Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Theorem 10.1. If \( G \) is \( \mathbf{P} \) -generic over \( M \) then \( M\left\lbrack G\right\rbrack \) is a standard transitive model of \( {ZF} \) that has the same order type as \( M \) . For any formula \( \varphi \) of \( {\mathcal{L}}_{0} \) | \[ M\left\lbrack G\right\rbrack \vDash \varphi \leftrightarrow h\left( \left\lbrack \varphi \right\rbrack \right) = 1 \] \[ \leftrightarrow \llbracket \varphi \rrbracket \in F \] \[ \leftrightarrow \left\lbrack \varphi \right\rbrack \cap G \neq 0. \] Furthermore if \( M \) satisfies the \( {AC} \) so does \( M\left\lbr... | Yes |
Theorem 10.4. Let \( k,{k}_{1},{k}_{2} \in V \) and \( t,{t}_{1},{t}_{2} \) be constant terms.\n\n1. \( p \Vdash \neg \varphi \leftrightarrow \left( {\forall q \leq p}\right) \neg \left( {q \Vdash \varphi }\right) \) . | Proof. The proofs of most of these statements are obvious from the definition:\n\n1. \( p \Vdash \neg \varphi \leftrightarrow p \in \llbracket \varphi \rrbracket \)\n\n\[ \leftrightarrow \left( {\forall q \leq p}\right) \left\lbrack {q \notin \llbracket \varphi \rrbracket }\right\rbrack \]\n\n\[ \leftrightarrow \left( ... | No |
Theorem 10.6. If \( \langle M,\mathbf{P}\rangle \) is a setting for forcing then\n\n\[ p \Vdash \varphi \leftrightarrow \left( {\forall {G}^{\prime }}\right) \left\lbrack {{G}^{\prime }\text{ is }\mathbf{P}\text{-generic over }M \land p \in {G}^{\prime } \rightarrow M\left\lbrack {G}^{\prime }\right\rbrack \vDash \varp... | Proof. Using the one-to-one correspondence between P-generic sets over \( M \) and \( M \) -complete homomorphisms from \( \mathbf{B} \) into 2 we need only show:\n\n\( p \Vdash \varphi \leftrightarrow \left( {\forall {h}^{\prime }}\right) \left\lbrack {{h}^{\prime } : \left| \mathbf{B}\right| \rightarrow \left| \mathb... | Yes |
Theorem 10.7. \( q \leq p \land p \Vdash \varphi \rightarrow q \Vdash \varphi \) . | Proof. Obvious from the definition. | No |
Theorem 10.9. If \( G \) is \( \mathbf{P} \) -generic over \( M \) and \( S = \{ p \in P \mid p \Vdash \varphi \} \) is dense then \( M\left\lbrack G\right\rbrack \vDash \varphi \) . | Proof. \( S = \llbracket \varphi \rrbracket \) is regular open in \( \mathbf{P} \) since \( \llbracket \varphi \rrbracket \in B \) . Then, since \( S \) is dense, \( S = {S}^{-0} = 1 \) . Therefore \( \llbracket \varphi \rrbracket = 1 \) . | No |
Theorem 10.11. If \( G \) is P-generic over \( M \), if \( p \in G \) and if \( S \in M \) is dense beneath \( p \) then \( G \cap S \neq 0 \) . | Proof. Under the given hypothesis if\n\n\[ \n{S}^{\prime } = S \cup \{ q \in P \mid \neg \operatorname{Comp}\left( {p, q}\right) \} \n\]\n\nthen \( {S}^{\prime } \in M \) and \( {S}^{\prime } \) is dense, hence \( G \cap {S}^{\prime } \neq 0 \) . But any two elements of \( G \) are compatible, hence \( G \cap S \neq 0 ... | Yes |
Theorem 10.12. If \( G \) is \( \mathbf{P} \)-generic over \( M \) and \( p \in G \) then\n\n\[ p \Vdash \left( {\exists x}\right) \varphi \left( x\right) \rightarrow \left( {\exists q \leq p}\right) \left( {\exists t \in T}\right) \left( {q \in G \land q \Vdash \varphi \left( t\right) }\right) . \] | Proof.\n\n\[ p \Vdash \left( {\exists x}\right) \varphi \left( x\right) \leftrightarrow p \in \mathop{\sum }\limits_{{t \in T}}\llbracket \varphi \left( t\right) \rrbracket = {\left( \mathop{\bigcup }\limits_{{t \in T}}\llbracket \varphi \left( t\right) \rrbracket \right) }^{-0} \]\n\n\[ \leftrightarrow \left\lbrack p\... | Yes |
Lemma 1. \( {a}_{1} \subseteq \omega \land {a}_{2} \subseteq \omega \land {a}_{1} \neq {a}_{2} \rightarrow \widetilde{G}\left( {a}_{1}\right) \neq \widetilde{G}\left( {a}_{2}\right) \) . | Proof. Without loss of generality we may assume\n\n\[ \left( {\exists n \in \omega }\right) \left\lbrack {n \in {a}_{1} \land n \notin {a}_{2}}\right\rbrack . \]\n\nThen \( \langle \{ n\} ,0\rangle \in \widetilde{G}\left( {a}_{1}\right) \) but \( \langle \{ n\} ,0\rangle \notin \widetilde{G}\left( {a}_{2}\right) \) . | Yes |
Lemma 2. If \( G \) is \( \mathbf{P} \) -generic over \( M \) then \( \widetilde{G}\left( {\widetilde{a}\left( G\right) }\right) = G \) . | Proof. If \( G \) is \( \mathbf{P} \) -generic over \( M \) and \( p = \left\langle {{p}_{1},{p}_{2}}\right\rangle \in G \) then \( {p}_{1} \subseteq \widetilde{a}\left( G\right) \) and \( {p}_{2} \subseteq \omega - \widetilde{a}\left( G\right) \). For if \( n \in {p}_{2} \) and \( n \in \widetilde{a}\left( G\right) \)... | Yes |
Lemma 3. If \( {G}_{1},{G}_{2} \) are each \( \mathbf{P} \) -generic over \( M \) then\n\n\[ \widetilde{a}\left( {G}_{1}\right) = \widetilde{a}\left( {G}_{2}\right) \leftrightarrow {G}_{1} = {G}_{2}. \] | Proof. Lemmas 1 and 2. | No |
Theorem 11.2. If \( a \in M \) and \( a \subseteq \omega \) then \( \widetilde{G}\left( a\right) \) is not \( \mathbf{P} \) -generic over \( M \) . | Proof. If \( a \in M \) and \( a \subseteq \omega \) then \( \widetilde{G}\left( a\right) \in M \) . If \( S = P - \widetilde{G}\left( a\right) \) then \( S \in M \) . For each \( p = \left\langle {{p}_{1},{p}_{2}}\right\rangle \in P \) there exists an \( n \in \omega \) such that \( n \notin {p}_{1} \) and \( n \notin... | Yes |
Theorem 11.3. If \( G \) is \( \mathbf{P} \) -generic over \( M \) then \( M\left\lbrack G\right\rbrack \) is a standard transitive model of \( {ZF} + {AC} + {GCH} + V \neq L \) . | Proof. If \( a = \widetilde{a}\left( G\right) \) then \( M\left\lbrack G\right\rbrack = M\left\lbrack a\right\rbrack \) and hence, by Theorem 11.2 and Lemma \( 2, a \notin M \) . Therefore \( M\left\lbrack a\right\rbrack \) is not a model of \( V = L \) .\n\nSince \( a \subseteq \omega \subseteq L,{L}_{a} = L\left\lbra... | No |
Theorem 11.5. \( {\pi }^{cc}G \) is \( \mathbf{P} \) -generic over \( M,{\widetilde{\pi }}^{cc}F \) is the \( M \) -complete ultrafilter for \( \mathbf{B} \) and \( h \circ {\widetilde{\pi }}^{-1} \) is the \( M \) -complete homomorphism of \( \mathbf{B} \) into 2 corresponding to \( {\pi }^{cc}G \) . Furthermore \( M\... | Proof. \( S \subseteq P \) is dense iff \( {\pi }^{\alpha }S \) is dense. Therefore\n\n\[ S \cap {\pi }^{ii}G \neq 0 \leftrightarrow {\pi }^{-1}\left( S\right) \cap G \neq 0.\]\n\nThus \( {\pi }^{\prime \prime }G \) is \( \mathbf{P} \) -generic over \( M \) . The ultrafilter corresponding to \( {\pi }^{\prime \prime }G... | Yes |
Theorem 11.7. If \( G \) is \( \mathbf{P} \) -generic over \( M \) then in \( M\left\lbrack G\right\rbrack \) there is no well-ordering of \( \mathcal{P}\left( \omega \right) \) definable in \( {\mathcal{L}}_{0}\left( {C\left( M\right) }\right) \) . | Proof. If \( \varphi \) is a formula of \( {\mathcal{L}}_{0}\left( {C\left( M\right) }\right) \) defining a well-ordering of \( \mathcal{P}\left( \omega \right) \) in \( M\left\lbrack G\right\rbrack \), then\n\n\[ \exists p \in G,\;p{ \Vdash }^{\prime \prime }\varphi \text{ well-orders }\mathcal{P}\left( \omega \right)... | No |
Theorem 11.8. If \( \langle M,\mathbf{P}\rangle \) is a setting for forcing, if \( M \) is a model of the \( {AC} \) that satisfies the countable chain condition on \( \mathbf{P} \) i.e., \[ \left( {\forall S \subseteq P}\right) \left\lbrack {\left( {\forall {p}_{1},{p}_{2} \in S}\right) \left\lbrack {{p}_{1} \neq {p}_... | Proof. Since \( M \subseteq M\left\lbrack G\right\rbrack \) and \( O{n}^{M} = O{n}^{M\left\lbrack G\right\rbrack } \), every cardinal in \( M\left\lbrack G\right\rbrack \) is a cardinal in \( M \) because \ | No |
If \( \langle M,\mathbf{P}\rangle \) is a setting for forcing, if \( G \) is \( \mathbf{P} \) -generic over \( M \), if \( M \) satisfies\n\n\[ \n\left( {\forall S \subseteq P}\right) \left\lbrack {\left( {\forall {p}_{1},{p}_{2} \in S}\right) \left\lbrack {{p}_{1} \neq {p}_{2} \rightarrow \neg \operatorname{Comp}\left... | We first show that \( \lambda \) is a cardinal in \( M\left\lbrack G\right\rbrack \) . Otherwise\n\n\[ \n\left( {\exists f \in M\left\lbrack G\right\rbrack }\right) \left( {\exists {\lambda }_{0} < \lambda }\right) \left\lbrack {f : {\lambda }_{0}\xrightarrow[]{\text{ onto }}\lambda }\right\rbrack .\n\]\n\nThen, as in ... | Yes |
Theorem 11.10. \( P \) satisfies the c.c.c. in \( M \) . | Proof. We show by induction on \( n \) that\n\n(i) \( S \subseteq P \land S \in M \land \left( {\forall {p}_{1},{p}_{2} \in S}\right) \left\lbrack {{p}_{1} \neq {p}_{2} \rightarrow \neg \operatorname{Comp}\left( {{p}_{1},{p}_{2}}\right) }\right\rbrack \)\n\n\[ \land \left( {\forall p \in S}\right) \left\lbrack {\overli... | Yes |
Theorem 12.3. For every formula \( \varphi \) , \[ \left( {\forall t \in T}\right) \left( {\forall p \in \mathop{\prod }\limits_{{i \in \omega }}^{w}{P}_{i}}\right) \left( {\exists m}\right) \left( {\forall \pi }\right) \left\lbrack {\pi \in {\mathbf{G}}_{m} \rightarrow \pi \left( t\right) = t \land \pi \left( \varphi ... | Proof. Take \( m \) to be the maximum of all \( i \in \omega \) for which \( {F}_{i} \) occurs in \( \varphi \) or \( t \) or \( {p}_{i} \) is not \( {1}_{i} \) . Then \( m \) has the required properties. | Yes |
In \( N, S \) is an infinite subset of \( P\left( \omega \right) \), and yet \( S \) contains no countable subset. In particular, \( P\left( \omega \right) \) is not well-ordered in \( N \) and hence the \( {AC} \) does not hold in \( N \) . | Recall that by Lemmas 2 and 3, pages 106-107,\n\n\[ \n{a}_{i} = {a}_{j} \leftrightarrow {F}_{i} = {F}_{j}\text{ for }i, j < \omega , \n\]\n\nand\n\n\[ \n\widetilde{G}\left( {a}_{i}\right) = {F}_{i}\;\text{ for }\;i \in \omega . \n\]\n\nFirst we prove that \( S \) is infinite by showing that\n\n1. \( i, j < \omega \land... | Yes |
Theorem 12.5. If \( G \) is \( \mathbf{P} \) -generic over \( M \) and\n\n\[ \n\left( {\forall S \subseteq P}\right) \left\lbrack {\overline{\bar{S}} \leq \omega \land \operatorname{Comp}\left( S\right) \rightarrow \left( {\exists p \in P}\right) \left( {\forall q \in S}\right) \left\lbrack {p \leq q}\right\rbrack }\ri... | Proof. Let \( {D}_{M\left\lbrack G\right\rbrack }\left( t\right) \) be an \( \omega \) -sequence of ordinals in \( M\left\lbrack G\right\rbrack \) . By Theorem 10.9, it suffices to show that \( \{ p \in P \mid p \Vdash V\left( t\right) \} \) is dense, i.e.,\n\n\[ \n\left( {\forall p \in P}\right) \left( {\exists q \leq... | No |
Theorem 12.6. If \( {G}_{1} \) is \( {\mathbf{P}}_{1} \) -generic over \( M \) and \( {G}_{2} \) is \( {\mathbf{P}}_{2} \) -generic over \( M\left\lbrack {G}_{1}\right\rbrack \), then \( {G}_{1} \times {G}_{2} \) is \( \mathbf{P} \) -generic over \( M \) . | Proof. Assume the hypothesis of the theorem and let \( S \) be an element of \( M \) that is dense in \( \mathbf{P} \) . Define\n\n\[ \n{S}_{2} \triangleq \left\{ {{p}_{2} \in {P}_{2} \mid \left( {{G}_{1} \times \left\{ {p}_{2}\right\} }\right) \cap S \neq 0}\right\} .\n\]\n118 Claim: \( {S}_{2} \) is dense in \( {P}_{... | Yes |
Theorem 12.7. If \( G \) is \( \mathbf{P} \) -generic over \( M \), then there exists a \( {G}_{1} \) which is \( {\mathbf{P}}_{1} \) -generic over \( M \) and a \( {G}_{2} \) which is \( {\mathbf{P}}_{2} \) -generic over \( M\left\lbrack {G}_{1}\right\rbrack \) such that \( G = {G}_{1} \times {G}_{2} \) | Proof. Let \( G \) be \( \mathbf{P} \) -generic over \( M \) and define\n\n\[ \n{G}_{1} \triangleq \left\{ {{p}_{1} \in {P}_{1} \mid \left\langle {{p}_{1},1}\right\rangle \in G}\right\} \n\]\n\n\[ \n{G}_{2} \triangleq \left\{ {{p}_{2} \in {P}_{2} \mid \left\langle {1,{p}_{2}}\right\rangle \in G}\right\} \n\]\n\nThen \(... | Yes |
Theorem 13.4. For \( u, v \in {V}^{\left( \mathbf{B}\right) } \), 1. \( \llbracket u = v\rrbracket = \llbracket v = u\rrbracket \). 2. \( \llbracket u = u\rrbracket = 1 \). 3. \( x \in \mathcal{D}\left( u\right) \rightarrow u\left( x\right) \leq \llbracket x \in u\rrbracket \). | Proof. 1. The definition of \( \llbracket u = v\rrbracket \) is symmetric in \( u \) and \( v \). 2 and 3 are proved by induction on rank \( \left( u\right) \). Let \( x \in \mathcal{D}\left( u\right) \). Then \[ \llbracket x \in u\rrbracket = \mathop{\sum }\limits_{{y \in \mathcal{D}\left( u\right) }}u\left( y\right) ... | Yes |
Theorem 13.5. Let \( u,{u}^{\prime }, v,{v}^{\prime }, w \in {V}^{\left( \mathbf{B}\right) } \) . Then for each \( \alpha \n\n1. \( \left\lbrack {\operatorname{rank}\left( u\right) < \alpha }\right\rbrack \land \left\lbrack {\operatorname{rank}\left( {u}^{\prime }\right) < \alpha }\right\rbrack \land \left\lbrack {\ope... | Proof. (By induction on \( \alpha \) ).\n\n1. If \( \operatorname{rank}\left( u\right) < \alpha ,\operatorname{rank}\left( {u}^{\prime }\right) < \alpha \), and \( \operatorname{rank}\left( v\right) \leq \alpha \), then\n\n\[ \n\llbracket u = {u}^{\prime }\rrbracket \cdot \llbracket u \in v\rrbracket = \mathop{\sum }\l... | Yes |
Theorem 13.9. \( {V}_{\alpha }^{\left( \mathbf{B}\right) } \) satisfies the Axiom of Extensionality. | Proof. Let \( u, v \in {V}_{\alpha }^{\left( \mathbf{B}\right) } \) . Then\n\n\[ \llbracket \left( {\forall x}\right) \left\lbrack {x \in u \leftrightarrow x \in v}\right\rbrack {\rrbracket }_{\alpha } = \mathop{\prod }\limits_{{x \in {V}_{\alpha }\left( \mathbf{B}\right) }}\llbracket x \in u \rightarrow x \in v{\rrbra... | Yes |
Theorem 13.10. If \( u \in {V}_{\alpha + 1}^{\left( \mathrm{B}\right) } \) then \( u \) is defined over \( {V}_{\alpha }^{\left( \mathrm{B}\right) } \), i.e., | Proof. Let \( u \) and \( v \) be in \( {V}_{\alpha + 1}^{\left( \mathrm{B}\right) } \) . \n\n\[ \llbracket v \in u\rrbracket = \mathop{\sum }\limits_{{x \in \mathcal{D}\left( u\right) }}u\left( x\right) \cdot \llbracket v = x\rrbracket \]\n\n\[ \leq \mathop{\sum }\limits_{{x \in \mathcal{D}\left( u\right) }}\llbracket... | Yes |
Theorem 13.11. For every formula \( \varphi \) of \( {\mathcal{L}}_{0} \) , \[ \left( {\forall {a}_{1},\ldots ,{a}_{n} \in {V}_{\alpha }^{\left( \mathbf{B}\right) }}\right) \left( {\exists b \in {V}_{\alpha + 1}^{\left( \mathbf{B}\right) }}\right) \left( {\forall a \in {V}_{\alpha }^{\left( \mathbf{B}\right) }}\right) ... | Proof. Let \( {a}_{1},\ldots ,{a}_{n} \in {V}_{\alpha }^{\left( \mathbf{B}\right) } \) and define \( b : {V}_{\alpha }^{\left( \mathbf{B}\right) } \rightarrow B \) by \[ b\left( a\right) = {\left\lbrack \varphi \left( a,{a}_{1},\ldots ,{a}_{n}\right) \right\rbrack }_{\alpha }\;\text{ for }\;a \in {V}_{\alpha }^{\left( ... | Yes |
Theorem 13.13. For \( u \in {V}^{\left( \mathbf{B}\right) } \) , 1. \( \llbracket \left( {\exists x \in u}\right) \varphi \left( x\right) \rrbracket = \mathop{\sum }\limits_{{x \in \mathcal{D}\left( u\right) }}u\left( x\right) \cdot \llbracket \varphi \left( x\right) \rrbracket \) . 2. \( \llbracket \left( {\forall x \... | Proof. For \( u \in {V}^{\left( \mathbf{B}\right) } \) , \[ \llbracket \left( {\exists x \in u}\right) \varphi \left( x\right) \rrbracket = \mathop{\sum }\limits_{{{x}^{\prime } \in {V}^{\left( \mathrm{B}\right) }}}\llbracket {x}^{\prime } \in u\rrbracket \cdot \llbracket \varphi \left( {x}^{\prime }\right) \rrbracket ... | Yes |
Theorem 13.15. Let \( \mathbf{B} \) be a complete subalgebra of the complete Boolean algebra \( {\mathbf{B}}^{\prime } \) . Then\n\n1. \( {V}^{\left( \mathbf{B}\right) } \subseteq {V}^{\left( {\mathbf{B}}^{\prime }\right) } \) .\n\n2. \( u, v \in {V}^{\left( \mathbf{B}\right) } \rightarrow \llbracket u \in v{\rrbracket... | Proof. (By induction)\n\n1. Obvious, since any function into \( B \) is also a function into \( {B}^{\prime } \) .\n\n2. Follows from the fact that \( \Pi \) and \( \sum \), over values in \( \mathbf{B} \), are the same in \( \mathbf{B} \) and \( {\mathbf{B}}^{\prime } \) respectively. | No |
Theorem 13.17. For \( x, y \in V \) , 1. \( x \in y \leftrightarrow \llbracket \check{x} \in \check{y}\rrbracket = \mathbf{1} \land x \notin y \leftrightarrow \llbracket \check{x} \in \check{y}\rrbracket = \mathbf{0} \) , 2. \( x = y \leftrightarrow \llbracket \check{x} = \check{y}\rrbracket = \mathbf{1} \land x \neq y... | Proof. 1 and 2 are proved simultaneously by induction from Definition 13.3. Proving this is in fact a very good exercise that we leave to the reader. In order to prove 3, let \( u \in {V}^{\left( 2\right) } \) and assume as induction hypothesis (i) \( \left( {\forall x \in \mathcal{D}\left( u\right) }\right) \left( {\e... | No |
Theorem 13.18. Let B be a complete subalgebra of the complete Boolean algebra \( {\mathbf{B}}^{\prime } \) and \( \varphi \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) be a formula in which every quantifier is bounded (i.e., of the form \( \exists x \in y \) or \( \forall x \in y \) ). Then for \( {u}_{1},\ldots ,{u}_{n} ... | Proof. (By induction on the number of logical symbols in \( \varphi \) .) If \( \varphi \) is atomic, (i) is true by Theorem 13.15. The only nontrivial case is \[ \varphi \left( {u,{u}_{1},\ldots ,{u}_{n}}\right) = \left( {\exists x \in u}\right) \psi \left( {x, u,{u}_{1},\ldots ,{u}_{n}}\right) . \] Then for \( u,{u}_... | Yes |
Corollary 13.19. If \( \phi \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) is a bounded formula (i.e., a formula containing only bounded quantifiers), then for \( {u}_{1},\ldots ,{u}_{n} \in V \) , \[ \phi \left( {{u}_{1},\ldots ,{u}_{n}}\right) \leftrightarrow {\left\lbrack \phi \left( {\breve{u}}_{1},\ldots ,{\breve{u}}_... | Proof. Apply Theorem 13.18 to the Boolean algebra 2 which is a complete subalgebra of each Boolean algebra B and use Theorem 13.17. | No |
Theorem 13.20. \( {\mathbf{V}}^{\left( \mathbf{B}\right) } \) satisfies the Axiom of Infinity. | Proof. We have \( \breve{\omega } \in {V}^{\left( 2\right) } \subseteq {V}^{\left( \mathbf{B}\right) } \) and\n\n\[ \left( {\exists x}\right) \left\lbrack {x \in \omega }\right\rbrack \land \left( {\forall x \in \omega }\right) \left( {\exists y \in \omega }\right) \left\lbrack {x \in y}\right\rbrack \]\n\nis a bounded... | Yes |
Theorem 13.22. For \( u \in {V}^{\left( \mathbf{B}\right) },\llbracket \operatorname{Ord}\left( u\right) \rrbracket = \mathop{\sum }\limits_{{\alpha \in {On}}}\llbracket u = \alpha \rrbracket \) . | Proof. \( \llbracket u = \check{\alpha }\rrbracket = \llbracket u = \check{\alpha }\rrbracket \cdot \llbracket \operatorname{Ord}\left( \check{\alpha }\right) \rrbracket \leq \llbracket \operatorname{Ord}\left( u\right) \rrbracket \) by Corollary 13.7. Therefore \( \mathop{\sum }\limits_{{\alpha \in {On}}}\llbracket u ... | Yes |
1. \( \llbracket \left( {\exists u}\right) \left\lbrack {\operatorname{Ord}\left( u\right) \land \phi \left( u\right) }\right\rbrack \rrbracket = \mathop{\sum }\limits_{{\alpha \in {On}}}\llbracket \phi \left( \breve{\alpha }\right) \rrbracket \) . | Proof.\n\n\[ \llbracket \left( {\exists u}\right) \left\lbrack {\mathrm{{Ord}}\left( u\right) \; \land \;\phi \left( u\right) }\right\rbrack \rrbracket = \mathop{\sum }\limits_{{u \in {V}^{\left( \mathrm{B}\right) }}}\llbracket \mathrm{{Ord}}\left( u\right) \rrbracket \cdot \llbracket \phi \left( u\right) \rrbracket \]... | Yes |
Theorem 14.1. \( u, v \in {V}_{\alpha }^{\left( \mathbf{B}\right) } \rightarrow \llbracket u = v\rrbracket \cdot \llbracket M\left( u\right) {\rrbracket }_{\alpha } \leq \llbracket M\left( v\right) {\rrbracket }_{\alpha } \) . | Proof. For \( u, v \in {V}_{\alpha }^{\left( \mathbf{B}\right) } \) , \n\n\[ \llbracket u = v\rrbracket \cdot \llbracket M\left( u\right) {\rrbracket }_{\alpha } = \mathop{\sum }\limits_{{k \in R\left( \alpha \right) }}\llbracket u = v\rrbracket \llbracket u = \check{k}\rrbracket \]\n\n\[ \leq \mathop{\sum }\limits_{{k... | Yes |
Theorem 14.2. Let \( u \in {V}_{\alpha + 1}^{\left( \mathrm{B}\right) } \) and \( k \in V \) . Then\n\n1. \( \alpha \leq \operatorname{rank}\left( k\right) \rightarrow \llbracket \check{k} \in u\rrbracket = \mathbf{0} \) .\n\n2. \( \alpha < \operatorname{rank}\left( k\right) \rightarrow \llbracket \check{k} = u\rrbrack... | Proof. (By induction on \( \alpha \) .)\n\n1. Let \( \alpha \leq \operatorname{rank}\left( k\right) \) . Since \( \mathcal{D}\left( u\right) \subseteq {V}_{\alpha }^{\left( \mathbf{B}\right) }\n\n\[ \llbracket \check{k} \in u\rrbracket = \mathop{\sum }\limits_{{x \in \mathcal{D}\left( u\right) }}u\left( x\right) \cdot ... | Yes |
Theorem 14.6. If \( \\mathbf{B} \) is nonatomic and \( S \\subseteq B = \\left| \\mathbf{B}\\right| \), then\n\n\[ \n\\mathop{\\prod }\\limits_{{b \\in S}}b \\cdot \\mathop{\\prod }\\limits_{{b \\in B - S}}\\left( {-b}\\right) = \\mathbf{0}.\n\] | Proof. Suppose \( \\mathop{\\prod }\\limits_{{b \\in S}}b \\cdot \\mathop{\\prod }\\limits_{{b \\in B - S}}\\left( {-b}\\right) \\neq \\mathbf{0} \) . Then\n\n\[ \n\\left( {\\forall b \\in B}\\right) \\neg \\left\\lbrack {b \\in S \\land \\neg b \\in S}\\right\\rbrack\n\]\n\nand\n\n\[ \n\\left( {\\forall b \\in B}\\rig... | Yes |
Theorem 14.7. If \( {b}_{0} \) is an atom in \( \mathbf{B} \) and \( S = \left\{ {b \in B \mid {b}_{0} \leq b}\right\} \), then \( \llbracket F = S\rrbracket = {b}_{0}. \) | Proof. From the definition of \( S \) and the fact that \( {b}_{0} \) is an atom it is easily shown that\n\n\[ b \in S \leftrightarrow - b \in B - S.\]\n\nSince, by Theorem 13.16, \( \llbracket \check{x} = \check{b}\rrbracket = \mathbf{0} \) unless \( x = b \) it follows from Theorem 13.2 that \( \llbracket \check{b} \... | Yes |
Theorem 14.8. \( \llbracket F \subseteq \breve{B}\rrbracket = 1 \) . | Proof. \( \llbracket F \subseteq \check{B}\rrbracket = \mathop{\prod }\limits_{{b \in B}}\left( {F\left( \check{b}\right) \Rightarrow \llbracket \check{b} \in \check{B}\rrbracket }\right) = \mathbf{1} \) . | Yes |
Theorem 14.9. If \( \mathbf{B} \) is nonatomic, then \( \llbracket M\left( F\right) \rrbracket = \mathbf{0} \) . | Proof. \( \llbracket M\left( F\right) \rrbracket = \mathop{\sum }\limits_{{k \in V}}\llbracket F = \check{k}\rrbracket \)\n\n\( \llbracket F = \check{k}\rrbracket \neq \mathbf{0} \rightarrow \mathbf{0} < \llbracket F = \check{k}\rrbracket \cdot \llbracket F \subseteq \check{B}\rrbracket \; \) by Theorem 14.8\n\n\[ \rig... | Yes |
Theorem 14.10. \( M \) is transitive in \( {V}^{\left( \mathbf{B}\right) } \), i.e., | Proof. Let \( u \in {V}^{\left( \mathbf{B}\right) } \) . Because of Theorem 13.13.2 it suffices to show that\ni.e.,\n(i) \( \left( {\forall x \in \mathcal{D}\left( u\right) }\right) \left\lbrack {u\left( x\right) \cdot \llbracket M\left( u\right) \rrbracket \leq \llbracket M\left( x\right) \rrbracket }\right\rbrack \) ... | No |
Theorem 14.11. \( \\left( {\\forall x \\in {V}^{\\left( \\mathbf{B}\\right) }}\\right) \\left( {\\exists k \\in V}\\right) \\left\\lbrack {\\llbracket M\\left( x\\right) \\rrbracket \\leq \\llbracket x \\in \\check{k}\\rrbracket }\\right\\rbrack \) . | Proof. Let \( x \\in {V}^{\\left( \\mathbf{B}\\right) } \) . Then \( x \\in {V}_{\\alpha + 1}^{\\left( \\mathbf{B}\\right) } \) for some \( \\alpha \) . Choose \( k = R\\left( {\\alpha + 1}\\right) \) . Then\n\n\( \\llbracket M\\left( x\\right) \\rrbracket = \\mathop{\\sum }\\limits_{{{k}_{0} \\in V}}\\llbracket x = {\... | Yes |
Theorem 14.12. \( M \) is almost universal in \( {V}^{\left( \mathbf{B}\right) } \), i.e., \[ \left( {\forall u \in {V}^{\left( \mathbf{B}\right) }}\right) \left\lbrack {\llbracket u \subseteq M \rightarrow \left( {\exists y \in M}\right) \left\lbrack {u \subseteq y}\right\rbrack \rrbracket = 1}\right\rbrack . | Proof. Let \( u \in {V}^{\left( \mathbb{B}\right) } \) . By Theorem 14.11, for each \( x \in \mathcal{D}\left( u\right) \) there exists a \( {k}_{x} \) such that \[ \llbracket M\left( x\right) \rrbracket \leq \llbracket x \in {\check{k}}_{x}\rrbracket . \] (From the proof of Theorem 14.11 we see that we can take \[ {k}... | Yes |
Theorem 14.14. For \( u, v \in {V}^{\left( \mathbf{B}\right) } \), 1. \( \llbracket \{ u, v{\} }^{\left( \mathbf{B}\right) } = \{ u, v\} \rrbracket = \mathbf{1} \). 2. \( \llbracket \{ u{\} }^{\left( \mathbf{B}\right) } = \{ u\} \rrbracket = \mathbf{1} \). 3. \( \llbracket \langle u, v{\rangle }^{\left( \mathbf{B}\righ... | Proof. 1. It is sufficient to prove \[ \llbracket \left( {\forall x}\right) \left\lbrack {x = u \vee x = v \leftrightarrow x \in \{ u, v{\} }^{\left( \mathbf{B}\right) }}\right\rbrack \rrbracket = \mathbf{1}. \] But this follows from the fact that for all \( x \in {V}^{\left( \mathbf{B}\right) } \) \[ \llbracket x \in ... | No |
Theorem 14.15. \( {\left\{ {\check{k}}_{1},{\check{k}}_{2}\right\} }^{\left( \mathbf{B}\right) } = {\left\{ {k}_{1},{k}_{2}\right\} }^{ \smile } \) and hence\n\n\[ \llbracket \left\{ {{\check{k}}_{1},{\check{k}}_{2}}\right\} = \left\{ {{k}_{1},{k}_{2}}\right\} {}^{ \sim }\rrbracket = \mathbf{1}. \] | Proof. Obvious from the definitions and Theorem 14.14.1. | No |
1. \( \llbracket \left( {\exists x}\right) \left( {\exists y}\right) \left\lbrack {\{ x, y{\} }^{\left( \mathrm{B}\right) } = {\check{k}}_{1} \land \varphi \left( {x, y}\right) }\right\rbrack \rrbracket = \mathop{\sum }\limits_{{\left\{ {{k}_{2},{k}_{3}}\right\} = {k}_{1}}}\llbracket \varphi \left( {{\check{k}}_{2},{\c... | 1. In \( {ZF} \) we have\n\n\[ \left( {\exists x}\right) \left( {\exists y}\right) \left\lbrack {\{ x, y\} = {\check{k}}_{1} \land \varphi \left( {x, y}\right) }\right\rbrack \leftrightarrow \left( {\exists x \in {\check{k}}_{1}}\right) \left( {\exists y \in {\check{k}}_{1}}\right) \left\lbrack {\{ x, y\} = {\check{k}}... | Yes |
Theorem 14.18. \( \\llbracket {On} \\subseteq M\\rrbracket = 1 \) . | Proof. Let \( u \\in {V}^{\\left( \\mathbf{B}\\right) } \) . Then\n\n\[ \n\\llbracket \\operatorname{Ord}\\left( u\\right) \\rrbracket = \\mathop{\\sum }\\limits_{{\\alpha \\in {On}}}\\llbracket u = \\alpha \\rrbracket \\;\\text{ by Theorem 13.22 }\n\]\n\n\[ \n\\leq \\mathop{\\sum }\\limits_{{k \\in V}}\\llbracket u = ... | Yes |
Theorem 14.19. \( \left\lbrack {{\mathcal{F}}_{i}\left( {{\check{k}}_{1},{\check{k}}_{2}}\right) = {\mathcal{F}}_{i}{\left( {k}_{1},{k}_{2}\right) }^{ \vee }}\right\rbrack = \mathbf{1} \) for \( i = 1,\ldots ,8 \) . | Proof. By Theorem 14.15 (for \( i = 1 \) ) and the following lemmas. | No |
Lemma 1. \( \left\lbrack {{\check{k}}_{1} \cap E = {\left( {k}_{1} \cap E\right) }^{ \smile }}\right\rbrack = \mathbf{1} \) where \( E = \{ \langle x, y\rangle \mid x \in y\} \) | Proof. Let \( u \in {V}^{\left( \mathbf{B}\right) } \) . Then\n\n\[ \llbracket u \in \left( {{\check{k}}_{1} \cap E}\right) \rrbracket = \llbracket \left( {\exists x}\right) \left( {\exists y}\right) \left\lbrack {\langle x, y\rangle \in {\check{k}}_{1}\land \langle x, y\rangle = u \land x \in y}\right\rbrack \rrbracke... | Yes |
Lemma 2. \( \llbracket \left( {{\check{k}}_{1} - {\check{k}}_{2}}\right) = {\left( {k}_{1} - {k}_{2}\right) }^{ \smile }\rrbracket = \mathbf{1} \) | Proof.\n\n\[ \n{\check{k}}_{1} - {\check{k}}_{2} = \left\{ {\langle \check{k},\mathbf{1}\rangle \mid k \in {k}_{1}}\right\} - \left\{ {\langle \check{k},\mathbf{1}\rangle \mid k \in {k}_{2}}\right\} \n\] \n\n\[ \n= \left\{ {\langle \check{k},1\rangle \mid k \in {k}_{1} - {k}_{2}}\right\} = {\left( {k}_{1} - {k}_{2}\rig... | Yes |
Lemma 3. \( \left\lbrack {\left( {{\check{k}}_{1} \sqcap {\check{k}}_{2}}\right) = {\left( {k}_{1} \sqcap {k}_{2}\right) }^{ \smile }}\right\rbrack = 1 \) . | Proof. For \( u \in {V}^{\left( \mathbf{B}\right) } \) ,\n\n\[ \llbracket u \in \left( {{\check{k}}_{1} \vdash {\check{k}}_{2}}\right) \rrbracket = \llbracket \left( {\exists x}\right) \left( {\exists y}\right) \left\lbrack {\langle x, y\rangle \in {\check{k}}_{1} \land u = \langle x, y\rangle \land x \in {\check{k}}_{... | Yes |
Lemma 4. \( \left\lbrack {{\breve{k}}_{1} \cap {\breve{k}}_{2} = {\left( {k}_{1} \cap {k}_{2}\right) }^{ \vee }}\right\rbrack = 1 \) . | Proof.\n\n\[ \n{\check{k}}_{1} \cap {\check{k}}_{2} = \left\{ {\langle \check{k},\mathbf{1}\rangle \mid k \in {k}_{1}}\right\} \cap \left\{ {\langle \check{k},\mathbf{1}\rangle \mid k \in {k}_{2}}\right\} \n\] \n\n\[ \n= \left\{ {\langle \check{k},1\rangle \mid k \in {k}_{1} \cap {k}_{2}}\right\} \n\] \n\n\[ \n= {\left... | Yes |
Lemma 5. \( \left\lbrack {\mathcal{D}\left( {\check{k}}_{1}\right) = \mathcal{D}{\left( {k}_{1}\right) }^{ \smile }}\right\rbrack = 1 \) . | Proof. Let \( u \in {V}^{\left( \mathbf{B}\right) } \) . \n\n\[ \llbracket u \in \mathcal{D}\left( {\check{k}}_{1}\right) \rrbracket = \llbracket \left( {\exists x}\right) \left( {\exists y}\right) \left\lbrack {\langle x, y\rangle \in {\check{k}}_{1} \land u = x}\right\rbrack \rrbracket \]\n\n\[ = \mathop{\sum }\limit... | Yes |
Lemma 6. \( \left\lbrack {{\left( {\check{k}}_{1}\right) }^{-1} = {\left( {k}_{1}{}^{-1}\right) }^{ \smile }}\right\rbrack = 1 \) . | Proof. For \( u \in {V}^{\left( \mathbf{B}\right) } \) , \n\n\[ \llbracket u \in {\left( {\check{k}}_{1}\right) }^{-1}\rrbracket = \llbracket \left( {\exists x}\right) \left( {\exists y}\right) \left\lbrack {\langle x, y\rangle \in {\check{k}}_{1} \land u = \langle y, x\rangle }\right\rbrack \rrbracket \]\n\n\[ = \math... | Yes |
Lemma 7. \[\left\lbrack \left\{ {\langle x, y, z\rangle \mid \langle x, z, y\rangle \in {\check{k}}_{1}\} = \left\{ {\langle x, y, z\rangle \mid \langle x, z, y\rangle \in {k}_{1}}\right\} }\right\} \right\rbrack = \mathbf{1}.\] | Proof. (By the same method as before.) Let \( u \in {V}^{\left( \mathbf{B}\right) } \) . Then\n\n\[ \left\lbrack \left\lbrack {u \in \{ \langle x, y, z\rangle \mid \langle x, z, y\rangle \in {\check{k}}_{1}\} }\right\rbrack \right\rbrack \]\n\n\[ = \llbracket \left( {\exists x, w}\right) \left\lbrack {\langle x, w\rang... | Yes |
Lemma 8. \( \llbracket {\mathcal{F}}_{8}\left( {{\check{k}}_{1},{\check{k}}_{2}}\right) = {\mathcal{F}}_{8}{\left( {k}_{1},{k}_{2}\right) }^{ \smile }\rrbracket = \mathbf{1} \) . | Remark. This completes the proof of Theorem 14.19. | No |
Theorem 14.20. For \( i = 1,\ldots ,8 \) , \n\n\[ \n\left( {\forall u, v \in {V}^{\left( \mathbf{B}\right) }}\right) \left\lbrack {\llbracket M\left( u\right) \land M\left( v\right) \rightarrow M\left( {{\mathcal{F}}_{i}\left( {u, v}\right) }\right) \rrbracket = \mathbf{1}}\right\rbrack . \n\] | Proof. Let \( u, v \in {V}^{\left( \mathbf{B}\right) } \) . Then \n\n\[ \n\llbracket u = {\check{k}}_{1}\rrbracket \cdot \llbracket u = {\check{k}}_{2}\rrbracket \leq \llbracket {\mathcal{F}}_{i}\left( {u, v}\right) = {\mathcal{F}}_{i}\left( {{\check{k}}_{1},{\check{k}}_{2}}\right) \rrbracket \;\text{by Corollary 13.7}... | Yes |
Theorem 14.21. In \( {\mathbf{V}}^{\left( \mathbf{B}\right) }, M \) is a standard transitive model of \( {ZF} \) containing all the ordinals, i.e., | \[ \llbracket \left( {\forall u, v}\right) \left\lbrack {M\left( u\right) \land v \in u \rightarrow M\left( v\right) }\right\rbrack \rrbracket = \mathbf{1} \] \[ \llbracket \left( {\forall u}\right) \left\lbrack {\operatorname{Ord}\left( u\right) \rightarrow M\left( u\right) }\right\rbrack \rrbracket = \mathbf{1}. \] \... | Yes |
Theorem 14.22. Suppose that \( N \) is a countable standard transitive model of \( {ZF} + {AC} \) such that \( \mathbf{B} \in N \) . Let \( \mathbf{P} \) be the partial order structure associated with \( \mathbf{B} \) (thus \( \langle N,\mathbf{P}\rangle \) is a setting for forcing). Then for any set \( {G}_{0} \) whic... | Proof. Given \( {G}_{0} \) which is \( \mathbf{P} \) -generic over \( N \) and \( {h}_{0} : \left| \mathbf{B}\right| \rightarrow \left| \mathbf{2}\right| \) where \( {h}_{0} \) is associated with \( {G}_{0} \) in the familiar way, define \( h : {\left( {V}^{\left( \mathbf{B}\right) }\right) }^{N} \rightarrow V \) by in... | Yes |
Corollary 14.23. Suppose that \( \langle N, \in ,\mathbf{B}\rangle \) is elementarily equivalent to \( \langle V, \in ,\mathbf{B}\rangle \) where \( N \) is transitive, countable and \( \mathbf{B} \in N \) . Let \( \mathbf{P} \) be the partial order structure related to \( \mathbf{B} \) . Then for every sentence \( \va... | Proof. If \( \llbracket \varphi \rrbracket = \mathbf{1} \) in \( {\mathbf{V}}^{\left( \mathbf{B}\right) },\llbracket \varphi \rrbracket = \mathbf{1} \) in \( {\left( {\mathbf{V}}^{\left( \mathbf{B}\right) }\right) }^{N} \) and the conclusion follows from (i) of Theorem 14.22. Conversely, if \( \llbracket \varphi \rrbra... | Yes |
Theorem 14.24. \( \llbracket V = M\left\lbrack F\right\rbrack \rrbracket = 1 \). | Proof. Let \( \mathbf{N} = \langle N,\epsilon \rangle \) be a countable transitive model of \( {ZF} + {AC} \) such that \( \mathbf{B} \in N \) and\n\n\( \langle N, \in ,\mathbf{B}\rangle \) is elementarily equivalent to \( \langle V, \in ,\mathbf{B}\rangle \) .\n\n(The existence of such an \( \mathbf{N} \) can be prove... | Yes |
Theorem 14.25. (Using the Axiom of Choice in \( V \) .) \( \llbracket {AC}\rrbracket = 1 \), i.e., the \( {AC} \) holds in \( {\mathbf{V}}^{\left( \mathbf{B}\right) } \) . | Proof. Choose \( N \) and \( \mathbf{B} \) as in Corollary 14.23 (again we need the system \( {GB} + \) Mathematical Induction). Then \( N \) satisfies the \( {AC} \), and so does \( N\left\lbrack G\right\rbrack \) for every \( G \) which is \( \mathbf{P} \) -generic over \( N \) . Hence \( \llbracket {AC}\rrbracket = ... | Yes |
Theorem 15.2. If \( u, v \in {V}^{\left( \mathbf{B}\right) } \), if \( \mathcal{D}\left( u\right) \subseteq S \) and \( \mathcal{D}\left( v\right) \subseteq S \) where \( S \subseteq {V}^{\left( \mathbf{B}\right) } \) , then \( \llbracket u = v\rrbracket = \mathop{\prod }\limits_{{w \in S}}\llbracket w \in u \leftright... | Proof. \( \llbracket u = v\rrbracket \leq \mathop{\prod }\limits_{{w \in S}}\llbracket w \in u \cdot \cdot w \in v\rrbracket \) follows from the Axioms of Equality. On the other hand,\n\n\[ \mathop{\prod }\limits_{{w \in S}}\llbracket w \in u \rightarrow w \in v\rrbracket \leq \mathop{\prod }\limits_{{w \in \mathcal{I}... | Yes |
Theorem 15.3. \( t \in {T}_{\alpha } \rightarrow j\left( t\right) \in {V}_{\alpha }^{\left( \mathbf{B}\right) } \) | Proof. (By induction on \( \alpha \) .) If \( t = \underline{k} \) and \( \underline{k} \in {T}_{\alpha } \), then \( k \in R\left( \alpha \right) \), i.e., \( k \in R\left( {\beta + 1}\right) \) for some \( \beta < \alpha \) . Therefore\n\n\[ j\left( \underline{k}\right) = \check{k} = \left\{ {\left\langle {{\check{k}... | Yes |
Theorem 15.4. If \( {t}_{1},{t}_{2} \) are constant terms and \( \varphi \) is a limited formula, then \( \llbracket \varphi \rrbracket = {D}_{j}\left( \varphi \right) \) . In particular,\n\n1. \( \llbracket {t}_{1} = {t}_{2}\rrbracket = \llbracket j\left( {t}_{1}\right) = j\left( {t}_{2}\right) \rrbracket \) .\n\n2. \... | Proof. (By induction on Ord ( \( \varphi \) ).)\n\n1. Let \( \beta = \max \left( {\rho \left( {t}_{1}\right) ,\rho \left( {t}_{2}\right) }\right) \) . Then\n\n\[ \llbracket {t}_{1} = {t}_{2}\rrbracket = \mathop{\prod }\limits_{{t \in {T}_{\beta }}}\llbracket t \in {t}_{1} \leftrightarrow t \in {t}_{2}\rrbracket \;\text... | Yes |
Corollary 15.5. \( \left( {\exists \alpha }\right) \left\lbrack {\int \left( {{\widehat{x}}_{n}{}^{\alpha }F\left( {{x}_{n}{}^{\alpha }}\right) }\right) = F}\right\rbrack = 1 \) . | Proof. Choose \( \alpha = \operatorname{rank}\left( B\right) \) and use 5. | Yes |
Theorem 15.6. Under the assumptions of Theorem 14.22, \( h\left( {j\left( t\right) }\right) = \) \( {D}_{N\left\lbrack {G}_{0}\right\rbrack }\left( t\right) \) for each constant term \( t \) in the relative sense of \( N \) . (Cp. the relativization of \( V\left\lbrack {F}_{0}\right\rbrack \) to \( N \) discussed on pa... | Proof. (By induction on \( \rho \left( t\right) \) .)\n\n\[ h\left( {j\left( {t}_{1}\right) }\right) \in h\left( {j\left( t\right) }\right) \leftrightarrow {h}_{0}\left( \left\lbrack \left\lbrack {j\left( {t}_{1}\right) \in j\left( t\right) }\right\rbrack \right\rbrack \right) = 1\;\text{by Theorem 14.22} \]\n\n\[ \rig... | Yes |
Theorem 15.8. The mapping \( j : V\left\lbrack {F}_{0}\right\rbrack \rightarrow {V}^{\left( \mathbf{B}\right) } \) (of Definition 15.1) is elementary. | Proof. Let \( N \) be a countable standard transitive model of \( {ZF} \) such that \( {\mathbf{B}}^{N} \in N \) and \( \left\langle {N, \in ,{\mathbf{B}}^{N}}\right\rangle \) is an elementary substructure of \( \langle V, \in ,\mathbf{B}\rangle \) . (The existence of such an \( N \) can be proved in \( {GB} + \) mathe... | Yes |
Theorem 16.1. Suppose \( \left\{ {{u}_{i} \mid i \in I}\right\} \subseteq {V}^{\left( \mathbf{B}\right) } \) and \( \left( {\forall i \in I}\right) \left\lbrack {\mathcal{D}\left( {u}_{i}\right) \subseteq d}\right\rbrack \) for some \( d \subseteq {V}^{\left( \mathbb{B}\right) } \) . Then there is a family \( \left\{ {... | Proof. We extend the domain of \( {u}_{i} \) to \( d \) by defining \( {u}_{i}^{\prime } \in {V}^{\left( \mathbf{B}\right) } \) by \( \mathcal{D}\left( {u}_{i}^{\prime }\right) = \) \( d \) and \( \left( {\forall x \in d}\right) \left\lbrack {{u}_{i}^{\prime }\left( x\right) = \llbracket x \in {u}_{i}\rrbracket }\right... | Yes |
Corollary 16.3. Suppose \( \left\{ {{u}_{\xi } \mid \xi < \alpha }\right\} \subseteq {V}^{\left( \mathbf{B}\right) },\;\left\{ {{b}_{\xi } \mid \xi < \alpha }\right\} \subseteq B \) , \( d \subseteq {V}^{\left( \mathbf{B}\right) },\left( {\forall \xi < \alpha }\right) \left\lbrack {\mathcal{D}\left( {u}_{\xi }\right) =... | \[ \left( {\exists u \in {V}^{\left( \mathbf{B}\right) }}\right) \left\lbrack {\mathcal{D}\left( u\right) = d \land \left( {\forall \xi < \alpha }\right) \left\lbrack {{b}_{\xi } \leq \left\lbrack {u = {u}_{\xi }}\right\rbrack }\right\rbrack }\right\rbrack . \] | Yes |
Theorem 16.6.\n\n1. \( {U}_{\alpha }{}^{\left( \mathbf{B}\right) } \subseteq {V}_{\alpha }{}^{\left( \mathbf{B}\right) } \), in particular,\n\n\[ \left( {\forall u \in {U}_{\alpha }{}^{\left( \mathbf{B}\right) }}\right) \left( {\exists v \in {V}_{\alpha }{}^{\left( \mathbf{B}\right) }}\right) \left\lbrack {\llbracket u... | Proof. 1 and 2 are proved by induction on \( \alpha \) . 1 is obvious.\n\n2. Let \( v \in {V}_{\alpha + 1}^{\left( \mathbf{B}\right) } \) . Then \( \mathcal{D}\left( v\right) \subseteq {V}_{\alpha }^{\left( \mathbf{B}\right) } \).\n\nBy the induction hypothesis, there exists a function \( f : \mathcal{D}\left( v\right)... | Yes |
Theorem 16.10. If \( u \in {V}^{\left( \mathbf{B}\right) } \). \[ u\text{is extensional} \leftrightarrow \left( {\forall x \in \mathcal{D}\left( u\right) }\right) \left\lbrack {u\left( x\right) = \llbracket x \in u\rrbracket }\right\rbrack \text{.} \] | Proof. If \( u \) is extensional and \( x \in \mathcal{D}\left( u\right) \), then \[ u\left( x\right) \leq \llbracket x \in u\rrbracket = \mathop{\sum }\limits_{{y \in \mathcal{D}\left( u\right) }}u\left( y\right) \cdot \llbracket x = y\rrbracket \leq u\left( x\right) , \] Therefore \[ u\left( x\right) = \llbracket x \... | Yes |
Theorem 16.11. \[ \left( {\forall v \in {V}^{\left( \mathbf{B}\right) }}\right) \left( {\forall d \subseteq {\mathbf{V}}^{\left( \mathbf{B}\right) }}\right) \lbrack \mathcal{D}\left( v\right) \subseteq d \rightarrow \left( {\exists u}\right) \left\lbrack {u\text{ is extensional } \land d = \mathcal{D}\left( u\right) \l... | Proof. For \( v \in {V}^{\left( \mathbf{B}\right) } \) and \( d \subseteq {V}^{\left( \mathbf{B}\right) } \) such that \( \mathcal{D}\left( v\right) \subseteq d \) define \( u : d \rightarrow B \) by \[ \left( {\forall x \in d}\right) \left\lbrack {u\left( x\right) = \left\lbrack {x \in v}\right\rbrack }\right\rbrack .... | Yes |
Theorem 16.13. \( \\left( {\\forall v \\in {V}^{\\left( \\mathbf{B}\\right) }}\\right) \\left( {\\exists u \\in {V}^{\\left( \\mathbf{B}\\right) }}\\right) \\left\\lbrack {u\\text{is uniform} \\land \\llbracket u = v\\rrbracket = \\mathbf{1}}\\right\\rbrack \) . | Proof. If \( v \\in {V}^{\\left( \\mathbf{B}\\right) }, v \\in {V}_{\\alpha + 2}^{\\left( \\mathbf{B}\\right) } \) for some \( \\alpha \) . Then, by Theorem 16.6.2, \( \\llbracket v = {v}_{1}\\rrbracket = \\mathbf{1} \) and \( {v}_{1} \\in {U}_{\\alpha + 2}^{\\left( \\mathbf{B}\\right) } \) for some \( {v}_{1} \) . Sin... | Yes |
Theorem 16.14. Let \( u \) be uniform. If \( \left\{ {{x}_{i} \mid i \in I}\right\} \subseteq \mathcal{D}\left( u\right) \) and\n\n\[ \left\{ {{b}_{i} \mid i \in I}\right\} \subseteq B \]\n\nis a partition of unity, then\n\n\[ u\left( {\mathop{\sum }\limits_{{i \in I}}{b}_{i}{x}_{i}}\right) = \mathop{\sum }\limits_{{i ... | Proof. Let \( y = \mathop{\sum }\limits_{{i \in I}}{b}_{i}{x}_{i} \) . Since \( \mathcal{D}\left( u\right) \) is complete, \( y \) exists, and \( y \in \mathcal{D}\left( u\right) \) . Since \( \left\{ {{x}_{i} \mid i \in I}\right\} \subseteq \mathcal{D}\left( u\right) \),\n\n\[ u\left( y\right) = \llbracket y \in u\rrb... | Yes |
Theorem 16.16. \( \;\left( {\forall u \in {V}^{\left( \mathbf{B}\right) }}\right) \left\lbrack {\sup \left( u\right) = \llbracket \left( {\exists x}\right) \left\lbrack {x \in u}\right\rbrack \rrbracket }\right\rbrack \) . | Proof.\n\n\[ \llbracket \left( {\exists x}\right) \left\lbrack {x \in u}\right\rbrack \rrbracket = \llbracket \left( {\exists x \in u}\right) \left\lbrack {x = x}\right\rbrack \rrbracket = \mathop{\sum }\limits_{{x \in \mathcal{L}\left( u\right) }}u\left( x\right) \cdot \llbracket x = x\rrbracket \]\n\n\[ = \mathop{\su... | Yes |
Theorem 16.18. Let \( u \in {V}^{\left( \mathbb{B}\right) } \) be uniform and \( b \in B \) . Then\n\n\[ \n\{ x \in \mathcal{D}\left( u\right) \mid b \leq u\left( x\right) \} \n\]\n\nis complete. | Proof. Let \( \left\{ {{b}_{i} \mid i \in I}\right\} \) be a partition of unity, and \( \left\{ {{x}_{i} \mid i \in I}\right\} \subseteq \mathcal{D}\left( u\right) \) satisfying \( \left( {\forall i \in I}\right) \left\lbrack {b \leq u\left( {x}_{i}\right) }\right\rbrack \) . Since \( \mathcal{D}\left( u\right) \) is c... | Yes |
Theorem 16.19. If \( u \in {V}^{\left( \mathbb{B}\right) } \) is uniform, then \( \left( {\exists x \in \mathcal{D}\left( u\right) }\right) \left\lbrack {u\left( x\right) = \sup \left( u\right) }\right\rbrack \) . | Proof. If we do not require \( x \in \mathcal{D}\left( u\right) \), the theorem follows from the maximum principle. In fact, we can use the same argument:\n\nLet \( \left\langle {{x}_{\xi } \mid \xi < \alpha }\right\rangle \) be an enumeration of \( \mathcal{D}\left( u\right) \), i.e.,\n\n\[ \mathcal{D}\left( u\right) ... | Yes |
Theorem 16.22. Let \( v \in {V}^{\left( \mathrm{B}\right) } \) be extensional, \( u \in {V}^{\left( \mathrm{B}\right) } \), and \( b = \llbracket v \subseteq u\rrbracket \) . If \( {v}^{\prime } = b \cdot v \), then\n\n1. \( {v}^{\prime } \) is extensional,\n\n2. \( \llbracket {v}^{\prime } \subseteq u\rrbracket = 1 \)... | Proof.\n\n1. Is obvious.\n\n2. \( \llbracket {v}^{\prime } \subseteq u\rrbracket = \mathop{\prod }\limits_{{x \in \mathcal{D}\left( v\right) }}\left( {{v}^{\prime }\left( x\right) \Rightarrow \llbracket x \in u\rrbracket }\right) \n\n\[ = \mathop{\prod }\limits_{{x \in \mathcal{L}\left( v\right) }}\left( {\llbracket v ... | No |
Theorem 16.23. Let \( u, v \in {V}^{\left( \mathbf{B}\right) } \) be extensional and \( \llbracket v \subseteq u\rrbracket = \mathbf{1} \) . Define \( {v}^{\prime } \) by the conditions\n\n1. \( \mathcal{D}\left( {v}^{\prime }\right) = \mathcal{D}\left( u\right) \) .\n\n2. \( \left( {\forall x \in \mathcal{D}\left( u\r... | Proof. \( {v}^{\prime } \) is extensional and \( \llbracket {v}^{\prime } \subseteq v\rrbracket = \mathbf{1} \) by the definition of \( {v}^{\prime } \) . It remains to show that \( \left( {\forall x \in \mathcal{D}\left( v\right) }\right) \left\lbrack {\llbracket x \in v\rrbracket \leq \left\lbrack \left\lbrack {x \in... | Yes |
Theorem 16.25. If \( u, v \in {V}^{\left( \mathbf{B}\right) } \), and if\n\n\[ A = \left\{ {{v}^{\prime } \in {V}^{\left( \mathbf{B}\right) } \mid \mathcal{D}\left( {v}^{\prime }\right) = \mathcal{D}\left( u\right) \land \llbracket {v}^{\prime } \subseteq u\rrbracket = \mathbf{1}}\right\} \]\n\nthen\n\n\[ \llbracket v ... | Proof. If \( b = \mathop{\sum }\limits_{{{v}^{\prime } \in A}}\left\lbrack {v = {v}^{\prime }}\right\rbrack \) then from Theorem 16.24\n\n\[ \llbracket v \subseteq u\rrbracket \leq b. \]\n\nOn the other hand,\n\n\[ b \leq \mathop{\sum }\limits_{{{v}^{\prime } \in A}}\left\lbrack {{v}^{\prime } \subseteq u}\right\rbrack... | Yes |
Theorem 16.26. For \( u \in {V}^{\left( \mathbf{B}\right) } \)\n\n1. \( \left\lbrack {u \in R\left( \breve{\alpha }\right) }\right\rbrack = \mathop{\sum }\limits_{{v \in {V}_{\alpha }\left( \mathbf{B}\right) }}\llbracket u = v\rrbracket = \left\lbrack {u \in {V}_{\alpha }{}^{\left( \mathbf{B}\right) }\times \{ \mathbf{... | Proof. (By induction on \( \alpha \) .)\n\n1. \( \llbracket u \in R\left( \breve{\alpha }\right) \rrbracket = \llbracket \left( {\exists \xi < \breve{\alpha }}\right) \lbrack u \subseteq R(\xi \lbrack \rbrack \)\n\n\[ = \mathop{\sum }\limits_{{\xi < \alpha }}\left\lbrack {u \subseteq R\left( \check{\xi }\right) }\right... | Yes |
Theorem 16.27. If \( u \) is definite, \( \llbracket \mathcal{P}\left( u\right) = {B}^{\mathcal{D}\left( u\right) } \times \{ \mathbf{1}\} \rrbracket = \mathbf{1} \) . | Proof. Note that \( {B}^{\mathcal{D}\left( u\right) } \subseteq {V}^{\left( \mathbf{B}\right) } \) and \( {B}^{\mathcal{D}\left( u\right) } \) is a set. Let \( x \in {V}^{\left( \mathbf{B}\right) } \) and \( {A}_{u} = \left\{ {v \in {V}^{\left( \mathbf{B}\right) } \mid \mathcal{D}\left( v\right) = \mathcal{D}\left( u\r... | Yes |
Theorem 16.28. Let \( u, v \in {V}^{\left( \mathrm{B}\right) } \) be definite and uniform. If \( f \in {V}^{\left( \mathrm{B}\right) } \) and \( \llbracket f : u \rightarrow v\rrbracket = \mathbf{1} \), then there exists a (real) function \( \varphi : \mathcal{D}\left( u\right) \rightarrow \mathcal{D}\left( v\right) \)... | Proof. Since \( \llbracket \left( {\forall x \in u}\right) \left( {\exists y \in v}\right) \left\lbrack {f\left( x\right) = y}\right\rbrack \rrbracket = 1 \), and \( u \) and \( v \) are definite\n\n\[ \left( {\forall x \in \mathcal{D}\left( u\right) }\right) \left\lbrack {\mathop{\sum }\limits_{{y \in \mathcal{D}\left... | Yes |
Theorem 17.1. If \( \alpha \) is not a cardinal, then \( \llbracket \neg \) Card \( \left( \breve{\alpha }\right) \rrbracket = 1 \) . | \[ \text{Proof.}\neg \text{Card}\left( \alpha \right) \rightarrow \left( {\exists f}\right) \left( {\exists \beta < \alpha }\right) \left\lbrack {f : \beta \rightarrow \alpha \land \mathcal{W}{\left( f\right) }^{ * } = \alpha }\right\rbrack \text{.} \]\n\nTherefore \( \neg \) Card \( \left( \alpha \right) \rightarrow \... | Yes |
Theorem 17.4. Let \( \gamma \) be an infinite cardinal and suppose that \( \mathbf{B} \) satisfies the \( \gamma \) -chain condition. If \( \alpha > \gamma \) is a cardinal, then \( \llbracket \operatorname{Card}\left( \breve{\alpha }\right) \rrbracket = \mathbf{1} \) . | Proof. As in the proof of Theorem 17.2, suppose that [Card (č)] ≠ 1 for some cardinal \( \alpha > \gamma \), then defining \( b \) as before, we have for some \( \beta < \alpha \), and \( f \in {V}^{\left( \mathbf{B}\right) } \n\n\[ \n\text{i)}b \leq \mathop{\prod }\limits_{{\eta < \alpha }}\mathop{\sum }\limits_{{\xi ... | Yes |
Theorem 17.8. \( \left( {\forall u \in {V}^{\left( \mathbf{B}\right) }}\right) \left\lbrack {\lbrack \text{Const}\left( u\right) \rbrack = \mathop{\sum }\limits_{{x \in L}}\llbracket u = \check{x}\rrbracket }\right\rbrack \) . | Proof. For \( u \in {V}^{\left( \mathbf{B}\right) } \) , \n\n\[ \llbracket \text{Const}\left( u\right) \rrbracket = \llbracket \left( {\exists v}\right) \left\lbrack {\text{Ord}\left( v\right) \land u = F\left( v\right) }\right\rbrack \rrbracket \]\n\n\[ = \mathop{\sum }\limits_{{\alpha \in {On}}}\llbracket u = F\left(... | Yes |
Theorem 18.1. \( \llbracket \mathcal{P}{\left( \omega \right) }^{ \vee } \subseteq \mathcal{P}\left( \check{\omega }\right) \rrbracket = 1 \) | Proof.\n\n\[ \left( {\forall s \subseteq \omega }\right) \left\lbrack {\llbracket \breve{s} \subseteq \breve{\omega }\rrbracket = 1}\right\rbrack \]\n\n\[ \left( {\forall s \subseteq \omega }\right) \left\lbrack {\llbracket \breve{s} \in \mathcal{P}\left( \breve{\omega }\right) \rrbracket = 1}\right\rbrack \]\n\n\[ \ll... | No |
Theorem 18.2. B satisfies the \( \left( {\omega ,2}\right) \) -DL iff \( \llbracket \mathcal{P}{\left( \omega \right) }^{ \smile } = \mathcal{P}\left( \check{\omega }\right) \rrbracket = 1 \) . | Proof. Assume that \( \mathbf{B} \) satisfies the \( \left( {\omega ,2}\right) \) -DL. We need only show that \( \begin{Vmatrix}{\mathcal{P}\left( \check{\omega }\right) \subseteq \mathcal{P}{\left( \omega \right) }^{ \vee }}\end{Vmatrix} = \mathbf{1} \) . Therefore let \( t \in {B}^{\mathcal{L}\left( \check{\omega }\r... | Yes |
Theorem 18.4. B satisfies the \( \\left( {\\alpha ,2}\\right) \) -DL iff \( \\mathbf{B} \) satisfies the restricted \( \\left( {\\alpha ,2}\\right) \) -DL. | Proof. We can assume that \( \\alpha \) is an infinite cardinal. Suppose\n\n\[ \n\\left\\{ {{a}_{ij} \\mid i \\in I \\land j < 2}\\right\\} \\subseteq B \n\]\n\nwhere \( \\bar{I} = \\alpha \) . Define \( T = \\left( {I\\times \\{ 0\\} }\\right) \\cup \\left( {I\\times \\{ 1\\} }\\right) \) . Then \( \\bar{T} = \\alpha ... | Yes |
Theorem 18.5. Let \( I, J \) be sets. Then the following conditions are equivalent:\n\n1. For all families \( \left\{ {{b}_{ij} \mid i \in I \land j \in J}\right\} \subseteq B \) ,\n\n\[ \mathop{\prod }\limits_{{i \in I}}\mathop{\sum }\limits_{{j \in J}}{b}_{ij} = \mathop{\sum }\limits_{{f \in {J}^{I}}}\mathop{\prod }\... | Proof. We need only show that 3 implies 1. Therefore let\n\n\[ \left\{ {{b}_{ij} \mid i \in I \land j \in J}\right\} \subseteq B. \]\n\nLet\n\n\[ b = \mathop{\prod }\limits_{{i \in I}}\mathop{\sum }\limits_{{j \in J}}{b}_{ij} - \mathop{\sum }\limits_{{f \in {2}^{I}}}\mathop{\prod }\limits_{{i \in J}}{b}_{{if}\left( i\r... | Yes |
Theorem 19.1. Let \( \mathbf{B} \) be a complete Boolean algebra which does not satisfy the \( \left( {\omega ,2}\right) \) -DL. Then \( \llbracket V \neq L\rrbracket = \mathbf{1} \) in \( {\mathbf{V}}^{\left( \mathbf{B}\right) } \) . | Proof. (By the method of forcing.) Let \( M \) be a countable transitive model of \( {ZF} \) such that \( \left\langle {M, \in ,{\mathbf{B}}^{M}}\right\rangle \) is elementary equivalent to \( \langle V, \in ,\mathbf{B}\rangle \), let \( G \) be \( \mathbf{P} \) -generic over \( M \) (where \( \mathbf{P} \) is the part... | Yes |
Theorem 19.3. Let \( \pi \) be an automorphism of \( \mathbf{B}\left( {\pi \in \text{Aut}\left( \mathbf{B}\right) }\right) \) . Then \( \pi \) can be extended to an isomorphism \[ \pi : {V}^{\left( \mathbf{B}\right) } \rightarrow {V}^{\left( \mathbf{B}\right) } \] such that for every formula \( \varphi \) and \( {u}_{1... | Proof. Note that any automorphism \( \pi : \left| \mathbf{B}\right| \rightarrow \left| \mathbf{B}\right| \) is complete since \( \left( {\forall {b}_{1},{b}_{2} \in B}\right) \left\lbrack {{b}_{1} \leq {b}_{2} \leftrightarrow \pi \left( {b}_{1}\right) \leq \pi \left( {b}_{2}\right) }\right\rbrack \) . We define \( \pi ... | Yes |
Theorem 19.4. Let \( \mathbf{P} \) be the partial order structure used in the proof of the independence of \( V = L \) (Definition 11.1) and let \( \mathbf{B} \) be the complete Boolean algebra of regular open sets of \( \mathbf{P} \) . Then \( \mathbf{0} \) and \( \mathbf{1} \) are the only elements of \( B \) which a... | Proof. Let \( b \in B \) and \( \mathbf{0} < b < \mathbf{1} \) . Then there exist \( p \) and \( q \) such that \( \left\lbrack p\right\rbrack \subseteq b \) and \( \left\lbrack q\right\rbrack \subseteq {}^{ - }b \) . By Theorem 11.6, \( \exists \pi \in \) Aut \( \left( \mathbf{P}\right) \) such that \( \pi \left( p\ri... | Yes |
Theorem 19.6. B satisfies the c.c.c. and therefore \( \mathbf{B} \) is complete. | Proof. Let \( S \subseteq B \) be a set of mutually disjoint elements. We have to show that \( \bar{S} \leq \omega \) . Therefore we can assume that \( 0 \notin S \) . Let \( {S}_{n} = \{ b \mid b \in S \land \) \( m\left( b\right) \geq 1/n\} \) for \( n \in \omega \) . Since the elements of \( S \) are mutually disjoi... | Yes |
Theorem 20.2. If \( \mathbf{B} \) satisfies the c.c.c. and \( {cf}\left( {\omega }_{\alpha }\right) > \omega \), then \( \mathbf{B} \) satisfies the \( \left( {\omega ,{\omega }_{\alpha }}\right) \) -WDL. | Proof. Let \( \left\{ {{b}_{n\xi } \mid n < \omega \land \xi < {\omega }_{\alpha }}\right\} \subseteq B \) . Then by the c.c.c., for each \( n \in \omega \) there exists a countable set \( {C}_{n} \subseteq B \) such that\n\n\[ \mathop{\sum }\limits_{{\xi < {\omega }_{\alpha }}}{b}_{n\xi } = \sup {C}_{n} \]\n\nDefine \... | Yes |
Theorem 20.4. B satisfies the \( \left( {\omega ,\omega }\right) \) -WDL iff\n\n\[ \llbracket \left( {\forall g}\right) \left\lbrack {\text{if}g : \check{\omega } \rightarrow \check{\omega }\text{then}\left( {\exists f \in {\left( {\omega }^{\omega }\right) }^{ \smile }}\right) \left( {\forall n \in \omega }\right) \le... | Proof. Assume that \( \mathbf{B} \) satisfies the \( \left( {\omega ,\omega }\right) \) -WDL. Let \( g \in {V}^{\left( \mathbf{B}\right) } \) and define\n\n\[ b = \llbracket g : \check{\omega } \rightarrow \check{\omega }\rrbracket \]\n\n\[ {b}_{nm} = \llbracket g\left( \breve{n}\right) = \breve{m}\rrbracket \]\n\nfor ... | Yes |
Theorem 20.6. Every measure algebra \( \mathbf{B} \) satisfies the \( \left( {\omega ,\omega }\right) \) -WDL. | Proof. Let \( \left\{ {{b}_{n, k} \mid n, k < \omega }\right\} \subseteq B \) . Then for every real \( \varepsilon > 0 \) \n\n\[ \n\left( {\forall n < \omega }\right) \left( {\exists l \in \omega }\right) \left\lbrack {m\left( {\mathop{\sum }\limits_{{k < \omega }}{b}_{nk} - \mathop{\sum }\limits_{{k \leq l}}{b}_{nk}}\... | Yes |
Theorem 20.7. The Boolean algebra of all regular open sets in \( {\omega }^{\omega } \) does not satisfy the \( \left( {\omega ,\omega }\right) \) -WDL. | Proof. Define \( {b}_{nm} = \left\{ {p \in {\omega }^{\omega } \mid p\left( n\right) = m}\right\} \) for \( n, m < \omega \) . Then \( {b}_{nm} \) is clopen and therefore it is regular open. Obviously,\n\n\[ \mathop{\prod }\limits_{{n < \omega }}\mathop{\sum }\limits_{{m < \omega }}{b}_{nm} = \mathbf{1} \]\n\nbut\n\n\[... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.