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Theorem 1.58. If \( A \neq 0 \) and \( \forall a \in A,{I}_{a} \) is an ideal in the Boolean algebra B then \( \mathop{\bigcap }\limits_{{a \in A}}{I}_{a} \) is an ideal in B. | Proof. Left to the reader. | No |
Theorem 1.60. If \( I \) is the ideal in the Boolean algebra \( \mathbf{B} \) that is generated by \( A \subseteq \left| \mathbf{B}\right|, A \neq 0 \) then \( b \in I \) iff \( b \in \left| \mathbf{B}\right| \) and\n\n\[ \left( {\exists {b}_{1},\ldots ,{b}_{n} \in A}\right) \left\lbrack {b \leq {b}_{1} + \cdots + {b}_... | Proof. If \( J = \left\{ {b \in \left| \mathbf{B}\right| \mid \left( {\exists {b}_{1}\cdots {b}_{n} \in A}\right) \left\lbrack {b \leq {b}_{1} + \cdots + {b}_{n}}\right\rbrack }\right\} \) then \( J \) is an ideal in \( \mathbf{B} \) (details are left to the reader) and \( A \subseteq J \) . Thus \( I \subseteq J \) . ... | No |
Theorem 1.61. If \( I \) is a proper ideal in the Boolean algebra \( \mathbf{B} \), if \( \ulcorner a \notin I \) and if \( {I}_{a} \) is the ideal in \( \mathbf{B} \) generated by \( I \cup \{ a\} \) then \( {I}_{a} \) is a proper ideal. | Proof. If \( \mathbf{1} \in {I}_{a} \) then from Theorem 1.60\n\n\[ \left( {\exists b \in I}\right) \left\lbrack {1 \leq b + a}\right\rbrack \text{.} \]\n\nThen\n\n\[ - a = - a\mathbf{1} \leq - {ab} \in I. \]\n\nThis is a contradiction. Hence \( {I}_{a} \) is a proper ideal. | Yes |
Theorem 1.62. Every proper ideal \( I \) in a Boolean algebra B can be extended to a maximal ideal i.e., there exists a maximal ideal \( J \) in \( \mathbf{B} \) for which \( I \subseteq J \) . | Proof. For any well ordering \( R \) of \( \left| \mathbf{B}\right| \) we define\n\n\( {I}_{0} = I \) .\n\n\( {I}_{\alpha + 1} = \) ideal generated by \( {I}_{\alpha } \cup \{ a\} \), where \( a \) is the first element of \( \left| \mathbf{B}\right| \) for which \( a \notin {I}_{\alpha } \land - a \notin {I}_{\alpha } ... | Yes |
Theorem 2.3. If \( \mathbf{P} = \langle P, \leq \rangle \) is a partial order structure, if \( p \) is a minimal element of \( P \) and if \( G = \{ q \in P \mid p \leq q\} \) then \( G \) is \( \mathbf{P} \) -generic over \( A \) (in the strong sense). | Proof. If \( {q}_{1},{q}_{2} \in G \) then \( p \leq {q}_{1} \) and \( p \leq {q}_{2} \) . Therefore \( G \) is compatible. If \( {q}_{1} \in G,{q}_{2} \in P \) and \( {q}_{1} \leq {q}_{2} \) then since \( p \leq {q}_{1} \) we have \( p \leq {q}_{2} \), hence \( {q}_{2} \in G \) . If \( S \in A \) and \( S \) is a dens... | Yes |
Theorem 2.4. If \( \mathbf{P} \in M, M \) a standard transitive model of \( {ZF} \), and if \( G \) is \( \mathbf{P} \) -generic over \( M \) then \( G \) is \( \mathbf{P} \) -generic over \( M \) in the strong sense. | Proof. If \( a, b \in G \) and\n\n\[ S = \{ c \in P \mid \left\lbrack {c \leq a \land c \leq b}\right\rbrack \vee \left\lbrack {\neg \operatorname{Comp}\left( {c, a}\right) \land \neg \operatorname{Comp}\left( {c, b}\right) }\right\rbrack \} \]\n\nthen \( \forall c \in P \)\n\n1. \( \left( {\exists x \leq c}\right) \le... | Yes |
Theorem 2.5. Let \( \\mathbf{P} = \\langle P, \\leq \\rangle \) be a partial order structure with \( \\mathbf{P} \\in A \) and let \( G \\subseteq P \) . Suppose that for all \( S \n\n1. \( S \\in A \\land S \\subseteq P \\rightarrow S \\cup \\{ p \\mid \\left\\lbrack p\\right\\rbrack \\cap S = 0\\} \\in A \) .\n\nThen... | Proof. \( \\;\\left( {3 \\rightarrow 2}\\right) \) . If \( S \\in A \) and \( S \) is a dense subset of \( P \) then\n\n\\[ \n\\left( {\\forall p}\\right) \\left\\lbrack {\\left\\lbrack p\\right\\rbrack \\cap S \\neq 0}\\right\\rbrack \\text{.} \n\\] \n\nTherefore by \( 3,\\left( {\\exists p \\in G}\\right) \\left\\lbr... | Yes |
Theorem 2.6. If \( \mathbf{P} = \langle P, \leq \rangle \) is a partial order structure and if \( A \) is countable then every member of \( P \) is contained in some subset of \( P \) that is P-generic over \( A \) in the strong sense. | Proof. Since \( A \) is countable we can enumerate the elements of \( A \) that are dense subsets of \( P \) :\n\n\[ \n{S}_{1},{S}_{2},\ldots \n\]\n\nIf \( a \in P \) then\n\n\[ \n\left( {\exists {p}_{1} \in {S}_{1}}\right) \left\lbrack {{p}_{1} \leq a}\right\rbrack ,\;\left( {\text{Since }{S}_{1}\text{ is dense in }P}... | Yes |
Theorem 2.13. If \( {F}_{1} \) and \( {F}_{2} \) are ultrafilters on a Boolean algebra \( \mathbf{B} \) then\n\n1. \( {F}_{1} \subseteq {F}_{2} \rightarrow {F}_{1} = {F}_{2} \) .\n\n2. \( {F}_{1} \neq {F}_{2} \rightarrow \left\lbrack {{F}_{1} - {F}_{2} \neq 0}\right\rbrack \land \left\lbrack {{F}_{2} - {F}_{1} \neq 0}\... | Proofs. Left to the reader. | No |
Theorem 2.15. If \( \mathbf{B} \) is an \( M \) -complete Boolean algebra and \( \mathcal{P}\left( \left| \mathbf{B}\right| \right) \cap M \) is countable then for each \( b \neq \mathbf{0} \) in \( \left| \mathbf{B}\right| \) there exists an \( M \) -complete homomorphism \( f \) from \( \mathbf{B} \) onto \( \mathbf{... | Proof. If \( S \in \mathcal{P}\left( \left| \mathbf{B}\right| \right) \cap M \) and if \( b \in \left| \mathbf{B}\right| \) with \( b \neq \mathbf{0} \) then, from the Rasiowa-Sikorski Theorem, there exists a homomorphism \( f \) from \( \mathbf{B} \) into \( \mathbf{2} \) such that \( f\left( b\right) = 1 \) and \( f ... | No |
Theorem 2.17. If \( \mathbf{P} = \langle P, \leq \rangle \) is a partial order structure, if \( \mathbf{B} \) is the Boolean algebra of regular open subsets of \( P \), if \( F \) is a proper \( M \) -complete ultrafilter in \( {\mathbf{B}}^{M} \), and if \( G = \left\{ {p \in P \mid {\left\lbrack p\right\rbrack }^{-0}... | Proof. Clearly \( G \subseteq P \) . If \( p, q \in G \) then \( {\left\lbrack p\right\rbrack }^{-0} \in F \) and \( {\left\lbrack q\right\rbrack }^{-0} \in F \) . But\n\n\[{\left\lbrack p\right\rbrack }^{-0}{\left\lbrack q\right\rbrack }^{-0} = {\left\lbrack p\right\rbrack }^{-0} \cap {\left\lbrack q\right\rbrack }^{-... | Yes |
Theorem 2.18. If \( G \) is \( \mathbf{P} \) -generic over \( M \) then \( G \) is a maximal compatible subset of \( P \) . | Proof. If there exists a \( p \notin G \) such that \( G \cup \{ p\} \) is compatible and if\n\n\[ S = \left\lbrack p\right\rbrack \cup \{ q \mid \neg \operatorname{Comp}\left( {p, q}\right) \}\]\n\nit is easily established that \( S \) is dense in \( P \) . Indeed if \( q \in P \) either \( q \) is compatible with \( ... | Yes |
Theorem 2.19. If \( \mathbf{P} = \langle P, \leq \rangle \in M \), if \( \mathbf{B} \) is the Boolean algebra of regular open subsets of \( P \), if \( G \) is \( \mathbf{P} \) -generic over \( M \), and if\n\n\[ F = \left\{ {b \in \left| \mathbf{B}\right| \cap M \mid b = {b}^{-0} \land b \cap G \neq 0}\right\} \]\n\n\... | Proof.\n\n\[ p \in G \rightarrow {\left\lbrack p\right\rbrack }^{-0} \cap G \neq 0 \]\n\n\[ \rightarrow p \in {G}^{\prime } \]\n\ni.e., \( G \subseteq {G}^{\prime } \) . Since, by Theorems 2.16 and \( {2.17},{G}^{\prime } \) is \( \mathbf{P} \) -generic over \( M \) it follows from Theorem 2.18 that \( G = {G}^{\prime ... | Yes |
Theorem 2.20. If \( \mathbf{B} \) is a natural Boolean algebra, if \( F \) is a proper \( M \) - complete ultrafilter for \( {\mathbf{B}}^{M} \), and if\n\n\[ G = \left\{ {p \mid {\left\lbrack p\right\rbrack }^{-0} \in F}\right\} \]\n\n\[ {F}^{\prime } = \left\{ {b \in \left| \mathbf{B}\right| \cap M \mid b = {b}^{-0} ... | Proof.\n\n\[ b \in {F}^{\prime } \rightarrow b = {b}^{-0} \land b \cap G \neq 0 \]\n\n\[ \rightarrow \left( {\exists p \in G}\right) \left\lbrack {p \in b}\right\rbrack \]\n\n\[ \rightarrow \left( {\exists p}\right) \left\lbrack {{\left\lbrack p\right\rbrack }^{-0} \in F \land {\left\lbrack p\right\rbrack }^{-0} \leq b... | Yes |
Theorem 3.3. If \( I \neq 0 \) and \( {\mathbf{B}}_{a} \) is a subalgebra of the Boolean algebra \( \mathbf{B} = \langle B, + , \cdot , - ,\mathbf{0},\mathbf{1}\rangle \) for \( a \in I \) then \( {\mathbf{B}}^{\prime } = \left\langle {\mathop{\bigcap }\limits_{{a \in I}}\left| {\mathbf{B}}_{a}\right| ,+,\cdot ,-,\math... | Proof. Left to the reader. | No |
Theorem 3.6. If \( \langle X, T\rangle \) is a topological space, if\n\n\[ \n{A}_{0} = T \cup \{ X - a \mid a \in T\}\]\n\n\[ \n{A}_{\alpha + 1} = \left\{ {a \mid \left( {\exists f \in {\left( {A}_{\alpha }\right) }^{\omega }}\right) \left\lbrack {\left\lbrack {a = \mathop{\bigcup }\limits_{{i < \omega }}f\left( i\righ... | Proof. Clearly each element in \( {A}_{0} \) is a Borel set. If \( {A}_{\alpha } \) is a collection of Borel sets then so is \( {A}_{\alpha + 1} \) . If for \( \beta < \alpha ,{A}_{\beta } \) is a collection of Borel sets then\n\n\[ \n\mathop{\bigcup }\limits_{{\beta < \alpha }}{A}_{\beta }\n\]\n\nis a collection of Bo... | Yes |
Theorem 3.8. If \( X \) is a topological space and \( A \subseteq X \) then\n\n1. \( A \) is open implies \( {A}^{ - } - A \) is meager.\n\n2. \( A \) is closed implies \( A - {A}^{0} \) is meager. | Proof.\n\n\[ \text{1.}{\left( {A}^{ - } - A\right) }^{0} = {\left\lbrack {A}^{ - } \cap \left( X - A\right) \right\rbrack }^{-0} \subseteq {A}^{-0} \cap {\left( X - A\right) }^{-0} \]\n\n\[ = {A}^{-0} \cap \left( {X - {A}^{0 - }}\right) \]\n\nsince \( A \) is open\n\n\[ \subseteq {A}^{-0} \cap \left( {X - {A}^{-0}}\rig... | Yes |
Theorem 3.9. 1. The collection of all meager sets in a topological space \( X \) is a proper \( \sigma \) -ideal in the natural algebra on \( \mathcal{P}\left( X\right) \). | Proof. Left to the reader. | No |
Theorem 3.10. If \( B \) is a Borel set of the topological space \( X \) then there exists an open set \( G \) and meager sets \( {N}_{1} \) and \( {N}_{2} \) such that\n\n\[ B = \left( {G + {N}_{1}}\right) - {N}_{2} \]\n\ni.e. every Borel set has the property of Baire. | Proof. If \( B \) is open then \( B = \left( {B + 0}\right) - 0 \) . If \( B \) is closed\n\n\[ B = \left\lbrack {{B}^{0} + \left( {B - {B}^{0}}\right) }\right\rbrack - 0. \]\n\nThus in the notation of Theorem 3.6, the result holds for each element of \( {A}_{0} \) . If it holds for each element of \( {A}_{\alpha } \) ... | Yes |
Corollary 3.11. If \( B \) is a Borel set of the topological space \( \langle X, T\rangle \) then there exists a regular open set \( G \) and meager sets \( {N}_{1} \) and \( {N}_{2} \) such that\n\n\[ B = \left( {G + {N}_{1}}\right) - {N}_{2} \] | Proof. By Theorem 3.10 there exists an open set \( G \) and meager sets \( {N}_{1} \) and \( {N}_{2} \) such that \( B = \left( {G + {N}_{1}}\right) - {N}_{2} \) . But\n\n\[ G = {G}^{-0} - \left( {{G}^{-0} - G}\right) . \]\n\nHence\n\n\[ B = \left( {{G}^{-0} + {N}_{1}}\right) - \left\lbrack {\left( {{G}^{-0} - G - {N}_... | Yes |
Theorem 3.14. If the topological space \( \langle X, T\rangle \) is a Hausdorff space then\n\n\[ \left( {\forall a, b \in X}\right) \left\lbrack {a \neq b \rightarrow \left( {\exists N\left( a\right) }\right) \left\lbrack {b \notin N{\left( a\right) }^{ - }}\right\rbrack }\right\rbrack \text{.} \] | Proof. By definition of a Hausdorff space\n\n\[ \left( {\exists N\left( a\right) }\right) \left( {\exists {N}^{\prime }\left( b\right) }\right) \left\lbrack {N\left( a\right) \cap {N}^{\prime }\left( b\right) = 0}\right\rbrack . \]\n\nTherefore \( b \notin N{\left( a\right) }^{ - } \) . | Yes |
Every compact set in a Hausdorff space is closed. | 1. Let \( A \) be a compact set in a Hausdorff space \( \langle X, T\rangle \) . If \( b \in {A}^{ - } - A \) then by Theorem 3.14\n\n\[ \left( {\forall a \in A}\right) \left( {\exists N\left( a\right) }\right) \left\lbrack {b \notin N{\left( a\right) }^{ - }}\right\rbrack . \]\n\nSince \( A \subseteq \bigcup \left\{ {... | Yes |
Theorem 3.17. The topological space \( \langle X, T\rangle \) is compact iff for each collection \( S \) of closed sets with the finite intersection property\n\n\[ \bigcap \left( S\right) \neq 0\text{.} \] | Proof. (By contradiction.) Suppose that \( \langle X, T\rangle \) is a compact topological space and there exists a collection of closed sets \( S \) with the finite intersection property but for which \( \cap \left( S\right) = 0 \) . Then\n\n\[ X - 0 = X - \bigcap \left( S\right) = \mathop{\bigcup }\limits_{{A \in S}}... | Yes |
Theorem 3.18. If \( \langle X, T\rangle \) is a topological space, if \( {X}^{\prime } \subseteq X \) and \( {T}^{\prime } = \left\{ {{X}^{\prime } \cap N \mid N \in T}\right\} \) then \( \left\langle {{X}^{\prime },{T}^{\prime }}\right\rangle \) is a topological space. Furthermore\n\n1. \( \left\langle {{X}^{\prime },... | Proof. Left to the reader. | No |
Theorem 3.20. If \( \langle X, T\rangle \) is a topological space, if \( {X}^{\prime } \subseteq X \), if \( {T}^{\prime } \) is the relative topology on \( {X}^{\prime } \) induced by \( T \), if \( B \) is a base for \( T \) and\n\n\[ \n{B}^{\prime } = \left\{ {{X}^{\prime } \cap N \mid N \in B}\right\} \n\]\n\nthen ... | Proof. Left to the reader. | No |
Theorem 3.21. If \( \langle X, T\rangle \) is a topological space, if \( {X}^{\prime } \subseteq X \), and if \( {T}^{\prime } \) is the relative topology on \( {X}^{\prime } \) induced by \( T \) then\n\n1. \( A \) is an open set in \( T \) implies \( A \cap {X}^{\prime } \) is an open set in \( {T}^{\prime } \)\n\n2.... | Proof. 1. If \( A \) is open in \( T \) then\n\n\[ \left( {\forall a \in A \cap {X}^{\prime }}\right) \left( {\exists N\left( a\right) \in T}\right) \left\lbrack {N\left( a\right) \subseteq A}\right\rbrack .\n\]\n\nThen \( N\left( a\right) \cap {X}^{\prime } \in {T}^{\prime } \) and \( a \in N\left( a\right) \cap {X}^{... | Yes |
Theorem 3.22. If \( \langle X, T\rangle \) is a locally compact Hausdorff space then for each open set \( A \) and each \( a \in A \) there exists an open set \( B \) such that\n\n\[ a \in B \land {B}^{ - } \subseteq A. \] | Proof. If \( A \) is an open set in \( X \) and \( a \in A \) then since \( \langle X, T\rangle \) is locally compact \( \exists N\left( a\right), N{\left( a\right) }^{ - } \) is compact. If\n\n\[ M = {\left( N{\left( a\right) }^{ - } \cap A\right) }^{0} \]\n\nthen \( {M}^{ - } \) is also compact. If\n\n\[ {T}^{\prime ... | Yes |
Theorem 3.23. (The Baire Category Theorem.) Every open meager set in a locally compact Hausdorff space is empty. | Proof. If \( B \) is an open meager set in the locally compact Hausdorff space \( \langle X, T\rangle \) then there exists an \( \omega \) -sequence of nowhere dense sets\n\n\[ \n{A}_{0},{A}_{1},\ldots \n\]\n\nsuch that\n\n\[ \nB = \mathop{\bigcup }\limits_{{\alpha < \omega }}{A}_{\alpha } \n\]\n\nIf \( B \neq 0 \) the... | Yes |
Theorem 3.24. If \( \mathbf{B} \) is the Boolean \( \sigma \) -algebra of all Borel sets in the locally compact Hausdorff space \( \langle X, T\rangle \) and if \( I \) is the \( \sigma \) -ideal of all meager Borel sets then \( \mathbf{B}/I \) is isomorphic to \( {\mathbf{B}}^{\prime } \), the complete Boolean algebra... | Proof. If\n\n\[ F\left( G\right) = G/I,\;G \in \left| {\mathbf{B}}^{\prime }\right| \]\n\nthen \( F\left( {G}_{1}\right) = F\left( {G}_{2}\right) \leftrightarrow {G}_{1} - {G}_{2} \in I \land {G}_{2} - {G}_{1} \in I \) . Then \( {G}_{1} - {G}_{2}^{ - } \) is meager and open. Thus, by the Baire Category Theorem\n\n\[ {G... | Yes |
Theorem 3.26. If \( X \) is a topological space with a countable base then the Boolean algebra of regular open sets in \( X \) satisfies the countable chain condition. | Proof. If \( {U}_{1},{U}_{2},\ldots \) is a countable base and if \( S \) is a pairwise disjoint subset of \( \left| \mathbf{B}\right| \) then since the elements of \( S \) are open it follows that\n\n\[ \left( {\forall A \in S}\right) \left( {\exists n < \omega }\right) \left\lbrack {{U}_{n} \subseteq A}\right\rbrack ... | Yes |
Theorem 3.27. If \( \mathbf{B} \) satisfies the c.c.c. then for each subset \( E \) of \( \left| \mathbf{B}\right| \) there exists a countable subset \( D \) of \( E \) such that \( D \) and \( E \) have the same set of upper bounds. | Proof. If \( I \) is the ideal generated by \( E \) then \( E \subseteq I \) . Consequently every upper bound for \( I \) is an upper bound for \( E \) . Conversely\n\n\[ \left( {\forall b \in I}\right) \left( {\exists {b}_{1},\ldots ,{b}_{n} \in E}\right) \left\lbrack {b \leq {b}_{1} + \cdots + {b}_{n}}\right\rbrack .... | Yes |
Theorem 3.28. Every Boolean \( \sigma \) -algebra \( \mathbf{B} \) satisfying the c.c.c. is complete. | Proof. By Theorem 3.27 if \( E \subseteq \left| \mathbf{B}\right| \) then there exists a countable subset \( D \) of \( E \) such that \( D \) and \( E \) have the same set of upper bounds. Since \( D \) is countable and \( \mathbf{B} \) is a \( \sigma \) -algebra\n\n\[ \mathop{\sum }\limits_{{b \in D}}b \]\n\nexists. ... | No |
Theorem 3.30. If \( A \neq 0 \) and \( \forall a \in A,\left\langle {{X}_{a},{T}_{a}}\right\rangle \) is a topological space then\n\n\[ \n{T}^{\prime } = \left\{ {B \subseteq \mathop{\prod }\limits_{{a \in A}}{X}_{a} \mid \left( {\exists n \in \omega }\right) \left( {\exists \sigma \in {A}^{n}}\right) \left( {\forall i... | Proof. If \( A \neq 0 \) then \( \exists b \in A \) . If \( f \in \mathop{\prod }\limits_{{a \in A}}{X}_{a} \) then \( f\left( b\right) \in {X}_{b} \) and hence \( \exists N\left( {f\left( b\right) }\right) \in {T}_{b} \) . Then\n\n\[ \nf \in \left\{ {g \in \mathop{\prod }\limits_{{a \in A}}{X}_{a} \mid g\left( b\right... | Yes |
Theorem 3.32. If \( A \neq 0 \) and \( \forall a \in A,\left\langle {{X}_{a},{T}_{a}}\right\rangle \) is a topological space then\n\n\[ \left( {\forall a \in A}\right) \left( {\forall C \subseteq \mathop{\prod }\limits_{{a \in A}}{T}_{a}}\right) \left\lbrack {{p}_{a}\left( C\right) \in {T}_{a}}\right\rbrack . \] | Proof. Left to the reader. | No |
Theorem 3.33. (Tychonoff’s Theorem.) If \( A \neq 0 \) and \( \forall a \in A,\left\langle {{X}_{a},{T}_{a}}\right\rangle \) is a compact topological space then the product topological space \( \mathop{\prod }\limits_{{a \in A}}{X}_{a} \), is also compact. | Proof. Let \( S \) be a collection of closed subsets of \( \mathop{\prod }\limits_{{a \in A}}{X}_{a} \), with the finite intersection property and\n\n\[ T \triangleq \left\{ {B \subseteq \mathcal{P}\left( {\mathop{\prod }\limits_{{a \in A}}{X}_{a}}\right) \mid S \subseteq B \land B\text{ has the finite intersection pro... | Yes |
Theorem 3.34. If \( A \neq 0 \) and \( \forall a \in A,\left\langle {{X}_{a},{T}_{a}}\right\rangle \) is a Hausdorff space then the product topology on \( \mathop{\prod }\limits_{{a \in A}}{X}_{a} \) is Hausdorff. | Proof. If \( f, g \in \mathop{\prod }\limits_{{a \in A}}{X}_{a} \) and \( f \neq g \) then \( \left( {\exists b \in A}\right) \left\lbrack {f\left( b\right) \neq g\left( b\right) }\right\rbrack \) . Since \( \left\langle {{X}_{b},{T}_{b}}\right\rangle \) is a Hausdorff space \( \exists N\left( {f\left( b\right) }\right... | Yes |
Theorem 4.2. If \( \\mathbf{B} \) is a complete Boolean algebra and\n\n\[ \n\\left\\{ {{b}_{ij} \\mid i \\in I \\land j \\in J}\\right\\} \\subseteq \\left| \\mathbf{B}\\right| \n\]\n\nthen\n\n\[ \n\\mathop{\\sum }\\limits_{{f \\in {JI}}}\\mathop{\\prod }\\limits_{{i \\in I}}{b}_{i, f\\left( i\\right) } \\leq \\mathop{... | Proof.\n\n\[ \n\\left( {\\forall f \\in {J}^{I}}\\right) \\left\\lbrack {{b}_{i, f\\left( i\\right) } \\leq \\mathop{\\sum }\\limits_{{j \\in J}}{b}_{ij}}\\right\\rbrack \n\]\n\n\[ \n\\left( {\\forall f \\in {J}^{I}}\\right) \\left\\lbrack {\\mathop{\\prod }\\limits_{{i \\in I}}{b}_{i, f\\left( i\\right) } \\leq \\math... | Yes |
Theorem 4.3. If \( \mathbf{B} \) is the complete natural Boolean algebra of all subsets of \( A \neq 0 \) then \( \mathbf{B} \) satisfies the complete distributive law. | Proof. If \( {b}_{ij} \subseteq A \) for \( i \in I \) and \( j \in J \) then\n\n\[ b \in \mathop{\bigcap }\limits_{{i \in I}}\mathop{\bigcup }\limits_{{j \in J}}{b}_{ij} \leftrightarrow \left( {\forall i \in I}\right) \left( {\exists j \in J}\right) \left\lbrack {b \in {b}_{ij}}\right\rbrack \]\n\n\[ \leftrightarrow \... | Yes |
Theorem 5.5. Let \( \mathbf{B} = \langle B, + , \cdot , - ,\mathbf{0},\mathbf{1}\rangle \) be a Boolean algebra and \( \mathbf{P} = \langle P, \leq \rangle \) be its associated partial order structure. If \( F \) is a nonempty subset of \( \left| \mathbf{B}\right| \) then \( F \) is a proper filter for the Boolean alge... | Proof. Let \( F \) be a proper filter for \( \mathbf{B} \) . Then \( \mathbf{0} \notin F \) i.e., \( F \subseteq B - \{ \mathbf{0}\} \) . If \( x, y \in F \) then \( {xy} \in F \) and hence there exists a \( z \in F \), namely \( {xy} \), such that \( z \leq x \) and \( z \leq y \) . Thus \( F \) is a filter for \( \ma... | Yes |
Theorem 5.7. \( \langle \mathbf{F},\mathbf{T}\rangle \) is a \( {T}_{1} \) -space. | Proof. First of all we shall show that \( \langle \mathbf{F},\mathbf{T}\rangle \) is a topological space. From Definition 5.6 it is clear that 0 and \( \mathbf{F} \) are each open. Let \( {G}_{1} \) and \( {G}_{2} \) be open sets and let \( F \in {G}_{1} \cap {G}_{2} \) . Then there exist \( p \) and \( {p}^{\prime } \... | Yes |
Theorem 5.8. Let \( \mathbf{P} = \langle P, \leq \rangle \) be the partial order structure associated with the Boolean algebra B. Then \( \langle \mathbf{F},\mathbf{T}\rangle \) is a Hausdorff space. | Proof. Suppose not. Then there would exist distinct \( {F}_{1},{F}_{2} \in \mathbf{F} \) such that\n\n\[ \left( {\forall {p}_{1} \in {F}_{1}}\right) \left( {\forall {p}_{2} \in {F}_{2}}\right) \left( {\exists F \in \mathbf{F}}\right) \left\lbrack {{p}_{1} \in F \land {p}_{2} \in F}\right\rbrack . \]\n\nThen \( \left( {... | Yes |
Theorem 5.10. If \( {G}_{1} \) and \( {G}_{2} \) are open subsets of \( P \) and \( \mathbf{F} \) respectively then\n\n1. \( {G}_{1} \subseteq {G}_{1}^{*\Delta } \) .\n\n2. \( {G}_{2} \subseteq {G}_{2}^{\Delta * } \) . | Proof.\n\n1. \( a \in {G}_{1} \rightarrow \left\lbrack a\right\rbrack \subseteq {G}_{1} \)\n\n\[ \n\rightarrow N\left( a\right) \subseteq {G}_{1}^{ * } \n\]\n\n\( \rightarrow \left\lbrack a\right\rbrack \subseteq {G}_{1}^{*\Delta }\; \) (since \( {G}_{1}^{ * } \) is an open subset of \( \mathbf{F} \) )\n\n\[ \n\rightar... | Yes |
Theorem 5.11. 1. If \( {G}_{1} \) and \( {G}_{2} \) are open subsets of \( P \) then\n\n\[ \n{G}_{1} \subseteq {G}_{2} \rightarrow {G}_{1}^{ * } \subseteq {G}_{2}^{ * } \n\] | Proof. Left to the reader. | No |
Theorem 5.12. If \( G \) is an open subset of \( \mathbf{F} \) and \( \left\lbrack a\right\rbrack \subseteq {G}^{\Delta } \), then \( N\left( a\right) \subseteq G \) . | Proof.\n\n\[ \left\lbrack a\right\rbrack \subseteq {G}^{\Delta } \rightarrow a \in {G}^{\Delta } \]\n\n\[ \rightarrow \left( {\exists b}\right) \left\lbrack {N\left( b\right) \subseteq G \land a \in \left\lbrack b\right\rbrack }\right\rbrack \]\n\n\[ \rightarrow \left( {\exists b \geq a}\right) \left\lbrack {N\left( b\... | Yes |
Theorem 5.13. If \( G \) is a regular open subset of \( P \) then \( {G}^{*\Delta } = G \) . | Proof.\n\n\[ a \in {G}^{*\Delta } \rightarrow \left\lbrack a\right\rbrack \subseteq {G}^{*\Delta } \]\n\n\[ \rightarrow N\left( a\right) \subseteq {G}^{ * } \]\n\n\[ \rightarrow \left( {\forall F}\right) \left\lbrack {a \in F \rightarrow F \in {G}^{ * }}\right\rbrack \]\n\n\[ \rightarrow \left( {\forall F}\right) \left... | Yes |
Theorem 5.14. If \( G \) is an open subset of \( \mathbf{F} \) then \( {G}^{\Delta * } = G \) . | Proof.\n\n\[ F \in {G}^{\Delta * } \rightarrow \left( {\exists a \in F}\right) \left\lbrack {\left\lbrack a\right\rbrack \subseteq {G}^{\Delta }}\right\rbrack \]\n\n\[ \rightarrow \left( {\exists a \in F}\right) \left\lbrack {N\left( a\right) \subseteq G}\right\rbrack \]\n\n\[ \rightarrow F \in G\text{.} \]\n\nTherefor... | No |
Theorem 5.15. Let \( {G}_{1} \) and \( {G}_{2} \) be open sets of a topological space \( \langle X, T\rangle \) . If for each regular open set \( H \)\n\n\[ \n{G}_{1} \cap H = 0 \rightarrow {G}_{2} \cap H = 0 \n\]\n\nthen \( {G}_{2} \subseteq {G}_{1}{}^{-0} \) . | Proof. If \( H = {\left( X - {G}_{1}\right) }^{0} \) then \( H \) is regular open. If \( {G}_{1} \cap H = 0 \) then \( {G}_{2} \cap H = 0 \) and hence\n\n\[ \n{G}_{2} \cap {H}^{ - } = 0. \n\]\n\nTherefore\n\n\[ \n{G}_{2} \subseteq X - {H}^{ - } = {G}_{1}{}^{-0}. \n\] | Yes |
Theorem 5.16. 1. If \( {G}_{1} \) is an open subset of \( P \) then\n\n\[ \n{G}_{1}^{ * } = 0 \rightarrow {G}_{1} = 0.\n\] | Proof. Left to the reader. | No |
If \( G \) is a regular open subset of \( \mathbf{F} \) then \( {G}^{\Delta } \) is regular open. | 1. Let \( {G}_{1} = {\left( {G}^{\Delta }\right) }^{-0} \) . Then\n\n\[ G = {G}^{\Delta * } \subseteq {\left( {G}^{\Delta }\right) }^{-0 * } = {G}_{1}^{ * }.\]\n\nIf \( {G}_{2} \) is regular open and \( G \cap {G}_{2} = 0 \) then\n\n\[ {\left( {G}^{\Delta } \cap {G}_{2}^{\Delta }\right) }^{ * } \subseteq {G}^{\Delta * ... | Yes |
Theorem 5.18. If \( \\mathbf{P} = \\langle P, \\leq \\rangle \) is a partial order structure, then the Boolean algebra \( \\mathbf{B} \) of regular open subsets of \( P \) is isomorphic to the Boolean algebra of regular open subsets of \( \\mathbf{F} \) . | Proof. The mapping \( * \) is a one-to-one, order preserving mapping from the first algebra onto the second. | No |
Lemma 5.22. If \( \mathbf{P} \) is fine, then for each \( p \in P \)\n\n\[{\left\lbrack p\right\rbrack }^{-0} = \left\lbrack p\right\rbrack \text{.} \] | Proof. We have only to show \( {\left\lbrack p\right\rbrack }^{-0} \subseteq \left\lbrack p\right\rbrack \) . Let \( q \in P \) such that \( q \notin \left\lbrack p\right\rbrack \) , i.e., \( q \nleq p \) . Then, by Definition 5.21, \( \left( {\exists r \in P}\right) \left\lbrack {r \leq q\land \neg \text{Comp}\left( {... | Yes |
Theorem 6.4. If \( \varphi \) is a closed formula of the language \( \mathcal{L} \) then \( \varphi \) is satisfied in every B-valued structure iff \( \varphi \) is logically valid. | Proof. \( \varphi \) is logically valid iff \( \varphi \) is satisfied in every 2-valued structure. Since 2 is a complete subalgebra of \( \mathbf{B} \) every 2-valued structure is a \( \mathbf{B} \) -valued structure. Conversely if \( \varphi \) is not satisfied by some B-valued structure A i.e., \[ \llbracket \varphi... | Yes |
Theorem 6.6. Every logically valid sentence of a first order language \( \mathcal{L} \) with equality is satisfied by a B-valued structure of \( \mathcal{L} \) . | In particular\n\n\[ \llbracket {c}_{1} = {c}_{1}^{\prime }\rrbracket \cdots \llbracket {c}_{n} = {c}_{n}^{\prime }\rrbracket \llbracket \varphi \left( {{c}_{1},\ldots ,{c}_{n}}\right) \rrbracket \leq \llbracket \varphi \left( {{c}_{1}^{\prime },\ldots ,{c}_{n}^{\prime }}\right) \rrbracket . \] | No |
Theorem 6.9. If \( \left\langle {{b}_{i} \mid i \in I}\right\rangle \) is a partition of unity, if\n\n\[ \left( {\forall i \in I}\right) \left\lbrack {{b}_{i} \leq \left\lbrack {a = {a}_{i}}\right\rbrack }\right\rbrack \]\n\nand\n\n\[ \left( {\forall i \in I}\right) \left\lbrack {{b}_{i} \leq \left\lbrack {{a}^{\prime ... | \[ \text{Proof.}{b}_{i} \leq \llbracket a = {a}_{i}\rrbracket \left\lbrack {{a}^{\prime } = {a}_{i}}\right\rbrack \leq \llbracket a = {a}^{\prime }\rrbracket, i \in I \]\n\n\[ 1 = \mathop{\sum }\limits_{{i \in I}}{b}_{i} \leq \left\lbrack {a = {a}^{\prime }}\right\rbrack \leq 1. \]\n\nHence \( \llbracket a = {a}^{\prim... | Yes |
Theorem 7.5. If \( \mathbf{A} = \left\langle {A,{\bar{R}}_{0},\ldots ,{\bar{R}}_{n},{\bar{c}}_{0},\ldots ,{\bar{c}}_{m}}\right\rangle \) is a transitive structure for \( \mathcal{L} \), where \( A \) is a set, then there is a wff \( \psi \) of \( {\mathcal{L}}_{0} \) such that for every closed wff \( \varphi \) of \( \... | We define a formula \( {\psi }_{0}\left( {f,\mathbf{A}}\right) \) in the language \( {\mathcal{L}}_{0} \) that formalizes the notion\n\n\ | No |
Theorem 7.7. If \( A \) is a class then for each formula \( \varphi \) of \( \mathcal{L} \) with free variables \( {a}_{1},\ldots ,{a}_{k} \) there is a formula \( \psi \) of \( {\mathcal{L}}_{0} \) for which\n\n\[ \left( {\forall {a}_{1},\ldots ,{a}_{k} \in A}\right) \left\lbrack {\mathbf{A} \vDash \varphi \left( {{a}... | Proof. For each \( {a}_{1},\ldots ,{a}_{k} \in A \)\n\n\[ \mathbf{A} \vDash \varphi \left( {{a}_{1},\ldots ,{a}_{k}}\right) \leftrightarrow \bar{\varphi }\left( {{a}_{1},\ldots ,{a}_{k}}\right) \]\n\nwhere \( \bar{\varphi }\left( {{a}_{1},\ldots ,{a}_{k}}\right) \) is \( {\varphi }^{A}\left( {{a}_{1},\ldots ,{a}_{k}}\r... | Yes |
Theorem 7.11.\n\n1. \( {M}_{\alpha } \in {M}_{\alpha + 1} \) .\n\n2. \( {M}_{\alpha } \in M \) . | Proof. Theorems 7.9 and 3 above. | No |
Theorem 7.12. \( a \subseteq M \rightarrow \left( {\exists \alpha }\right) \left\lbrack {a \subseteq {M}_{\alpha }}\right\rbrack \) . | Proof. From the Axiom Schema of Replacement it follows that\n\n\[ \left( {\exists \alpha }\right) \left\lbrack {\alpha = \mathop{\bigcup }\limits_{{x \in a}}{\mu }_{\beta }\left( {x \in {M}_{\beta }}\right) }\right\rbrack \]\n\nthen \( a \subseteq {M}_{\alpha } \) . | Yes |
Theorem 7.15. If \( {F}_{1},\ldots ,{F}_{n} \) are semi-normal functions then\n\n\[ \left( {\forall \alpha }\right) \left( {\exists \beta > \alpha }\right) \left\lbrack {\beta = {F}_{1}\left( \beta \right) = \cdots = {F}_{n}\left( \beta \right) }\right\rbrack . | Proof. We define an \( \omega \) -sequence \( \left\langle {{\alpha }_{m} \mid m \in \omega }\right\rangle \) by recursion:\n\n\[ {\alpha }_{1} = \alpha + 1,{\alpha }_{2} = {F}_{1}\left( {\alpha }_{1}\right) ,\ldots ,{\alpha }_{n + 1} = {F}_{n}\left( {\alpha }_{n}\right) \]\n\n\[ {\alpha }_{k + i} = {F}_{i}\left( {\alp... | Yes |
Theorem 7.16. If \( \varphi \left( {{a}_{0},\ldots ,{a}_{n}}\right) \) is a formula of \( \mathcal{L} \) then\n\n\[ \left( {\forall \alpha }\right) \left( {\exists \beta \geq \alpha }\right) \left( {{\forall }^{\prime }{a}_{1},\ldots ,{a}_{n} \in {M}_{\alpha }}\right) \lbrack \mathbf{M} \vDash \left( {\exists x}\right)... | Proof. By Theorem 7.7.\n\n\[ \mathbf{M} \vDash \varphi \left( {a,{a}_{1},\ldots ,{a}_{n}}\right) \]\n\nis expressible by a formula of \( {\mathcal{L}}_{0} \) . Using the fact that \( {M}_{\alpha } \) is a set we can then define\n\n\[ \beta = \mathop{\sup }\limits_{{{a}_{1},\ldots ,{a}_{n} \in {M}_{\alpha }}}{\mu }_{{\b... | Yes |
Theorem 7.17. For each formula \( \varphi \left( {{a}_{0},\ldots ,{a}_{n}}\right) \) of \( \mathcal{L} \) there exists a seminormal function \( F \) such that\n\n\[ \left( {\forall \alpha }\right) \left( {\forall \beta }\right) \lbrack \beta = F\left( \alpha \right) \rightarrow \left( {\forall {a}_{1},\ldots ,{a}_{n} \... | Proof. From Theorem 7.16\n\n\[ \left( {\exists \beta \geq \alpha }\right) \left( {\forall {a}_{1},\ldots ,{a}_{n} \in {M}_{\alpha }}\right) \lbrack \mathbf{M} \vDash \left( {\exists x}\right) \varphi \left( {x,{a}_{1},\ldots ,{a}_{n}}\right) \leftrightarrow \left( {\exists a \in {M}_{\beta }}\right) \left\lbrack {\math... | Yes |
Theorem 7.19. For each formula \( \varphi \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) of \( \mathcal{L} \) there are finitely many semi-normal functions \( {F}_{1},\ldots ,{F}_{n} \) such that\n\n\[ \left( {\forall \beta }\right) \lbrack \beta = {F}_{1}\left( \beta \right) = \cdots = {F}_{m}\left( \beta \right) \rightar... | Proof. (By induction on the number of logical symbols in \( \varphi \) .) If \( \varphi \) is atomic the theorem follows from the fact that \( \left( {\forall \alpha }\right) \left\lbrack {{\mathbf{M}}_{\alpha } \subseteq \mathbf{M}}\right\rbrack \) . If \( \varphi \) is of the form \( \neg \psi \) or \( \psi \land \et... | Yes |
Theorem 7.21. M satisfies the Axiom of Separation. | Proof. If \( \varphi \left( {x,{y}_{1},\ldots ,{y}_{n}}\right) \) is a formula of \( \mathcal{L} \), if \( a,{a}_{1},\ldots ,{a}_{n} \in M \), and if\n\n\[ A \triangleq a \cup \left\{ {a,{a}_{1},\ldots ,{a}_{n}}\right\} \]\n\nthen \( A \) is a subset of \( M \) . Thus \( \left( {\exists \alpha }\right) \left\lbrack {A ... | Yes |
Theorem 7.22. M satisfies the Power Set Axiom. | Proof. If \( a \in M \) then \( \mathcal{P}\left( a\right) \cap M \) is a subset of \( M \) . Hence\n\n\[ \left( {\exists \alpha }\right) \left\lbrack {\mathcal{P}\left( a\right) \cap M \subseteq {M}_{\alpha } \in M}\right\rbrack .\n\]\n\nSince, by Theorem 7.21, \( \mathbf{M} \) satisfies the Axiom of Separation and \(... | Yes |
Theorem 7.23. M satisfies the Axiom Schema of Replacement. | Proof. If \( \varphi \left( {{a}_{0},\ldots ,{a}_{n}}\right) \) is a formula of \( \mathcal{L} \) such that\n\n\[ \left( {\forall {a}_{2},\ldots ,{a}_{n} \in M}\right) \left( {\forall x \in M}\right) \left( {\exists !y \in M}\right) \left\lbrack {\mathbf{M} \vDash \varphi \left( {x, y,{a}_{2},\ldots ,{a}_{n}}\right) }\... | Yes |
1. \( {L}_{K} \) is definable in \( {\mathcal{L}}_{0}\left( {\{ K\left( \right) \} }\right) \) . | 1. Obvious from Theorem 7.10 and the definition of \( {L}_{K} \) . | No |
Theorem 7.27. If \( M \) is a standard transitive model of \( {ZF} \) in the language \( {\mathcal{L}}_{0}\left( {\{ K\left( \right) \} }\right) \) and if \( {On} \subseteq M \), then \( {L}_{K} \subseteq M \) . | Proof. We prove by induction that \( {A}_{\alpha } \in M \) . Clearly \( {A}_{0} = 0 \in M \) . If \( {A}_{\alpha } \in M \) then \( {\bar{K}}_{\alpha } = {A}_{\alpha } \cap K \in M \), hence \( {A}_{\alpha } = \left\langle {{A}_{\alpha },{\bar{K}}_{\alpha }}\right\rangle \in M \) . By Theorem 7.10, \( {Df}\left( {\mat... | Yes |
Theorem 7.31. If \( \mathbf{A} \) is a structure and \( F \) a set of Skolem functions such that for every formula \( \left( {\exists x}\right) \varphi \left( {x,{a}_{1},\ldots ,{a}_{n}}\right) \) of the language of \( \mathbf{A} \) there exists in \( F \) a Skolem function for that formula with respect to \( \mathbf{A... | Proof. By induction on the number of logical symbols in \( \varphi \) . If \( \varphi \) is atomic or of the form \( \neg \psi \) or \( \psi \land \eta \) the conclusion is obvious. If \( \varphi \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) is \( \left( {\exists x}\right) \psi \left( {x,{a}_{1},\ldots ,{a}_{n}}\right) \)... | Yes |
Theorem 7.32. If \( A \) is a transitive set, if \( k \in A \) and if \( \langle A, \in, k\rangle \) is a model of \( {ZF} + V = {L}_{k} \) then \( \left( {\exists \alpha }\right) \left\lbrack {A = {A}_{\alpha }}\right\rbrack \) where \( {A}_{\alpha } \) is as in Definition 7.24. | Proof. Since \( {L}_{k} = \mathop{\bigcup }\limits_{{\alpha \in {On}}}{A}_{\alpha } \) and \( {A}_{\alpha } \) is absolute with respect to \( A \) for each \( \alpha \in A \cap {On}, A = \mathop{\bigcup }\limits_{{\alpha \in A \cap {On}}}{A}_{\alpha } \) . Furthermore, because \( A \) is transitive, \( A \cap {On} = \m... | Yes |
Theorem 7.33. If \( {k}_{0} \) is the transitive closure of \( k \) then\n\n\[ V = {L}_{k} \rightarrow \left( {\forall \alpha }\right) \left\lbrack {{\overline{\bar{k}}}_{0} \leq {\aleph }_{\alpha } \rightarrow {2}^{{\aleph }_{\alpha }} = {\aleph }_{\alpha + 1}}\right\rbrack . | Proof. If \( V = {L}_{k} \) then \( k \in {L}_{k} \) . Let \( F \) be a countable family of Skolem functions, with respect to \( {L}_{k} \), for all formulas of the language \( {\mathcal{L}}_{0}\left( {\{ k\left( \right) \} }\right) \) . If \( a \subseteq \aleph \alpha ,{\bar{k}}_{0} \leq {\aleph }_{\alpha } \) and\n\n... | Yes |
Theorem 7.38. 1. \( L\left\lbrack {K;F}\right\rbrack \) and \( L\left\lbrack F\right\rbrack \) are definable in \( {\mathcal{L}}_{0}\left( {\{ K\left( \right), F\left( \right) \} }\right) \) and \( {\mathcal{L}}_{0}\left( {\{ F\left( \right) \} }\right) \) respectively. | Proof. 1. Similar to Theorem 7.10 for a language with constants \( K\left( \right) \) and \( F\left( \right) \) . | No |
Theorem 7.39. If \( M \) is a standard transitive model of \( {ZF} \) in the language \( {\mathcal{L}}_{0}\left( {\{ a\left( \right) \} }\right) \) such that\n\n1. \( {On} \subseteq M \) ,\n\n2. \( a \in M \) ,\n\nthen \( L\left\lbrack a\right\rbrack \subseteq M \) . | Proof. If \( {a}_{0} \) is the transitive closure of \( a \) then since \( a \in M \) and \( M \) is a model of \( {ZF},{a}_{0} \in M \) and \( {a}_{0} \) is the transitive closure of \( a \) in \( M \) . Since \( M \) is transitive \( a \subseteq M \) and \( {a}_{0} \subseteq M \) . Also since the rank function is abs... | Yes |
Theorem 7.40. If \( a \) has a well ordering in \( L\left\lbrack a\right\rbrack \) then \( L\left\lbrack a\right\rbrack \) satisfies the \( {AC} \) . | Proof. The proof is similar to that of Theorem 7.28 and is left to the reader. | No |
Theorem 7.41. 1. \( a \subseteq L \rightarrow {L}_{a} = L\left\lbrack a\right\rbrack \) . 2. \( L\left\lbrack a\right\rbrack \vDash V = L\left\lbrack a\right\rbrack \) . | Proof. 1. If \( a \subseteq L \) then \( a \subseteq {L}_{a} \) since \( L \subseteq {L}_{a} \) . Therefore \( a \in {L}_{a} \) . Since \( L\left\lbrack a\right\rbrack \) is the least standard transitive model of \( {ZF} \) that contains \( a \) and all the ordinals as elements we have \( L\left\lbrack a\right\rbrack \... | No |
Theorem 8.5. If \( t \) is a term in \( \varphi \) such that \( {\operatorname{Ord}}^{1}\left( \varphi \right) = g\left( t\right) \) then \( t \) does not occur in any other abstraction term of \( \varphi \) . | Proof. If \( {t}_{1} = {\widehat{x}}^{\alpha }\psi \left( {x}^{\alpha }\right) \) is an abstraction term such that \( t \) occurs in \( {t}_{1} \) and \( {t}_{1} \) occurs in \( \varphi \), then \( t \) occurs in \( \psi \left( {x}^{\alpha }\right) \) . Hence by Definition 8.4\n\n\[ g\left( {t}_{1}\right) = {2\alpha } ... | Yes |
Theorem 8.6. \( t \in {T}_{\alpha } \rightarrow \operatorname{Ord}\left( {\varphi \left( t\right) }\right) < \operatorname{Ord}\left( {\left( {\forall {x}^{\alpha }}\right) \varphi \left( {x}^{\alpha }\right) }\right) \) . | Proof. If \( t \in {T}_{\alpha } \) then \( \rho \left( t\right) < \alpha \) . Hence\n\n\[ g\left( t\right) < {2\alpha } + 1 = g\left( {\forall {x}^{\alpha }}\right) . \]\n\nTherefore\n\n\[ {\operatorname{Ord}}^{1}\left( {\varphi \left( t\right) }\right) \leq {\operatorname{Ord}}^{1}\left( {\left( {\forall {x}^{\alpha ... | Yes |
Theorem 8.8. \( {t}_{1} \in {T}_{\alpha } \rightarrow \operatorname{Ord}\left( {t \simeq {t}_{1}}\right) < \operatorname{Ord}\left( {t \in {\widehat{x}}^{\alpha }\varphi \left( {x}^{\alpha }\right) }\right) \) . | Proof. If \( {t}_{1} \in {T}_{\alpha } \) we obtain, as in the proof of Theorem 8.6\n\n\[ \n{\operatorname{Ord}}^{1}\left( {t \simeq {t}_{1}}\right) \leq {\operatorname{Ord}}^{1}\left( {t \in {\widehat{x}}^{\alpha }\varphi \left( {x}^{\alpha }\right) }\right) .\n\]\n\nAgain we need only consider the case\n\n\[ \n{\alph... | Yes |
Theorem 8.10. \\[\n\\operatorname{rank}\\left( {k}_{1}\\right) < \\operatorname{rank}\\left( {k}_{2}\\right) \\rightarrow \\operatorname{Ord}\\left( {t \\simeq {\\underline{k}}_{1}}\\right) < \\operatorname{Ord}\\left( {t \\in {\\underline{k}}_{2}}\\right) .\n\\] | Proof. If \\( \\operatorname{rank}\\left( {k}_{1}\\right) < \\operatorname{rank}\\left( {k}_{2}\\right) \\) then\n\n\\[\n{\\operatorname{Ord}}^{1}\\left( {t \\simeq {\\underline{k}}_{1}}\\right) = \\max \\left( {g\\left( t\\right), g\\left( {\\underline{k}}_{1}\\right) }\\right) \\leq \\max \\left( {g\\left( t\\right),... | Yes |
Theorem 8.11. \( \\operatorname{rank}\\left( k\\right) \\leq \\rho \\left( t\\right) \\rightarrow \\operatorname{Ord}\\left( {t \\simeq k}\\right) < \\operatorname{Ord}\\left( {P\\left( t\\right) }\\right) \) . | Proof. \( {\\operatorname{Ord}}^{1}\\left( {P\\left( t\\right) }\\right) = g\\left( t\\right) = {\\operatorname{Ord}}^{1}\\left( {t \\simeq \\underline{k}}\\right) \), since \( \\operatorname{rank}\\left( k\\right) \\leq \\rho \\left( t\\right) \). \[ {\\operatorname{Ord}}^{2}\\left( {P\\left( t\\right) }\\right) = 1 \... | Yes |
Theorem 9.6. \( \left( {\forall a \in {M}_{\alpha }}\right) \left\lbrack {\lbrack \left\lbrack {\exists x \in a)\varphi \left( x\right) }\right\rbrack = \mathop{\sum }\limits_{{x \in {M}_{\alpha }}}\llbracket x \in a\rrbracket \llbracket \varphi \left( x\right) \rrbracket }\right\rbrack \) . | Proof. If \( a \in {M}_{\alpha } \) then \( a \in {M}_{\alpha + 1} \) and hence \( a \) is defined over \( {M}_{\alpha } \) . \n\n\[ \llbracket \left( {\exists x \in a}\right) \varphi \left( x\right) \rrbracket = \mathop{\sum }\limits_{{x \in M}}\llbracket x \in a\rrbracket \llbracket \varphi \left( x\right) \rrbracket... | Yes |
Theorem 9.7. \( \left( {\forall a \in {M}_{\alpha }}\right) \left\lbrack {\lbrack \left( {\forall x \in a}\right) \varphi \left( x\right) \rbrack = \mathop{\prod }\limits_{{x \in {M}_{\alpha }}}\left( {\llbracket x \in a\rrbracket \Rightarrow \llbracket \varphi \left( x\right) \rrbracket }\right) }\right\rbrack \) . | Remark. Theorem 9.7 follows from duality. | No |
Theorem 9.8. M satisfies the Axiom of Extensionality i.e., \n\n\\[ \n\\left( {\\forall a, b \\in M}\\right) \\left\\lbrack {\\lbrack \\lbrack \\left( {\\forall x}\\right) \\left\\lbrack {x \\in a \\leftrightarrow x \\in b}\\right\\rbrack \\rightarrow a = b\\rbrack = 1}\\right\\rbrack . \n\\] | Proof. If \\( a, b \\in M \\) then \\( \\left( {\\exists \\alpha }\\right) \\left\\lbrack {a \\in {M}_{\\alpha } \\land b \\in {M}_{\\alpha }}\\right\\rbrack \\) . Then from Theorem 9.7 \n\n\\[ \n\\llbracket \\left( {\\forall x}\\right) \\left\\lbrack {x \\in a \\leftrightarrow x \\in b}\\right\\rbrack \\rrbracket = \\... | Yes |
Theorem 9.9. M satisfies the Axiom of Unions i.e., \n\n\[ \n\\left( {\\forall a \\in M}\\right) \\llbracket \\left( {\\exists b}\\right) \\left( {\\forall x}\\right) \\left\\lbrack {x \\in b \\leftrightarrow \\left( {\\exists y \\in a}\\right) \\left\\lbrack {x \\in y}\\right\\rbrack }\\right\\rbrack = 1. \n\] | Proof. If \( a \\in {M}_{\\alpha } \) then \( \\exists b \\in {M}_{\\alpha + 1} \) such that \n\n\[ \n\\left( {\\forall {x}^{\\prime } \\in {M}_{\\alpha }}\\right) \\left\\lbrack {{\\left\\lbrack \\left( \\exists y \\in a\\right) \\left\\lbrack {x}^{\\prime } \\in y\\right\\rbrack \\right\\rbrack }_{{\\mathbf{M}}_{\\al... | Yes |
Theorem 9.11. M satisfies the Axiom of Infinity. | Proof. Left to the reader. | No |
Theorem 9.12. The function \( F : {On} \rightarrow B \) defined by\n\n\[ F\left( \beta \right) = \mathop{\sum }\limits_{{a \in {M}_{\beta }}}\left\lbrack {\varphi \left( a\right) }\right\rbrack \]\n\nis nondecreasing, with respect to the Boolean relation \( \leq \) of \( \mathbf{B} \), and it is continuous. | Proof. Obvious. | No |
Theorem 9.13. If \( F : {On} \rightarrow B \) is nondecreasing, then\n\n\[ \left( {\exists \beta }\right) \left( {\forall \alpha \geq \beta }\right) \left\lbrack {F\left( \alpha \right) = F\left( \beta \right) }\right\rbrack \]\n\ni.e., \( F \) is eventually constant. | Proof. If \( g\left( b\right) = {\mu }_{\beta }\left( {b \leq F\left( \beta \right) }\right), b \in B \) then \( g : B \rightarrow {On} \) . Since \( B \) is a set, \( \left( {\exists \beta }\right) \left\lbrack {\beta = \sup {g}^{cc}B}\right\rbrack \) . Then \( \left( {\forall \alpha \geq \beta }\right) \left\lbrack {... | Yes |
Theorem 9.14. \( \left( {\exists \beta }\right) \left\lbrack {\lbrack \left( {\exists y}\right) \varphi \left( y\right) \rbrack = \mathop{\sum }\limits_{{a \in {M}_{\beta }}}\llbracket \varphi \left( a\right) \rrbracket }\right\rbrack \) . | Proof. Theorems 9.12 and 9.13. | No |
Theorem 9.16. For each formula \( \varphi \) of \( \mathcal{L} \) there exists a semi-normal function \( F \) such that\n\n\[ \left( {\forall \alpha }\right) \left( {\forall \beta }\right) \left\lbrack {\beta = F\left( \alpha \right) \rightarrow \left( {\forall {a}_{1},\ldots ,{a}_{n} \in {M}_{\alpha }}\right) \left\lb... | Proof. If\n\n\[ F\left( \alpha \right) \triangleq {\mu }_{\gamma }\left( {\gamma \geq \alpha \land \left( {\forall {a}_{1},\ldots ,{a}_{n} \in {M}_{\alpha }}\right) \left\lbrack \left\lbrack {\left( {\exists y}\right) \varphi \left( {{a}_{1},\ldots ,{a}_{n}, y}\right) }\right\rbrack \right. }\right.\n\n\[ \left. \left.... | Yes |
Theorem 9.18. For each formula \( \varphi \) of \( \mathcal{L} \) there exist finitely many seminormal functions \( {F}_{1},\ldots ,{F}_{m} \) such that\n\n\[ \left( {\forall \beta }\right) \lbrack \beta = {F}_{1}\left( \beta \right) = \cdots = {F}_{m}\left( \beta \right) \rightarrow \left( {\forall {a}_{1},\ldots ,{a}... | Proof. Left to the reader. | No |
Theorem 9.20. If \( a \in {M}_{\alpha + 1} \) and \( \llbracket \left( {\forall x}\right) \left\lbrack {x \in b \leftrightarrow x \in a \land \varphi \left( {x,{a}_{1},\ldots ,{a}_{n}}\right) }\right\rbrack \rrbracket \) \( = \mathbf{1} \) then \( b \) is defined over \( {M}_{\alpha } \), i.e., every definable B-valued... | Proof. Under the hypothesis of the proposition\n\n\[ \llbracket c \in b\rrbracket = \llbracket c \in a\rrbracket \llbracket \varphi \left( {c,{a}_{1},\ldots ,{a}_{n}}\right) \rrbracket \]\n\n\[ = \mathop{\sum }\limits_{{{c}^{\prime } \in {M}_{\alpha }}}\left\lbrack {c = {c}^{\prime }}\right\rbrack \left\lbrack {{c}^{\p... | Yes |
Theorem 9.21. If \( {b}_{1},{b}_{2} \in M \) are defined over \( {M}_{\alpha } \) then\n\n\[ \n\left( {\forall {x}^{\prime } \in {M}_{\alpha }}\right) \left\lbrack {\left\lbrack {{x}^{\prime } \in {b}_{1}}\right\rbrack = \left\lbrack \left\lbrack {{x}^{\prime } \in {b}_{2}}\right\rbrack \right\rbrack }\right\rbrack \ri... | Proof.\n\n\[ \n\llbracket x \in {b}_{1}\rrbracket = \mathop{\sum }\limits_{{{x}^{\prime } \in {M}_{\alpha }}}\llbracket x = {x}^{\prime }\rrbracket \llbracket {x}^{\prime } \in {b}_{1}\rrbracket \n\]\n\n\[ \n= \mathop{\sum }\limits_{{{x}^{\prime } \in {M}_{\alpha }}}\left\lbrack {x = {x}^{\prime }}\right\rbrack \left\l... | Yes |
Theorem 9.22. \( \\left( {\\forall \\alpha }\\right) \\left( {\\exists \\beta }\\right) \\left( {\\forall a \\in M}\\right) \\left\\lbrack {a\\text{ is defined over }{M}_{\\alpha } \\rightarrow \\left( {\\exists b \\in {M}_{\\beta }}\\right) \\left\\lbrack {\\llbracket a = b\\rrbracket = 1}\\right\\rbrack }\\right\\rbr... | Proof. \( \\;{B}^{{M}_{\\alpha }} \) is a set. For \( s \\in {B}^{{M}_{\\alpha }} \) we define\n\n\( f\\left( s\\right) \\triangleq {\\mu }_{\\beta }\\left( {\\left( {\\exists a \\in {M}_{\\beta }}\\right) \\left\\lbrack {a\\text{ is defined over }{M}_{\\alpha } \\land s = \\left\\{ {\\langle x,\\llbracket x \\in a\\rr... | Yes |
Theorem 9.23. \( \left( {\forall \alpha }\right) \left( {\exists b \in M}\right) \left( {\forall a \in {M}_{\alpha }}\right) \left\lbrack {\llbracket a \in b\rrbracket = 1}\right\rbrack \) . | Proof. \( \left( {\exists b \in {M}_{\alpha + 1}}\right) \left( {\forall a \in {M}_{\alpha }}\right) \left\lbrack {\llbracket a \in b\rrbracket = \llbracket a = a{\rrbracket }_{{\mathbf{M}}_{\alpha }} = \mathbf{1}}\right\rbrack \) . | No |
Theorem 9.24. M satisfies the Axiom of Powers i.e., \n\n\\[ \n\\left( {\\forall a \\in M}\\right) \\left\\lbrack {\\lbrack \\lbrack \\left( {\\exists x}\\right) \\left\\lbrack {x = \\mathcal{P}\\left( a\\right) }\\right\\rbrack \\rbrack = 1}\\right\\rbrack . \n\\] \n | Proof. If \\( a \\in M \\) then \\( \\left( {\\exists \\alpha }\\right) \\left\\lbrack {a \\in {M}_{\\alpha }}\\right\\rbrack \\) . By Theorem 9.22 \n\n\\( \\left( {\\exists \\beta }\\right) \\left( {\\forall b}\\right) \\left\\lbrack {b\\text{ is defined over }{M}_{\\alpha } \\rightarrow \\left( {\\exists {b}^{\\prime... | Yes |
Theorem 9.28. \( \forall {t}_{1},{t}_{2} \in {T}_{\alpha } \). 1. \( {\left\lbrack {t}_{1} = {t}_{1}\right\rbrack }_{{\mathbf{T}}_{\alpha }} = \mathbf{1} \). 2. \( \llbracket {t}_{1} = {t}_{2}{\rrbracket }_{{\mathbf{T}}_{\alpha }} = \llbracket {t}_{2} = {t}_{1}{\rrbracket }_{{\mathbf{T}}_{\alpha }} \). | Proof. Obvious from Definition 9.27. | No |
Theorem 9.29. \( \forall {k}_{1},{k}_{2} \in R\left( \alpha \right) \). \[ {k}_{1} = {k}_{2} \rightarrow {\left\lbrack {k}_{1} = {k}_{2}\right\rbrack }_{{\mathbf{T}}_{\alpha }} = \mathbf{1} \] \[ {k}_{1} \neq {k}_{2} \rightarrow {\left\lbrack {k}_{1} = {k}_{2}\right\rbrack }_{{\mathbf{T}}_{\alpha }} = \mathbf{0}. \] | Proof. If \( {k}_{1} \neq {k}_{2} \) and \( \beta \triangleq \max \left( {\rho \left( {k}_{1}\right) ,\rho \left( {k}_{2}\right) }\right) \) then by symmetry we can assume \( \left( {\exists k}\right) \left\lbrack {k \in {k}_{1} \land k \notin {k}_{2}}\right\rbrack \). Then \( \rho \left( k\right) < \beta \), and \[ {\... | Yes |
Theorem 9.32. \( \forall {t}_{1},{t}_{2},{t}_{3} \in {T}_{\alpha } \)\n\n1. \( {\left\lbrack {t}_{1} = {t}_{2}\right\rbrack }_{{\mathbf{T}}_{\alpha }}{\left\lbrack {t}_{2} \in {t}_{3}\right\rbrack }_{{\mathbf{T}}_{\alpha }} \leq {\left\lbrack {t}_{1} \in {t}_{3}\right\rbrack }_{{\mathbf{T}}_{\alpha }} \). | Proof.\n\n1. \( \llbracket {t}_{1} = {t}_{2}\rrbracket \llbracket {t}_{2} \in {t}_{3}\rrbracket \leq \mathop{\sum }\limits_{{t \in {T}_{\alpha }}}\llbracket {t}_{1} = {t}_{2}\rrbracket \llbracket t = {t}_{2}\rrbracket \llbracket t \in {t}_{3}\rrbracket \)\n\n\[ \leq \mathop{\sum }\limits_{{t \in {T}_{\alpha }}}\left\lb... | Yes |
Theorem 9.33. \( {T}_{\alpha } \) satisfies the Axiom of Extensionality. | Proof. Obvious from Definition 9.27.6. | No |
Theorem 9.34. If \( t \in {T}_{\alpha + 1} \) then \( t \) is defined over \( {T}_{\alpha } \) . | Proof. See remark following Theorem 9.30. | No |
Theorem 9.35. If \( \varphi \) is a formula of \( \mathcal{L} \) then\n\n\[ \n\left( {\forall {t}_{1},\ldots ,{t}_{n} \in {T}_{\alpha }}\right) \left( {\exists {t}^{\prime } \in {T}_{\alpha + 1}}\right) \left( {\forall t \in {T}_{\alpha }}\right) \left\lbrack {{\left\lbrack \varphi \left( t,{t}_{1},\ldots ,{t}_{n}\righ... | Proof. If \( {t}^{\prime } \triangleq {\widehat{x}}^{\alpha }{\varphi }^{\alpha }\left( {{x}^{\alpha },{t}_{1},\ldots ,{t}_{n}}\right) \) where \( {\varphi }^{\alpha } \) is the formula obtained from \( \varphi \) by replacing \( \forall x \) by \( \forall {x}^{\alpha } \), then \( {t}^{\prime } \in {T}_{\alpha + 1} \)... | Yes |
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