |
Theorem 7.8. If \( {2}^{{\aleph }_{0}} = {\aleph }_{1} \), then a Ramsey ultrafilter exists. | Proof. Let \( {\mathcal{A}}_{\alpha },\alpha < {\omega }_{1} \), enumerate all partitions of \( \omega \) and let us construct an \( {\omega }_{1} \) -sequence of infinite subsets of \( \omega \) as follows: Given \( {X}_{\alpha } \), let \( {X}_{\alpha + 1} \subset {X}_{\alpha } \) be such that either \( {X}_{\alpha +... | Yes |
Theorem 7.10 (The Prime Ideal Theorem). Every ideal on B can be extended to a prime ideal. | The proof of the Prime Ideal Theorem uses the Axiom of Choice. It is known that the theorem cannot be proved without using the Axiom of Choice. However, it is also known that the Prime Ideal Theorem is weaker than the Axiom of Choice. | Yes |
Theorem 7.11 (Stone's Representation Theorem). Every Boolean algebra is isomorphic to an algebra of sets. | Proof. Let \( B \) be a Boolean algebra. We let\n\n(7.20)\n\n\[ S = \{ p : p\text{ is an ultrafilter on }B\} .\n\]\n\nFor every \( u \in B \), let \( {X}_{u} \) be the set of all \( p \in S \) such that \( u \in p \) . Let\n\n(7.21)\n\n\[ \mathcal{S} = \left\{ {{X}_{u} : u \in B}\right\} \]\n\nLet us consider the mappi... | Yes |
Lemma 7.12. The completion of a Boolean algebra \( B \) is unique up to isomorphism. | Proof. Let \( C \) and \( D \) be completions of \( B \) . We define an isomorphism \( \pi : C \rightarrow \) \( D \) by\n\n(7.23)\n\n\[ \pi \left( c\right) = \mathop{\sum }\limits^{D}\{ u \in B : u \leq c\} . \]\n\nTo verify that \( \pi \) is an isomorphism, one uses the fact that \( B \) is a dense subalgebra of both... | Yes |
Theorem 7.13. Every Boolean algebra has a completion. | Proof. We use a construction similar to the method of Dedekind cuts. Let \( A \) be a Boolean algebra. Let us call a set \( U \subset {A}^{ + } \) a cut if\n\n(7.24)\n\n\[ p \leq q\text{and}q \in U\text{implies}p \in U\text{.} \]\n\nFor every \( p \in {A}^{ + } \), let \( {U}_{p} \) denote the cut \( \{ x : x \leq p\} ... | No |
Theorem 7.15. If \( B \) is an infinite complete Boolean algebra, then \( \operatorname{sat}\left( B\right) \) is a regular uncountable cardinal. | Proof. Let \( \kappa = \operatorname{sat}\left( B\right) \) . It is clear that \( \kappa \) is uncountable. Let us assume that \( \kappa \) is singular; we shall obtain a contradiction by constructing a partition of size \( \kappa \) .\n\nFor \( u \in B, u \neq 0 \), let \( \operatorname{sat}\left( u\right) \) denote \... | Yes |
Lemma 8.2. If \( C \) and \( D \) are closed unbounded, then \( C \cap D \) is closed unbounded. | Proof. It is immediate that \( C \cap D \) is closed. To show that \( C \cap D \) is unbounded, let \( \alpha < \kappa \) . Since \( C \) is unbounded, there exists an \( {\alpha }_{1} > \alpha \) with \( {\alpha }_{1} \in C \) . Similarly there exists an \( {\alpha }_{2} > {\alpha }_{1} \) with \( {\alpha }_{2} \in D ... | Yes |
Theorem 8.3. The intersection of fewer than \( \kappa \) closed unbounded subsets of \( \kappa \) is closed unbounded. | Proof. We prove, by induction on \( \gamma < \kappa \), that the intersection of a sequence \( \left\langle {{C}_{\alpha } : \alpha < \gamma }\right\rangle \) of closed unbounded subsets of \( \kappa \) is closed unbounded. The induction step works at successor ordinals because of Lemma 8.2. If \( \gamma \) is a limit ... | Yes |
Lemma 8.4. The diagonal intersection of a \( \kappa \) -sequence of closed unbounded sets is closed unbounded. | Proof. Let \( \left\langle {{C}_{\alpha } : \alpha < \kappa }\right\rangle \) be a sequence of closed unbounded sets. It is clear from the definition that if we replace each \( {C}_{\alpha } \) by \( \mathop{\bigcap }\limits_{{\xi \leq \alpha }}{C}_{\xi } \), the diagonal intersection is the same. In view of Theorem 8.... | Yes |
Theorem 8.7 (Fodor). If \( f \) is a regressive function on a stationary set \( S \subset \kappa \), then there is a stationary set \( T \subset S \) and some \( \gamma < \kappa \) such that \( f\left( \alpha \right) = \gamma \) for all \( \alpha \in T \) . | Proof. Let us assume that for each \( \gamma < \kappa \), the set \( \{ \alpha \in S : f\left( \alpha \right) = \gamma \} \) is nonstationary, and choose a closed unbounded set \( {C}_{\gamma } \) such that \( f\left( \alpha \right) \neq \gamma \) for each \( \alpha \in S \cap {C}_{\gamma } \) . Let \( C = {\bigtriangl... | Yes |
Every stationary subset of \( {E}_{\omega }^{\kappa } \) is the union of \( \kappa \) disjoint stationary sets. | Let \( W \subset \{ \alpha < \kappa : \) cf \( \alpha = \omega \} \) be stationary. For every \( \alpha \in W \), we choose an increasing sequence \( \left\langle {{a}_{n}^{\alpha } : n \in \mathbf{N}}\right\rangle \) such that \( \mathop{\lim }\limits_{n}{a}_{n}^{\alpha } = \alpha \) . First we show that there is an \... | Yes |
Lemma 8.9. Let \( S \) be a stationary subset of \( \kappa \) and assume that every \( \alpha \in S \) is a regular uncountable cardinal. Then the set \( T = \{ \alpha \in S : S \cap \alpha \) is not a stationary subset of \( \alpha \} \) is stationary. | Proof. We prove that \( T \) intersects every closed unbounded subset of \( \kappa \) . Let \( C \) be closed unbounded. The set \( {C}^{\prime } \) of all limit points of \( C \) is also closed unbounded, and hence \( S \cap {C}^{\prime } \neq \varnothing \) . Let \( \alpha \) be the least element of \( S \cap {C}^{\p... | Yes |
Theorem 8.10 (Solovay). Let \( \kappa \) be a regular uncountable cardinal. Then every stationary subset of \( \kappa \) is the disjoint union of \( \kappa \) stationary subsets. | Proof. We follow the proof of Lemma 8.8 as much as possible. Let \( A \) be a stationary subset of \( \kappa \) . By Lemma 8.8, by the subsequent discussion and by Lemma 8.9, we may assume that the set \( W \) of all \( \alpha \in A \) such that \( \alpha \) is a regular cardinal and \( A \cap \alpha \) is not stationa... | Yes |
Lemma 8.11. If \( \kappa \) is regular and uncountable and if \( F \) is a normal filter on \( \kappa \) that contains all final segments \( \left\{ {\alpha : {\alpha }_{0} < \alpha < \kappa }\right\} \), then \( F \) contains all closed unbounded sets. | Proof. First we note that the set \( {C}_{0} \) of all limit ordinals is in \( F : {C}_{0} \) is the diagonal intersection of the sets \( {X}_{\alpha } = \{ \xi : \alpha + 1 < \xi < \kappa \} \) . Now let \( C \) be a closed unbounded set, and let \( C = \left\{ {{a}_{\alpha } : \alpha < \kappa }\right\} \) be its incr... | Yes |
Lemma 8.16. Assume that \( {\aleph }_{\alpha }^{{\aleph }_{1}} < {\aleph }_{{\omega }_{1}} \) for all \( \alpha < {\omega }_{1} \) . Let \( F \) be an almost disjoint family of functions\n\n\[ F \subset \mathop{\prod }\limits_{{\alpha < {\omega }_{1}}}{A}_{\alpha } \]\n\nsuch that the set\n\n(8.9)\n\n\[ \left\{ {\alpha... | Proof. We may as well assume that each \( {A}_{\alpha } \) is a set of ordinals and that \( {A}_{\alpha } \subset {\omega }_{\alpha } \) for all \( \alpha \) in some stationary subset of \( {\aleph }_{1} \) . Let\n\n\[ {S}_{0} = \left\{ {\alpha < {\omega }_{1} : \alpha \text{ is a limit ordinal and }{A}_{\alpha } \subs... | Yes |
Lemma 8.19.\n\n(i) \( A < \operatorname{Tr}\left( A\right) \) ,\n\n(ii) if \( A < B \) and \( B < C \) then \( A < C \) ,\n\n(iii) if \( A < B, A \simeq {A}^{\prime }{\;\operatorname{mod}\;{I}_{\mathrm{{NS}}}} \) and \( B \simeq {B}^{\prime }{\;\operatorname{mod}\;{I}_{\mathrm{{NS}}}} \) then \( {A}^{\prime } < {B}^{\p... | Thus \( < \) is a transitive relation on \( P\left( \kappa \right) /{I}_{\mathrm{{NS}}} \) . The next theorem shows that it is a well-founded partial ordering:\n\nTheorem 8.20 (Jech). | No |
Theorem 8.20 (Jech). The relation \( < \) is well-founded. | Proof. Assume to the contrary that there exist stationary sets such that \( {A}_{1} > {A}_{2} > {A}_{3}\ldots \) Therefore there exist closed unbounded sets \( {C}_{n} \) such that \( {A}_{n} \cap {C}_{n} \subset \operatorname{Tr}\left( {A}_{n + 1}\right) \) for \( n = 1,2,3,\ldots \) For each \( n \), let\n\n\[ \n{B}_... | Yes |
Theorem 8.22 (Jech). The closed unbounded filter on \( {P}_{\kappa }\left( A\right) \) is \( \kappa \) -complete. | Proof. This is a generalization of Theorem 8.3. First we proceed as in Lemma 8.2 and show that if \( C \) and \( D \) are closed unbounded then \( C \cap D \) is closed unbounded. Both proofs have straightforward generalizations from \( \left( {\kappa , < }\right) \) to \( \left( {{P}_{\kappa }\left( A\right) , \subset... | No |
Lemma 8.23. If \( \left\{ {{C}_{a} : a \in A}\right\} \) is a collection of closed unbounded subsets of \( {P}_{\kappa }\left( A\right) \) then its diagonal intersection is closed unbounded. | Proof. Let \( C = {\bigtriangleup }_{a \in A}{C}_{a} \) . First we show that \( C \) is closed. Let \( {x}_{0} \subset {x}_{1} \subset \) \( \ldots \subset {x}_{\xi } \subset \ldots ,\xi < \alpha \), be a chain in \( C \), with \( \alpha < \kappa \), and let \( x \) be its union. To show that \( x \in C \), let \( a \i... | Yes |
Theorem 8.24 (Jech). If \( f \) is a function on a stationary set \( S \subset {P}_{\kappa }\left( \lambda \right) \) and if \( f\left( x\right) \in x \) for every nonempty \( x \in S \), then there exist a stationary set \( T \subset S \) and some \( a \in A \) such that \( f\left( x\right) = a \) for all \( a \in T \... | Proof. The proof uses Lemma 8.23 and generalizes the proof of Theorem 8.7.\n\nLet us call a set \( D \subset {P}_{\kappa }\left( A\right) \) directed if for all \( x \) and \( y \) in \( D \) there is a \( z \in D \) such that \( x \cup y \subset z \) . | No |
Lemma 8.25. If \( C \) is a closed subset of \( {P}_{\kappa }\left( A\right) \) then for every directed set \( D \subset C \) with \( \left| D\right| < \kappa ,\bigcup D \in C \) . | Proof. By induction on \( \left| D\right| \) . Let \( \left| D\right| = \gamma, D = \left\{ {{x}_{\alpha } : \alpha < \gamma }\right\} \), and assume the lemma holds for every directed set of cardinality \( < \gamma \) . By induction on \( \alpha < \gamma \), let \( {D}_{\alpha } \) be a smallest directed subset of \( ... | Yes |
Lemma 8.26. For every closed unbounded set \( C \) in \( {P}_{\kappa }\left( A\right) \) there exists a function \( f : {\left\lbrack A\right\rbrack }^{ < \omega } \rightarrow {P}_{\kappa }\left( A\right) \) such that \( {C}_{f} \subset C \) . | Proof. By induction on \( \left| e\right| \) we find for each \( e \in {\left\lbrack A\right\rbrack }^{ < \omega } \) an infinite set \( f\left( e\right) \in C \) such that \( e \subset f\left( e\right) \) and that \( f\left( {e}_{1}\right) \subset f\left( {e}_{2}\right) \) whenever \( {e}_{1} \subset {e}_{2} \) . We w... | Yes |
Theorem 8.27 (Menas). Let \( A \subset B \) .\n\n(i) If \( S \) is stationary in \( {P}_{\kappa }\left( B\right) \), then \( S \mid A \) is stationary in \( {P}_{\kappa }\left( A\right) \) .\n\n(ii) If \( S \) is stationary in \( {P}_{\kappa }\left( A\right) \), then \( {S}^{B} \) is stationary in \( {P}_{\kappa }\left... | Proof. (i) holds because if \( C \) is a closed unbounded set in \( {P}_{\kappa }\left( A\right) \), then \( {C}^{B} \) is closed unbounded in \( {P}_{\kappa }\left( B\right) \) . For (ii), it suffices to prove that if \( C \) is closed unbounded in \( {P}_{\kappa }\left( B\right) \), then \( C \upharpoonright A \) con... | Yes |
Theorem 8.28 (Kueker). For every closed unbounded set \( C \subset {\left\lbrack A\right\rbrack }^{\omega } \) there is an operation \( F \) on \( A \) such that \( C \supset {C}_{F} = \left\{ {x \in {\left\lbrack A\right\rbrack }^{\omega } : x}\right. \) is closed under \( \left. F\right\} \) . | Proof. We may assume that \( A = \lambda \) is an infinite cardinal, and let \( C \) be a closed unbounded subset of \( {\left\lbrack \lambda \right\rbrack }^{\omega } \) . As in the proof of Lemma 8.26 there exists a function \( f : {\left\lbrack \lambda \right\rbrack }^{ < \omega } \rightarrow C \) such that \( e \su... | Yes |
Theorem 9.1 (Ramsey). Let \( n \) and \( k \) be natural numbers. Every partition \( \left\{ {{X}_{1},\ldots ,{X}_{k}}\right\} \) of \( {\left\lbrack \omega \right\rbrack }^{n} \) into \( k \) pieces has an infinite homogeneous set. | Proof. By induction on \( n \) . If \( n = 1 \), the theorem is trivial, so we assume that it holds for \( n \) and prove for \( n + 1 \) . Let \( F \) be a function from \( {\left\lbrack \omega \right\rbrack }^{n + 1} \) into \( \{ 1,\ldots, k\} \) . For each \( a \in \omega \), let \( {F}_{a} \) be the function on \(... | Yes |
Lemma 9.2. Let \( D \) be a nonprincipal ultrafilter on \( \omega \) . \( D \) is Ramsey if and only if for all natural numbers \( n \) and \( k \), every partition \( F : {\left\lbrack \omega \right\rbrack }^{n} \rightarrow \{ 1,\ldots, k\} \) has a homogeneous set \( H \in D \) . | Proof. First assume that \( D \) has the partition property stated in the lemma. Let \( \mathcal{A} \) be a partition of \( \omega \) such that \( A \notin D \) for all \( A \in \mathcal{A} \) ; we shall find \( X \in D \) such that \( \left| {X \cap A}\right| \leq 1 \) for all \( A \in \mathcal{A} \) . Let \( F : {\le... | Yes |
Lemma 9.3. For all \( \kappa \) ,\n\n\[ {2}^{\kappa } \nrightarrow {\left( \omega \right) }_{\kappa }^{2} \]\n\nIn other words, there is a partition of \( {2}^{\kappa } \) into \( \kappa \) pieces that does not have an infinite homogeneous set. | Proof. In fact, we find a partition that has no homogeneous set of size 3. Let \( S = \{ 0,1{\} }^{\kappa } \) and let \( F : {\left\lbrack S\right\rbrack }^{2} \rightarrow \kappa \) be defined by \( F\left( {\{ f, g\} }\right) = \) the least \( \alpha < \kappa \) such that \( f\left( \alpha \right) \neq g\left( \alpha... | Yes |
Lemma 9.4. For every \( \kappa \) ,\n\n\[ \n{2}^{\kappa } \rightarrow {\left( {\kappa }^{ + }\right) }^{2} \n\]\n\n(Thus the obvious generalization of Ramsey’s Theorem, namely \( {\aleph }_{1} \rightarrow \) \( {\left( {\aleph }_{1}\right) }_{2}^{2} \), is false.) | To construct a partition of \( {\left\lbrack {2}^{\kappa }\right\rbrack }^{2} \) that violates the partition property, let us consider the linearly ordered set \( \left( {P, < }\right) \) where \( P = \{ 0,1{\} }^{\kappa } \), and \( f < g \) if and only if \( f\left( \alpha \right) < g\left( \alpha \right) \) where \(... | Yes |
Lemma 9.5. The lexicographically ordered set \( \{ 0,1{\} }^{\kappa } \) has no increasing or decreasing \( {\kappa }^{ + } \) -sequence. | Proof. Assume that \( W = \left\{ {{f}_{\alpha } : \alpha < {\kappa }^{ + }}\right\} \subset \{ 0,1{\} }^{\kappa } \) is such that \( {f}_{\alpha } < {f}_{\beta } \) whenever \( \alpha < \beta \) (the decreasing case is similar). Let \( \gamma \leq \kappa \) be the least \( \gamma \) such that the set \( \left\{ {{f}_{... | Yes |
Theorem 9.7 (Dushnik-Miller). For every infinite cardinal \( \kappa \) , \[ \kappa \rightarrow {\left( \kappa ,\omega \right) }^{2} \] | Proof. Let \( \{ A, B\} \) be a partition of \( {\left\lbrack \kappa \right\rbrack }^{2} \) . For every \( x \in \kappa \), let \( {B}_{x} = \{ y \in \kappa \) : \( x < y \) and \( \{ x, y\} \in B\} \) . First let us assume that in every set \( X \subset \kappa \) of cardinality \( \kappa \) there exists an \( x \in X ... | Yes |
Lemma 9.9. Every weakly compact cardinal is inaccessible. | Proof. Let \( \kappa \) be a weakly compact cardinal. To show that \( \kappa \) is regular, let us assume that \( \kappa \) is the disjoint union \( \bigcup \left\{ {{A}_{\gamma } : \gamma < \lambda }\right\} \) such that \( \lambda < \kappa \) and \( \left| {A}_{\gamma }\right| < \kappa \) for each \( \gamma < \lambda... | Yes |
Lemma 9.13. If there exists a Suslin tree then there exists a normal Suslin tree. | Proof. Let \( T \) be a Suslin tree. \( T \) has height \( {\omega }_{1} \), and each level of \( T \) is countable. We first discard all points \( x \in T \) such that \( {T}_{x} = \{ y \in T : y \geq x\} \) is at most countable, and let \( {T}_{1} = \left\{ {x \in T : {T}_{x}}\right. \) is uncountable \( \} \) . Note... | Yes |
Theorem 9.18 (Shanin). Let \( W \) be an uncountable collection of finite sets. Then there exists an uncountable \( Z \subset W \) that is a \( \Delta \) -system. | Proof. Since \( W \) is uncountable, it is clear that uncountably many \( X \in W \) have the same size; thus we may assume that for some \( n,\left| X\right| = n \) for all \( X \in W \) . We prove the theorem by induction on \( n \) . If \( n = 1 \), the theorem is trivial. Thus assume that the theorem holds for \( n... | Yes |
Lemma 9.21. There exists an almost disjoint family of \( {2}^{{\aleph }_{0}} \) subsets of \( \omega \) . | Proof. Let \( S \) be the set of all finite \( 0 - 1 \) sequences: \( S = \mathop{\bigcup }\limits_{{n = 0}}^{\infty }\{ 0,1{\} }^{n} \) . For every \( f : \omega \rightarrow \{ 0,1\} \), let \( {A}_{f} \subset S \) be the set \( {A}_{f} = \{ s \in S : s \subset f\} = \{ f \mid n : n \in \omega \} \) . Clearly, \( {A}_... | Yes |
Lemma 9.23. For every regular cardinal \( \kappa \), there exists an almost disjoint family of \( {\kappa }^{ + } \) functions from \( \kappa \) to \( \kappa \) . | Proof. It suffices to show that given \( \kappa \) almost disjoint functions \( \left\{ {{f}_{\nu } : \nu < \kappa }\right\} \) , then there exists \( f : \kappa \rightarrow \kappa \) almost disjoint from all \( {f}_{\nu },\nu < \kappa \) ; this we do by diagonalization: Let \( f\left( \alpha \right) \neq {f}_{\nu }\le... | No |
Lemma 9.25. A Kurepa tree exists if and only if there exists a family \( \mathcal{F} \) of subsets of \( {\omega }_{1} \) such that:\n\n(9.12)\n\n(i) \( \left| \mathcal{F}\right| \geq {\aleph }_{2} \) ;\n\n(ii) for each \( \alpha < {\omega }_{1},\left| {\{ X \cap \alpha : X \in \mathcal{F}\} }\right| \leq {\aleph }_{0}... | Proof. (a) Let \( \left( {T,{ < }_{T}}\right) \) be a Kurepa tree. Since \( T \) has size \( {\aleph }_{1} \), we may assume that \( T = {\omega }_{1} \), and moreover that \( \alpha < \beta \) whenever \( \alpha { < }_{T}\beta \) . If we let \( \mathcal{F} \) be the family of all \( {\omega }_{1} \) -branches of \( T ... | Yes |
Lemma 10.2. Let \( \kappa \) be the least cardinal with the property that there is a nonprincipal \( \sigma \) -complete ultrafilter on \( \kappa \), and let \( U \) be such an ultrafilter. Then \( U \) is \( \kappa \) -complete. | Proof. Let \( U \) be a \( \sigma \) -complete ultrafilter on \( \kappa \), and let us assume that \( U \) is not \( \kappa \) -complete. Then there exists a partition \( \left\{ {{X}_{\alpha } : \alpha < \gamma }\right\} \) of \( \kappa \) such that \( \gamma < \kappa \), and \( {X}_{\alpha } \notin U \) for all \( \a... | Yes |
Lemma 10.4. Every measurable cardinal is inaccessible. | Proof. We have just given an argument why a measurable cardinal is regular. Let us show that measurable cardinals are strong limit cardinals. Let \( \kappa \) be measurable, and let us assume that there exists \( \lambda < \kappa \) such that \( {2}^{\lambda } \geq \kappa \) ; we shall reach a contradiction.\n\nLet \( ... | Yes |
Lemma 10.6. Let \( \mu \) be a measure on \( S \), and let \( {I}_{\mu } \) be the ideal of null sets. If \( {I}_{\mu } \) is \( \kappa \) -complete, then \( \mu \) is \( \kappa \) -additive. | Proof. Let \( \gamma < \kappa \), and let \( {X}_{\alpha },\alpha < \gamma \), be disjoint subsets of \( S \) . Since the \( {X}_{\alpha } \) are disjoint, at most countably many of them have positive measure. Thus let us write\n\n\[ \left\{ {{X}_{\alpha } : \alpha < \gamma }\right\} = \left\{ {{Y}_{n} : n = 0,1,2,\ldo... | Yes |
Corollary 10.10. If \( \kappa \) is a real-valued measurable cardinal, then either \( \kappa \) is measurable or \( \kappa \leq {2}^{{\aleph }_{0}} \) . | More generally, if \( \kappa \) carries a \( \kappa \) -complete \( \sigma \) -saturated ideal, then either \( \kappa \) is measurable or \( \kappa \leq {2}^{{\aleph }_{0}} \) . The measure \( \nu \) obtained in Lemma 10.9(i) is atomless; this follows from the fact that \( \kappa \leq {2}^{{\aleph }_{0}} \) and Lemma 1... | Yes |
Lemma 10.12. An Ulam matrix exists. | Proof. For each \( \xi < {\omega }_{1} \), let \( {f}_{\xi } \) be a function on \( \omega \) such that \( \xi \subset \operatorname{ran}\left( {f}_{\xi }\right) \) . Let us define \( {A}_{\alpha, n} \) for \( \alpha < {\omega }_{1} \) and \( n < \omega \) by\n\n(10.13)\n\n\[ \xi \in {A}_{\alpha, n}\;\text{ if and only... | Yes |
Lemma 10.13. There is no nontrivial \( \sigma \) -additive measure on \( {\omega }_{1} \) . More generally, there is no \( \sigma \) -complete \( \sigma \) -saturated ideal on \( {\omega }_{1} \) . | Proof. Let \( \left\{ {{A}_{\alpha, n} : \alpha < {\omega }_{1}, n < \omega }\right\} \) be an Ulam matrix. Assuming that we have a measure on \( {\omega }_{1} \), there is for each \( \alpha \) some \( n = {n}_{\alpha } \) such that \( {A}_{\alpha, n} \) has positive measure (because of (10.12)(ii)). Hence there exist... | Yes |
Lemma 10.14. If \( \kappa = {\lambda }^{ + } \), then there is no \( \kappa \) -complete \( \sigma \) -saturated ideal on \( \kappa \) . | Proof. For each \( \xi < {\lambda }^{ + } \), we let \( {f}_{\xi } \) be a function on \( \lambda \) such that \( \xi \subset \operatorname{ran}\left( {f}_{\xi }\right) \), and let\n\n\[ \xi \in {A}_{\alpha ,\eta }\;\text{ if and only if }\;{f}_{\xi }\left( \eta \right) = \alpha . \]\n\nThen \( \left\{ {{A}_{\alpha ,\e... | Yes |
Corollary 10.15. Every real-valued measurable cardinal is weakly inaccessible. | Lemma 10.14 completes the proof of Theorem 10.1: If there is a \( \sigma \) -additive nontrivial measure on \( S \), then either the measure has an atom \( A \) and we can construct a two-valued measure on \( S \) via a \( \sigma \) -complete nonprincipal ultrafilter on \( A \), and then \( \left| S\right| \geq \) the ... | Yes |
Lemma 10.16. If there exists a \( \kappa \) -scale, then \( \kappa \) is not a real-valued measurable cardinal. | Proof. Let \( {f}_{\alpha },\alpha < \kappa \), be a \( \kappa \) -scale. We define an \( \left( {{\aleph }_{0},{\aleph }_{0}}\right) \) -matrix of subsets of \( \kappa \) as follows: For \( n, k < \omega \), let\n\n(10.15)\n\n\[ \alpha \in {A}_{n, k}\;\text{ if and only if }\;{f}_{\alpha }\left( n\right) = k\;\left( {... | Yes |
If there is a measure on \( {2}^{{\aleph }_{0}} \), then \( {2}^{{\aleph }_{0}} > {\aleph }_{1} \). | Proof. If \( {2}^{{\aleph }_{0}} = {\aleph }_{1} \), then there exists an \( {\omega }_{1} \) -scale; a scale \( \left\langle {{f}_{\alpha } : \alpha < {\omega }_{1}}\right\rangle \) is constructed by transfinite induction to \( {\omega }_{1} \):\n\nLet \( \left\{ {{g}_{\alpha } : \alpha < {\omega }_{1}}\right\} \) enu... | Yes |
Lemma 10.18. Every measurable cardinal is weakly compact. | Proof. Let \( \kappa \) be a measurable cardinal. To show that \( \kappa \) is weakly compact, it suffices to prove the tree property. Let \( \left( {T, < }\right) \) be a tree of height \( \kappa \) with levels of size \( < \kappa \) . We consider a nonprincipal \( \kappa \) -complete ultrafilter \( U \) on \( T \) . ... | Yes |
Lemma 10.19. If \( D \) is a normal measure on \( \kappa \), then every set in \( D \) is stationary. | Proof. By Lemma 8.11, every closed unbounded set is in \( D \), and the lemma follows. | No |
Every measurable cardinal carries a normal measure. If \( U \) is a nonprincipal \( \kappa \) -complete ultrafilter on \( \kappa \) then there exists a function \( f : \kappa \rightarrow \kappa \) such that \( {f}_{ * }\left( U\right) = \left\{ {X \subset \kappa : {f}_{-1}\left( X\right) \in U}\right\} \) is a normal m... | Let \( U \) be a nonprincipal \( \kappa \) -complete ultrafilter on \( \kappa \) . For \( f \) and \( g \) in \( {\kappa }^{\kappa } \) , let\n\n\[ f \equiv g\;\text{ if and only if }\;\{ \alpha < \kappa : f\left( \alpha \right) = g\left( \alpha \right) \} \in U.\]\n\nIt is easily seen that \( \equiv \) is an equivalen... | Yes |
Lemma 10.21. Every measurable cardinal is a Mahlo cardinal. | Proof. Let \( \kappa \) be a measurable cardinal. We shall show that the set of all inaccessible cardinals \( \alpha < \kappa \) is stationary. As \( \kappa \) is strong limit, the set of all strong limit cardinals \( \alpha < \kappa \) is closed unbounded, and it suffices to show that the set of all regular cardinals ... | No |
Theorem 10.22. Let \( \kappa \) be a measurable cardinal, let \( D \) be a normal measure on \( \kappa \), and let \( F \) be a partition of \( {\left\lbrack \kappa \right\rbrack }^{ < \omega } \) into less than \( \kappa \) pieces. Then there exists a set \( H \in D \) homogeneous for \( F \) . Hence every measurable ... | Proof. Let \( D \) be a normal measure on \( \kappa \), and let \( F \) be a partition of \( {\left\lbrack \kappa \right\rbrack }^{ < \omega } \) into fewer than \( \kappa \) pieces. It suffices to show that for each \( n = 1,2,\ldots \) , there is \( {H}_{n} \in D \) such that \( F \) is constant on \( {\left\lbrack {... | Yes |
Lemma 11.1. Let \( X \) be a Polish space. Then there exists a continuous mapping from \( \mathcal{N} \) onto \( X \) . | Proof. Let \( X \) be a complete separable metric space; we construct a mapping \( f \) of \( \mathcal{N} \) onto \( X \) as follows: It is easy to construct, by induction on the length of \( s \in {Seq} \), a collection \( \left\{ {{C}_{s} : s \in {Seq}}\right\} \) of closed balls such that \( {C}_{\varnothing } = X \... | Yes |
For each \( \alpha \geq 1 \) there exists a set \( U \subset {\mathcal{N}}^{2} \) such that \( U \) is \( {\mathbf{\sum }}_{\alpha }^{0} \) (in \( {\mathcal{N}}^{2} \) ), and that for every \( {\mathbf{\sum }}_{\alpha }^{0} \) set \( A \) in \( \mathcal{N} \) there exists some \( a \in \mathcal{N} \) such that\n\n\[ A ... | Proof. By induction on \( \alpha \) . To construct a universal open set in \( {\mathcal{N}}^{2} \), let \( {G}_{1},\ldots \) , \( {G}_{k},\ldots \) be an enumeration of all basic open sets in \( \mathcal{N} \), and let \( {G}_{0} = \varnothing \) . Let\n\n\[ \left( {x, y}\right) \in U\text{if and only if}x \in {G}_{y\l... | Yes |
Corollary 11.3. For every \( \alpha \geq 1 \), there is a set \( A \subset \mathcal{N} \) that is \( {\mathbf{\sum }}_{\alpha }^{0} \) but not \( {\mathbf{\Pi }}_{\alpha }^{0} \) . | Proof. Let \( U \subset {\mathcal{N}}^{2} \) be a universal \( {\mathbf{\sum }}_{\alpha }^{0} \) set. Let us consider the set\n\n(11.8)\n\n\[ A = \{ x : \left( {x, x}\right) \in U\} . \]\n\nClearly, \( A \) is a \( {\mathbf{\sum }}_{\alpha }^{0} \) set. If \( A \) were \( {\mathbf{\Pi }}_{\alpha }^{0} \), then its comp... | Yes |
Lemma 11.7. A set \( A \) in a Polish space is analytic if and only if \( A \) is the result of the operation \( \mathcal{A} \) applied to a family of closed sets. | Proof. First we show that if \( {F}_{s}, s \in {Seq} \), are closed sets in a Polish space \( X \) , then \( A = \mathcal{A}\left\{ {{F}_{s} : s \in {Seq}}\right\} \) is analytic. We have\n\n\[ x \in A \leftrightarrow \exists a \in \mathcal{N}x \in \mathop{\bigcap }\limits_{{n = 0}}^{\infty }{F}_{a \upharpoonright n} \... | Yes |
Lemma 11.8. For each \( n \geq 1 \), there exists a universal \( {\mathbf{\sum }}_{n}^{1} \) set in \( {\mathcal{N}}^{2} \) ; i.e., a set \( U \subset {\mathcal{N}}^{2} \) such that \( U \) is \( {\mathbf{\sum }}_{n}^{1} \) and that for every \( {\mathbf{\sum }}_{n}^{1} \) set \( A \) in \( \mathcal{N} \) there exists ... | Proof. Let \( h \) be a homeomorphism of \( \mathcal{N} \times \mathcal{N} \) onto \( \mathcal{N} \) . If \( n = 1 \), let \( V \) be a universal \( {\mathbf{\sum }}_{1}^{0} \) set; if \( n > 1 \), let \( V \) be, by the induction hypothesis, a universal \( {\mathbf{\sum }}_{n - 1}^{1} \) set. Let\n\n(11.13)\n\n\[ \lef... | Yes |
Corollary 11.9. For each \( n \geq 1 \), there is a set \( A \subset \mathcal{N} \) that is \( {\mathbf{\sum }}_{n}^{1} \) but not \( {\mathbf{\Pi }}_{n}^{1} \). | Proof. Let \( U \subset {\mathcal{N}}^{2} \) be a universal \( {\mathbf{\sum }}_{n}^{1} \) set and let\n\n\[ A = \{ x : \left( {x, x}\right) \in U\} \] | Yes |
For any set \( X \subset {\mathbf{R}}^{n} \) there exists a measurable set \( A \supset X \) with the property that whenever \( Z \subset A - X \) is measurable, then \( Z \) is null. | If \( {\mu }^{ * }\left( X\right) < \infty \), then because \( {\mu }^{ * }\left( X\right) = \inf \{ \mu \left( A\right) : A \) is measurable and \( A \supset X\} \), there is a measurable \( A \supset X \) such that \( \mu \left( A\right) = {\mu }^{ * }\left( X\right) \) ; clearly such an \( A \) will do. If \( {\mu }... | Yes |
Lemma 11.15. The sets having the Baire property form a \( \sigma \) -algebra; hence every Borel set has the Baire property. | If \( \mathcal{B} \) denotes the \( \sigma \) -algebra of Borel sets, and if we denote by \( \mathcal{C} \) the \( \sigma \) - algebra of sets with the Baire property, and if \( I \) is the \( \sigma \) -ideal of meager sets, we have \( \mathcal{B}/I = \mathcal{C}/I \) . Note that the algebra \( \mathcal{B}/I \) is \( ... | Yes |
Lemma 11.17. For any set \( S \) in a Polish space \( X \), there exists a set \( A \supset S \) that has the Baire property and such that whenever \( Z \subset A - S \) has the Baire property, then \( Z \) is meager. | Proof. Let us consider a fixed countable topology basis \( O \) for \( X \) . Let \( S \subset X \) . Let\n\n\[ D\left( S\right) = \{ x \in X : \text{ for every }U \in O\text{ such that }x \in U, U \cap S\text{ is not meager }\} .\n\]\n\nNote that the complement of \( D\left( S\right) \) is the union of open sets and h... | Yes |
Theorem 11.18(i) (measurability of analytic sets) is due to Luzin [1917]. | The present proof of (i) and (ii) follows Marczewski [1930a]. | No |
Lemma 12.2. If \( \left\{ {{\mathfrak{A}}_{i},{e}_{i, j} : i, j \in D}\right\} \) is a directed system of models, there exists a model \( \mathfrak{A} \), unique up to isomorphism, and elementary embeddings \( {e}_{i} \) : \( {\mathfrak{A}}_{i} \rightarrow \mathfrak{A} \) such that \( \mathfrak{A} = \mathop{\bigcup }\l... | Proof. Consider the set \( S \) of all pairs \( \left( {i, a}\right) \) such that \( i \in D \) and \( a \in {A}_{i} \), and define an equivalence relation on \( S \) by\n\n\[ \left( {i, a}\right) \equiv \left( {j, b}\right) \leftrightarrow \exists k\left( {i \leq k, j \leq k\text{ and }{e}_{i, k}\left( a\right) = {e}_... | Yes |
Corollary 12.4. An ultrapower of a model \( \mathfrak{A} \) is elementarily equivalent to \( \mathfrak{A} \) . | Proof. By Theorem 12.3(ii) we have \( {\operatorname{Ult}}_{U}\mathfrak{A} \vDash \sigma \) if and only if \( \{ x : \mathfrak{A} \vDash \sigma \} \) is either \( S \) or empty, according to whether \( \mathfrak{A} \vDash \sigma \) or not. | Yes |
Corollary 12.5. The canonical embedding \( j : \mathfrak{A} \rightarrow {\operatorname{Ult}}_{U}\mathfrak{A} \) is an elementary embedding. | Proof. Let \( a \in A \) . By Loś’s Theorem, \( {\operatorname{Ult}}_{U}\mathfrak{A} \vDash \varphi \left\lbrack {j\left( a\right) }\right\rbrack \) if and only if \( {\operatorname{Ult}}_{U}\mathfrak{A} \vDash \) \( \varphi \left\lbrack {c}_{a}\right\rbrack \) if and only if \( \mathfrak{A} \vDash \varphi \left\lbrack... | Yes |
Theorem 12.7 (Tarski). A truth definition does not exist. | Proof. Let us assume that there is a formula \( T\left( x\right) \) satisfying (12.17). Let\n\n\[ \n{\varphi }_{0},\;{\varphi }_{1},\;{\varphi }_{2},\;\ldots \n\]\n\nbe an enumeration of all formulas with one free variable. Let \( \psi \left( x\right) \) be the formula\n\n\[ \nx \in \omega \land \neg T\left( {\# \left(... | Yes |
Lemma 12.9. If \( M \) is a transitive class and \( \varphi \) is a \( {\Delta }_{0} \) -formula, then for all \( {x}_{1},\ldots ,{x}_{n}, \)\n\n(12.18)\n\n\[{\varphi }^{M}\left( {{x}_{1},\ldots ,{x}_{n}}\right) \leftrightarrow \varphi \left( {{x}_{1},\ldots ,{x}_{n}}\right) . | Proof. If \( \varphi \) is an atomic formula, then (12.18) holds. If (12.18) holds for \( \varphi \) and \( \psi \), then it holds for \( \neg \varphi ,\varphi \land \psi ,\varphi \vee \psi ,\varphi \rightarrow \psi \), and \( \varphi \leftrightarrow \psi \) .\n\nLet \( \varphi \) be the formula \( \left( {\exists u \i... | Yes |
Theorem 12.11. In ZF minus Regularity, \( {\sigma }^{V} \) holds for every axiom \( \sigma \) of \( \mathrm{{ZF}} \) . | Proof. We use absoluteness of \( {\Delta }_{0} \) -formulas and the fact that for every set \( x \) , if \( x \subset V \), then \( x \in V \) .\n\nExtensionality. The formula\n\n\[ \left( {\left( {\forall u \in X}\right) u \in Y \land \left( {\forall u \in Y}\right) u \in X}\right) \rightarrow X = Y \]\n\nis \( {\Delt... | No |
Theorem 12.12. The existence of inaccessible cardinals is not provable in ZFC. Moreover, it cannot be shown that the existence of inaccessible cardinals is consistent with \( \mathrm{ZFC} \) . | We shall prove the first assertion and invoke Gödel's Second Incompleteness Theorem to obtain the second part.\n\nFirst we prove (in ZFC):\n\nLemma 12.13. If \( \kap | No |
Lemma 12.13. If \( \kappa \) is an inaccessible cardinal, then \( {V}_{\kappa } \) is a model of ZFC. | Proof. The proof of all axioms of ZFC except Replacement is as in the proof of consistency of the Axiom of Regularity (see Exercises 12.7 and 12.8). To show that \( {V}_{\kappa } \vDash \) Replacement, it is enough to show:\n\n(12.21)\n\nIf \( F \) is a function from some \( X \in {V}_{\kappa } \) into \( {V}_{\kappa }... | No |
(i) Let \( \varphi \left( {{u}_{1},\ldots ,{u}_{n}, x}\right) \) be a formula. For each set \( {M}_{0} \) there exists a set \( M \supset {M}_{0} \) such that\n\n(12.23)\n\n\[ \n\text{if}\exists {x\varphi }\left( {{u}_{1},\ldots ,{u}_{n}, x}\right) \text{then}\left( {\exists x \in M}\right) \varphi \left( {{u}_{1},\ldo... | Proof. We shall give a detailed proof of (i). An obvious modification of the proof gives (ii); we leave that to the reader.\n\nNote that the operation \( H\left( {{u}_{1},\ldots ,{u}_{n}}\right) \) defined below plays the same role as Skolem functions in the Löwenheim-Skolem Theorem.\n\nLet us recall the definition (6.... | Yes |
Lemma 13.2. For every \( \alpha ,\alpha \subset {L}_{\alpha } \) (and \( {L}_{\alpha } \cap \) Ord \( = \alpha \) ). | Proof. By induction on \( \alpha \) . At stage \( \alpha + 1 \), we need to show that \( \alpha \in {L}_{\alpha + 1} \) , or that \( \alpha \) is a definable subset of \( {L}_{\alpha } \) . Since \( \alpha = \left\{ {x \in {L}_{\alpha } : x}\right. \) is an ordinal \( \} \) , and \ | No |
Theorem 13.3. \( L \) is a model of \( \mathrm{{ZF}} \). | Proof. We show that \( {\sigma }^{L} \) holds for every axiom \( \sigma \) of ZF. Since \( L \) is a transitive class, every \( {\Delta }_{0} \) formula is absolute for \( L \). \n\nExtensionality. \( L \) is transitive and therefore extensional. \n\nPairing. Given \( a, b \in L \), let \( c = \{ a, b\} \). Let \( \alp... | No |
Theorem 13.4 (Gödel's Normal Form Theorem). There exist operations \( {G}_{1},\ldots ,{G}_{10} \) such that if \( \varphi \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) is a \( {\Delta }_{0} \) formula, then there is a composition \( G \) of \( {G}_{1},\ldots ,{G}_{10} \) such that for all \( {X}_{1},\ldots ,{X}_{n} \) , (... | The operations \( {G}_{1},\ldots ,{G}_{10} \) will be defined below. Compositions of \( {G}_{1},\ldots ,{G}_{10} \) are called Gödel operations. | No |
Corollary 13.5. If \( M \) is a transitive class closed under Gödel operations then \( M \) satisfies \( {\Delta }_{0} \)-Separation. | Proof. Let \( \varphi \left( {u,{p}_{1},\ldots ,{p}_{n}}\right) \) be a \( {\Delta }_{0} \) formula, and let \( X,{p}_{1},\ldots ,{p}_{n} \in M \) . Let\n\n\[ Y = \left\{ {u \in X : \varphi \left( {u,{p}_{1},\ldots ,{p}_{n}}\right) }\right\} . \]\n\nBy Lemma 12.9 it suffices to show that \( Y \in M \), in order that \(... | Yes |
Lemma 13.7. If \( G \) is a Gödel operation then the property \( Z = G\left( {{X}_{1},\ldots }\right. \) , \( \left. {X}_{n}\right) \) can be written as a \( {\Delta }_{0} \) formula. | Proof. We show, by induction on the complexity of \( G \) (a composition of \( {G}_{1} \) , \( \ldots ,{G}_{10}) \) :\n\n(13.5)\n\n(i) \( u \in G\left( {X,\ldots }\right) \) is \( {\Delta }_{0} \).\n\n(ii) If \( \varphi \) is \( {\Delta }_{0} \), then so are \( \forall u \in G\left( {X,\ldots }\right) \varphi \) and \(... | Yes |
A transitive class \( M \) is an inner model of \( \mathrm{{ZF}} \) if and only if it is closed under Gödel operations and is almost universal, i.e., every subset \( X \subset M \) is included in some \( Y \in M \) . | As Gödel operations are absolute for transitive models, an inner model is necessarily closed under \( {G}_{1},\ldots ,{G}_{10} \) . If \( X \) is a subset of an inner model \( M \) , then \( X \subset {V}_{\alpha } \cap M \) for some \( \alpha \), and \( {V}_{\alpha } \cap M \) is in \( M \) because \( \alpha \in M \) ... | No |
Lemma 13.10. Let \( n \geq 1 \) .\n\n(i) If \( P, Q \) are \( {\sum }_{n} \) properties, then so are \( \exists {xP}, P \land Q, P \vee Q,\left( {\exists u \in x}\right) P \) , \( \left( {\forall u \in x}\right) P \) .\n\n(ii) If \( P, Q \) are \( {\Pi }_{n} \) properties, then so are \( \forall {xP}, P \land Q, P \vee... | Proof. Let us prove the lemma for \( n = 1 \) . The general case follows easily by induction.\n\n(i) Let\n\n\[ P\left( {x,\ldots }\right) \leftrightarrow \exists {z\varphi }\left( {z, x,\ldots }\right) ,\]\n\n\[ Q\left( {x,\ldots }\right) \leftrightarrow \exists {u\psi }\left( {u, x,\ldots }\right) \]\n\nwhere \( \varp... | Yes |
Lemma 13.12. Let \( n \geq 1 \), let \( G \) be a \( {\sum }_{n} \) function (on \( V \) ), and let \( F \) be defined by induction:\n\n\[ F\left( \alpha \right) = G\left( {F \upharpoonright \alpha }\right) . \]\n\nThen \( F \) is a \( {\sum }_{n} \) function on Ord. | Proof. Since \( {Ord} \) is a \( {\sum }_{0} \) class, it is enough to verify that the following expression is \( {\sum }_{n} \) :\n\n(13.10) \( y = F\left( \alpha \right) \) if and only if \( \exists f(f \) is a function \( \land \operatorname{dom}\left( f\right) = \alpha \)\n\n\[ \land \left( {\forall \xi < \alpha }\... | Yes |
Lemma 13.14. The function \( \alpha \mapsto {L}_{\alpha } \) is \( {\Delta }_{1} \) . | Proof. The function \( {L}_{\alpha } \) is defined by transfinite induction and so by Lemma 13.12 it suffices to show that the induction step is \( {\sum }_{1} \) . In view of Corollary 13.8 it suffices to verify that\n\n(13.11)\n\n\[ Y = \operatorname{cl}\left( M\right) \]\n\n(where cl denotes closure under Gödel oper... | Yes |
Theorem 13.16 (Gödel).\n\n(i) \( L \) satisfies the Axiom of Constructibility \( \left( {V = L}\right) \) .\n\n(ii) \( L \) is the smallest inner model of \( \mathrm{{ZF}} \) . | Proof. (i) For every \( x \in L,{\left( x\text{is constructible}\right) }^{L} \) if and only if \( x \) is constructible, and hence \ | No |
Lemma 13.19. The relation \( { < }_{L} \) is \( {\sum }_{1} \) and moreover, for every limit ordinal \( \delta \) and every \( y \in {L}_{\delta }, x{ < }_{L}y \) if and only if \( x \in {L}_{\delta } \) and \( \left( {{L}_{\delta }, \in }\right) \vDash x{ < }_{L}y \) . | Proof. It suffices to prove that the function, \( \alpha \mapsto { < }_{\alpha } \) which assigns to each \( \alpha \) the canonical well-ordering of \( {L}_{\alpha } \) is \( {\sum }_{1} \) .\n\nThe function \( \alpha \mapsto { < }_{\alpha } \) is defined by induction and thus it suffices to show that the induction st... | Yes |
Theorem 13.20 (Gödel). If \( V = L \) then \( {2}^{{\aleph }_{\alpha }} = {\aleph }_{\alpha + 1} \) for every \( \alpha \) . | Proof. We shall prove that if \( X \) is a constructible subset of \( {\omega }_{\alpha } \) then there exists a \( \gamma < {\omega }_{\alpha + 1} \) such that \( X \in {L}_{\gamma } \) . Therefore \( {P}^{L}\left( {\omega }_{\alpha }\right) \subset {L}_{{\omega }_{\alpha + 1}} \), and since \( \left| {L}_{{\omega }_{... | Yes |
Theorem 13.22. Let \( A \) be an arbitrary set.\n\n(i) \( L\left\lbrack A\right\rbrack \) is a model of \( \mathrm{{ZFC}} \).\n\n(ii) \( L\left\lbrack A\right\rbrack \) satisfies the axiom \( \exists X\left( {V = L\left\lbrack X\right\rbrack }\right) \).\n\n(iii) If \( M \) is an inner model of \( \mathrm{{ZF}} \) such... | Proof. The proof follows closely the corresponding proofs for \( L \), but some additional arguments are needed. | No |
Lemma 13.23. Let \( \bar{A} = A \cap L\left\lbrack A\right\rbrack \) . Then \( L\left\lbrack \bar{A}\right\rbrack = L\left\lbrack A\right\rbrack \) and moreover \( \bar{A} \in \) \( L\left\lbrack \bar{A}\right\rbrack \) . | Proof. We show by induction on \( \alpha \) that \( {L}_{\alpha }\left\lbrack \bar{A}\right\rbrack = {L}_{\alpha }\left\lbrack A\right\rbrack \) . The induction step is obvious if \( \alpha \) is a limit ordinal; thus assume that \( {L}_{\alpha }\left\lbrack \bar{A}\right\rbrack = {L}_{\alpha }\left\lbrack A\right\rbra... | Yes |
Lemma 13.25. There exists a definable well-ordering of the class \( {OD} \) (and a one-to-one definable mapping \( F \) of Ord onto OD). | Proof. Earlier we described how to construct from a given well-ordering of a set \( M \), a well-ordering of the set \( \operatorname{cl}\left( M\right) \) . For every \( \alpha \), the set \( \left\{ {{V}_{\beta } : \beta < \alpha }\right\} \) has an obvious well-ordering, which induces a well-ordering of \( \operator... | Yes |
Theorem 13.26. The class HOD is a transitive model of ZFC. | Proof. The class HOD is transitive, and it is easy to see that it is closed under Gödel operations. Thus to show that \( {HOD} \) is a model of ZF, it suffices to show that \( {HOD} \) is almost universal. For that, it is enough to verify that \( {V}_{\alpha } \cap {HOD} \in {HOD} \), for all \( \alpha \) . For any \( ... | Yes |
Theorem 13.27. Let \( M \) be an inner model of ZF. Then\n\n(i) \( {P}^{M}\left( X\right) = P\left( X\right) \cap M,{V}_{\alpha }^{M} = {V}_{\alpha } \cap M \) .\n\n(ii) If \( {\left| X\right| }^{M} = {\left| Y\right| }^{M} \) then \( \left| X\right| = \left| Y\right| \) .\n\n(iii) If \( \alpha \) is a cardinal then \(... | Concerning (vi), if \( \alpha < \kappa \), then since \( M \vDash \mathrm{{AC}} \), we must have either \( {\left( {2}^{\alpha }\right) }^{M} < \kappa \) or \( {\left( {2}^{\alpha }\right) }^{M} \geq \kappa \) and the latter is impossible since \( {2}^{\alpha } < \kappa \) . | Yes |
Theorem 13.28. Let \( M \) and \( N \) be transitive models of ZF and assume that the Axiom of Choice holds in \( M \) . If \( M \) and \( N \) have the same sets of ordinals, i.e., \( {P}^{M}\left( {\operatorname{Ord}}^{M}\right) = {P}^{N}\left( {\operatorname{Ord}}^{N}\right) \), then \( M = N \) . | Proof. We start with a rather trivial remark: \( M \) and \( N \) have the same sets of pairs of ordinals. To see this, use the absolute canonical one-to-one function \( \Gamma : {Ord} \times {Ord} \rightarrow {Ord} \) . If \( X \subset {Or}{d}^{2} \) and \( X \in M \), then \( \Gamma \left( X\right) \) is both in \( M... | Yes |
Let \( P \) be the following notion of forcing: The elements of \( P \) are finite \( 0 - 1 \) sequences \( \langle p\left( 0\right) ,\ldots, p\left( {n - 1}\right) \rangle \) and a condition \( p \) is stronger than \( q\left( {p < q}\right) \) if \( p \) extends \( q \) . Clearly, \( p \) and \( q \) are compatible i... | This example describes the simplest way of adjoining a new set of natural numbers to the ground model. A set \( A \subset \omega \) obtained this way is called a Cohen generic real. | Yes |
Lemma 14.4. If \( \left( {P, < }\right) \) is a partially ordered set and \( \mathcal{D} \) is a countable collection of dense subsets of \( P \), then there exists a \( \mathcal{D} \) -generic filter on \( P \) . In fact, for every \( p \in P \) there exists a \( \mathcal{D} \) -generic filter \( G \) on \( P \) such ... | Proof. Let \( {D}_{1},{D}_{2},\ldots \) be the sets in \( \mathcal{D} \) . Let \( {p}_{0} = p \), and for each \( n \), let \( {p}_{n} \) be such that \( {p}_{n} \leq {p}_{n - 1} \) and \( {p}_{n} \in {D}_{n} \) . The set\n\n\[ G = \left\{ {q \in P : q \geq {p}_{n}\text{ for some }n \in \mathbf{N}}\right\} \]\n\nis a \... | Yes |
Theorem 14.10. Let \( \\left( {P, < }\\right) \) be a separative partially ordered set. Then there is a complete algebra \( B \) such that:\n\n(i) \( P \\subset {B}^{ + } \) and \( < \) agrees with the partial ordering of \( B \) .\n\n(ii) \( P \) is dense in \( B \) .\n\nThe algebra \( B \) is unique up to isomorphism... | Proof. The proof is exactly the same as the proof of Theorem 7.13. \( B \) is the set of all regular cuts in \( P \) and separativity implies that every \( {U}_{p} \) (where \( p \\in P) \) is regular. | Yes |
Lemma 14.11. Let \( \left( {P, < }\right) \) be a partially ordered set. There exists a separative partially ordered set \( \left( {Q, \prec }\right) \) and a mapping \( h \) of \( P \) onto \( Q \) such that\n\n(14.5)\n\n(i) \( x \leq y \) implies \( h\left( x\right) \preccurlyeq h\left( y\right) \) ;\n\n(ii) \( x \) ... | Proof. Let us define the following equivalence relation on \( P \) :\n\n\( x \sim y \) if and only if \( \forall z \) ( \( z \) is compatible with \( \mathrm{x} \leftrightarrow z \) is compatible with \( \mathrm{y} \) ). Let \( Q = P/ \sim \) and let us define\n\n\[ \left\lbrack x\right\rbrack \preccurlyeq \left\lbrack... | Yes |
Lemma 14.13. (i) In the ground model \( M \), let \( Q \) be the separative quotient of \( P \) and let \( h \) map \( P \) onto \( Q \) such that (14.5) holds. If \( G \subset P \) is generic over \( M \) then \( h\left( G\right) \subset Q \) is generic over \( M \) . Conversely, if \( H \subset Q \) is generic over \... | Proof. The proof is an exercise in verifying definitions (Exercise 14.1 is useful here). | No |
Lemma 14.14. Let \( \mathfrak{A} \) be full. For any formula \( \varphi \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) ,\n\n\[ \mathfrak{A}/F \vDash \varphi \left( {\left\lbrack {a}_{1}\right\rbrack ,\ldots ,\left\lbrack {a}_{n}\right\rbrack }\right) \;\text{ if and only if }\begin{Vmatrix}{\varphi \left( {{a}_{1},\ldots ,... | Proof. (a) If \( \varphi \) is atomic, then (14.13) is true by definition.\n\n(b) If \( \varphi \) is a negation, conjunction, etc., we use the basic properties of an ultrafilter, and the definition of \( \parallel \parallel \) ; e.g., we use\n\n\[ \parallel \neg \psi \parallel \in F\;\text{ if and only if }\parallel \... | Yes |
Lemma 14.15. \( \parallel x = x\parallel = 1 \) for all \( x \in {V}^{B} \) . | Proof. By induction on \( \rho \left( x\right) \) . Clearly, it suffices to show that \( \parallel x \subset x\parallel = 1 \), i.e., we wish to show that \( x\left( t\right) \Rightarrow \parallel t \in x\parallel = 1 \) for all \( t \in \operatorname{dom}\left( x\right) \), or equivalently, that \( x\left( t\right) \l... | Yes |
Lemma 14.17. \( {V}^{B} \) is extensional. | Proof. Let \( X, Y \in {V}^{B} \) . By the definition of \( a \Rightarrow b \) we observe that if \( a \leq {a}^{\prime } \) , then \( \left( {{a}^{\prime } \Rightarrow b}\right) \leq \left( {a \Rightarrow b}\right) \) . Thus for any \( u \in {V}^{B} \) we have \( (\parallel u \in X\parallel \Rightarrow \parallel u \in... | Yes |
Lemma 14.18. If \( W \) is a set of pairwise disjoint elements of \( B \) and if \( {a}_{u} \) , \( u \in W \), are elements of \( {V}^{B} \), then there exists some \( a \in {V}^{B} \) such that \( u \leq \) \( \begin{Vmatrix}{a = {a}_{u}}\end{Vmatrix} \) for all \( u \in W \) . | Proof. Let \( D = \mathop{\bigcup }\limits_{{u \in W}}\operatorname{dom}\left( {a}_{u}\right) \), and for every \( t \in D \), let \( a\left( t\right) = \sum \left\{ {u \cdot {a}_{u}\left( t\right) }\right. \) : \( u \in W\} \) . Since the \( u \) ’s are pairwise disjoint, we have \( u \cdot a\left( t\right) = u \cdot ... | Yes |
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