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Lemma 16.13. \( {\mathcal{L}}_{2} \vDash \mathbf{B}{v}_{0} \land {v}_{1} \subseteq {v}_{0} \rightarrow \mathbf{B}{v}_{1} \) . | Proof. Assume that \( \mathbf{B}x \) and \( y \subseteq x \) . Thus \( \mathbf{J}x \) and \( \mathbf{C}x \) . Hence by \( {16.10},\mathbf{J}y \) . To check that \( \mathbf{C}y \), let \( z \) be given; we want to show that \( y \sim z \) has its usual sense. But clearly \( \forall w\left( {w \in y{ \cap }_{x}\left( {x ... | Yes |
Lemma 16.14. \( {\mathcal{L}}_{2} \vDash {\mathrm{{Cv}}}_{0} \rightarrow \operatorname{COp}\left( {{v}_{0},{v}_{1}}\right) \) . | Proof. Assume that \( \mathbf{C}x, y \) is arbitrary, \( z = \mathbf{{Op}}\left( {x, y}\right) \), and \( w \) is arbitrary; we want to show that \( z \sim w \) has its usual sense. If \( y \in w \), then \( \forall u(u \in x \sim w \) iff \( u \in z \) and \( u \notin w \) ), since \( \mathbf{C}x \) . If \( y \notin w... | No |
Lemma 16.19. \( {\mathcal{L}}_{3} \vDash \mathbf{J}{v}_{0} \land \mathbf{W}{v}_{0} \rightarrow \mathbf{{WU}}{v}_{0} \) . | Proof. Assume that \( \mathbf{J}x \) and \( \mathbf{W}x \) . To check \( \mathbf{W}\mathbf{U}x \), we consider separately the three conjuncts in the definition of \( \mathbf{W} \) . If \( y \in \mathbf{U}x \), then \( y \in x \) or \( y = x \) . In the first case, \( \neg \exists z\left( {y \in z \in y}\right) \) since... | Yes |
Lemma 16.20. \( {\mathcal{L}}_{3} \vDash \mathbf{\Omega }{v}_{0} \leftrightarrow \mathbf{\Omega }\mathbf{U}{v}_{0} \) . | Proof. First assume \( \mathbf{\Omega }x \) ; thus \( \mathbf{B}x \), Trans \( x,\forall y \in x \) (Trans \( y \) ), and \( \mathbf{W}x \) . Hence \( \mathbf{{BU}}x \) and \( \mathbf{{WU}}x \) by 16.15 and 16.19. Suppose \( z \in y \in \mathbf{U}x \) . Then \( z \in y \in x \) or \( z \in y = x \), so \( z \in x \) si... | Yes |
Lemma 16.21. \( {\mathcal{L}}_{3} \vDash \) Trans \( {v}_{0} \land \neg \left( {{v}_{0} \in {v}_{0}}\right) \land \mathbf{U}{v}_{0} \equiv \mathbf{U}{v}_{1} \rightarrow {v}_{0} \equiv {v}_{1} \) . | Proof. Assume that Trans \( x, x \notin x \), and \( \mathbf{U}x = \mathbf{U}y \) . Take any \( z \in x \) . Then \( z \in \mathbf{U}x = \mathbf{U}y \), so \( z \in y \) or \( z = y \) ; we shall show that \( z = y \) is impossible. Assume \( z = y \) ; thus \( y \in x \) . Since \( x \in \mathbf{U}x \), we have \( x \... | Yes |
Lemma 16.22. \( {\mathcal{L}}_{3} \vDash \Omega {v}_{0} \land {v}_{1} \in {v}_{0} \rightarrow \Omega {v}_{1} \) . | Proof. Assume \( \mathbf{\Omega }x \) and \( y \in x \) . Since Trans \( x \), it follows that \( y \subseteq x \) . Thus \( \mathbf{W}y \) by 16.18, and \( \mathbf{B}y \) by 16.13. Clearly Trans \( y \) . If \( z \in y \), then \( z \in x \) since Trans \( x \), and hence Trans \( z \) . Thus \( \mathbf{\Omega }y \) . | Yes |
Lemma 16.23. \( {\mathcal{L}}_{3} \vDash \) Trans \( {v}_{0} \land \mathrm{J}{v}_{0} \land \mathrm{W}{v}_{0} \land \neg \left( {{v}_{0} = 0}\right) \rightarrow \exists {v}_{1}\left( {{v}_{0} = \mathrm{U}{v}_{1}}\right) \) . | Proof. Assume Trans \( x,\mathbf{J}x,\mathbf{W}x \) and \( x \neq 0 \) . Since \( \mathbf{W}x \), choose \( y \in x \) so that \( \forall w \in x\left( {w \in y\text{or}w = y}\right) \) . We shall show that \( x = \mathbf{U}y \) ; the inclusion \( x \subseteq \mathbf{U}y \) has just been mentioned. If \( w \in y \), th... | Yes |
Lemma 16.25. \( {\mathcal{L}}_{4} \vDash \mathrm{s}{v}_{0} \equiv \mathrm{s}{v}_{1} \rightarrow {v}_{0} \equiv {v}_{1} \) . | Proof. Assume that \( \mathbf{s}x = \mathbf{s}y \) . If \( \mathbf{\Omega }x \) or \( \mathbf{\Omega }y \), then \( \mathbf{\Omega }x \) and \( \mathbf{\Omega }y \) by 16.24 and 16.20, and hence \( x = y \) by 16.21 . If neither \( \mathbf{\Omega }x \) nor \( \mathbf{\Omega }y \), then \( x = y \) by 16.24. | No |
Lemma 16.26. \( {\mathcal{L}}_{4} \vDash \neg 0 = \mathrm{s}{v}_{0} \) . | Proof. If \( \mathbf{\Omega }x \), then \( \mathbf{s}x = \mathbf{U}x \neq 0 \) by 16.24, since \( x \in \mathbf{U}x \) . If (not \( \mathbf{\Omega }x \) ), then \( \mathbf{s}x = x \), and \( x \neq 0 \) since \( \mathbf{\Omega }0 \) by 16.17. | No |
Lemma 16.27. \( {\mathcal{L}}_{4} \vDash \neg {v}_{0} = 0 \rightarrow \exists {v}_{1}\left( {{v}_{0} = \mathrm{s}{v}_{1}}\right) \) . | Proof. Assume that \( x \neq 0 \) . If \( \mathbf{\Omega }x \), the desired conclusion is clear by 16.23 and 16.20. If (not \( \mathbf{\Omega }x \) ), then \( x = \mathbf{s}x \) . | No |
Lemma 16.30. \( {\mathcal{L}}_{5} \vDash \mathbf{B}{v}_{0} \land \mathbf{D}{v}_{0} \leftrightarrow \mathbf{{BOp}}\left( {{v}_{0},{v}_{1}}\right) \land \mathbf{{DOp}}\left( {{v}_{0},{v}_{1}}\right) \) | Proof. Assume that \( \mathbf{B}x \) and \( \mathbf{D}x \), while \( y \) is arbitrary. Thus \( \mathbf{{BOp}}\left( {x, y}\right) \) by 16.15. To check \( \operatorname{\mathbf{D} \mathbf{O} \mathbf{p} }\left( {x, y}\right) \), let \( z \subseteq \operatorname{\mathbf{O} \mathbf{p} }\left( {x, y}\right) \) . Choose w ... | Yes |
Lemma 16.32. \( {\mathcal{L}}_{6} \vDash \mathbf{\Omega }{v}_{0} \rightarrow \mathbf{\sigma }\left( {{v}_{0},\mathbf{0},{v}_{0}}\right) \) . | Proof. Assume that \( \mathbf{\Omega }x \) . Then \( \{ \left( {0, x}\right) \} \) is easily seen to satisfy the necessary conditions on \( {v}_{3} \) in 16.31. | No |
Lemma 16.33. \( {\mathcal{L}}_{6} \vDash \sigma \left( {{v}_{0},{v}_{1},{v}_{2}}\right) \land \sigma \left( {{v}_{0},{v}_{1},{v}_{3}}\right) \rightarrow {v}_{2} \equiv {v}_{3} \) . | Proof. Assume that \( \mathbf{\sigma }\left( {x, y, z}\right) \) and \( \mathbf{\sigma }\left( {x, y, w}\right) \) . Let \( f \) and \( g \) be the functions mentioned in \( \mathbf{\sigma }\left( {x, y, z}\right) \) and \( \mathbf{\sigma }\left( {x, y, w}\right) \) respectively. Now we aim to prove\n\n(1)\n\n\[ \opera... | Yes |
Lemma 16.34. \( {\mathcal{L}}_{6} \vDash \sigma \left( {{v}_{0},{v}_{1},{v}_{2}}\right) \rightarrow \sigma \left( {{v}_{0},\mathbf{U}{v}_{1},\mathbf{U}{v}_{2}}\right) \) . | Proof. Assume \( \sigma \left( {x, y, z}\right) \), and let \( f \) be the function mentioned in \( \sigma \left( {x, y, z}\right) \) . Then \( g = \mathbf{{Op}}\left( {f,\left( {\mathbf{U}y,\mathbf{U}z}\right) }\right) \) shows that \( \mathbf{\sigma }\left( {x,\mathbf{U}y,\mathbf{U}z}\right) \), making use of 16.20 a... | Yes |
Lemma 16.35. \( {\mathcal{L}}_{6} \vDash \sigma \left( {{v}_{0},\mathbf{U}{v}_{1},{v}_{2}}\right) \rightarrow \exists {v}_{3}\left\lbrack {{v}_{2} = \mathbf{U}{v}_{3} \land \sigma \left( {{v}_{0},{v}_{1},{v}_{3}}\right) }\right\rbrack \) . | Proof. Assume that \( \mathbf{\sigma }\left( {x,\mathbf{U}y, z}\right) \), and let \( f \) be the function mentioned in \( \mathbf{\sigma }\left( {x,\mathbf{U}y, z}\right) \) . Since \( y \in \mathbf{{UU}}y \) and \( \mathbf{{Dmn}}\left( {f,\mathbf{{UU}}y}\right) \), we can choose \( w \) so that \( \left( {y, w}\right... | Yes |
Lemma 16.37. \( {\mathcal{L}}_{7} \vDash \neg \Omega \{ 1\} \) . | Proof. We have \( 0 \in 1 \in \{ 1\} \) but \( 0 \notin \{ 1\} \), so Trans \( \{ 1\} \) is false. Hence \( \mathbf{\Omega }\{ \mathbf{1}\} \) does not hold. | Yes |
Lemma 16.38. \( {\mathcal{L}}_{7} \vDash {v}_{0} + 0 = {v}_{0} \) | Proof. Let \( x \) be given. If \( \mathbf{\Omega }x \), then \( \mathbf{\sigma }\left( {x,0, x}\right) \) by \( {16.32} \), so \( x + 0 = x \) . If \( \mathbf{\Omega }x \) fails, then there is no \( z \) with \( \mathbf{\sigma }\left( {x,0, z}\right) \), and hence \( x + 0 = x \) is clear from \( {16.36}\left( {ii}\ri... | Yes |
Lemma 16.39. \( {\mathcal{L}}_{7} \vDash {v}_{0} + \mathrm{s}{v}_{1} = \mathrm{s}\left( {{v}_{0} + {v}_{1}}\right) \) . | Proof. Let \( x \) and \( y \) be given. If not \( \mathbf{\Omega }x \), then \( x + \mathbf{s}y = x \) and \( \mathbf{s}\left( {x + y}\right) = x \) . So assume that \( \mathbf{\Omega }x \) . If not \( \mathbf{\Omega }y \), then \( \mathbf{s}y = y \) and for no \( z \) do we have \( \sigma \left( {x, y, z}\right) \), ... | Yes |
Lemma 16.47. \( {\mathcal{L}}_{9} \vDash {v}_{0} \cdot \mathbf{s}{v}_{1} = {v}_{0} \cdot {v}_{1} + {v}_{0} \) . | Proof. Let \( x \) and \( y \) be given. We consider four cases.\n\nCase 1. \( \neg \mathbf{\Omega }x \) . Clearly \( \mathbf{s}y \neq 0 \), so \( x \cdot \mathbf{s}y = \{ 1\} \) . Also, \( x \cdot y = 0 \) or \( x \cdot y = \{ 1\} \) .\n\nSubcase 1. \( x \cdot y = 0 \) . Then by \( {16.36}, x \cdot y + x = \{ 1\} \), ... | Yes |
Theorem 16.51. The theory of one binary relation is finitely inseparable. | Proof. The theorem is to be interpreted as saying that the theory \( {\Gamma }^{\prime } \) in the language \( {\mathcal{L}}^{\prime } \) is finitely inseparable, where \( {\mathcal{L}}^{\prime } \) is a language with just one nonlogical constant, namely a binary relation symbol \( \mathbf{S} \), and \( {\Gamma }^{\pri... | No |
Theorem 16.52. If \( {\mathcal{L}}^{\prime } \) is a language with at least one relation symbol which is at least binary, then \( \left\{ {\varphi \in {\operatorname{Sent}}_{\mathcal{L}} : { \vDash }_{\varphi }}\right\} \) is finitely inseparable. | Proof. We interpret the theory of one binary relation into \( {\mathcal{L}}^{\prime } \), i.e., we apply 15.17 as in the proof of 16.51. The following list gives the required definitions. The hypotheses of 15.17 are then clearly satisfied.\n\n\( \mathcal{L} \) : language with a single binary relation symbol \( \mathbf{... | Yes |
Theorem 16.53. The theory of a binary operation is finitely inseparable. | Proof. Again we interpret the theory of a binary relation. The following list and construction outline the procedure (cf. the proofs of 16.51 and 16.52).\n\n\\( \\mathcal{L} \\) : language with a single binary relation symbol \\( \\mathbf{R} \\) .\n\n\\( {\\mathcal{L}}^{\\prime } \\) : language with a single binary ope... | Yes |
Theorem 16.54. If \( {\mathcal{L}}^{\prime } \) is a language with at least one operation symbol which is at least binary, then \( \left\{ {\varphi : \varphi \in {\operatorname{Sent}}_{\mathcal{L}},{ \vDash }_{\varphi }}\right\} \) is finitely inseparable. | The proof is similar to the proof of 16.52 . To take care of the remaining class of undecidable languages we need some auxiliary results which are interesting in themselves. | No |
Theorem 16.55. The theory of a symmetric binary relation is finitely inseparable. | Proof. The following list and diagram outline the proof:\n\n\\( \\mathcal{L} \\) : language with a single binary relation symbol \\( \\mathbf{R} \\) ;\n\n\\( \\Gamma = \\left\\{ {\\varphi \\in {\\operatorname{Sent}}_{\\mathcal{L}} : \\vDash \\varphi }\\right\\} \)\n\n\\( {\\Gamma }^{\\prime } = \\left\\{ {\\varphi \\in... | Yes |
Theorem 16.56. The theory of two equivalence relations over the universe is finitely inseparable. | Proof. We interpret a symmetric binary relation into this theory:\n\n\\( \\mathcal{L} \\) : language with a single binary relation symbol \\( \\mathbf{R} \\) ;\n\n\\( \\Gamma = \\left\\{ {\\varphi \\in {\\operatorname{Sent}}_{\\mathcal{L}} : \\Theta \\vDash \\varphi }\\right\\} \\), where \\( \\Theta = \\left\\{ {\\for... | Yes |
Theorem 16.57. The theory of two total functions is finitely inseparable. | Proof. We interpret the theory of 16.56 into this theory:\n\n\\( \\left( {{\\mathcal{L}}^{\\prime },{\\Gamma }^{\\prime }}\\right) \\) : as in the proof of 16.56.\n\n\\( {\\mathcal{L}}^{\\prime \\prime } \\) : language with two unary operation symbols \\( \\mathbf{O} \\) and \\( \\mathbf{P} \\) ;\n\n\\( {\\Gamma }^{\\p... | Yes |
Theorem 16.58. If \( \mathcal{L} \) is a language with at least two unary operation symbols, then \( \mathcal{L} \) is finitely inseparable. | In Theorems 16.52,16.54 and 16.58 we have shown that if \( \mathcal{L} \) satisfies any one of the following three conditions, then \( \left\{ {\varphi \in {\operatorname{Sent}}_{\mathcal{L}} : { \vDash }_{\varphi }}\right\} \) is finitely inseparable and hence undecidable:\n\n(1) \( \mathcal{L} \) has at least one rel... | No |
Theorem 17.2 (Löb). Let \( \Gamma \) be a strong theory in a language \( \mathcal{L} \). Assume that \( \Pr \) is a formula of \( \mathcal{L} \) such that \( \operatorname{Fv}\Pr \subseteq \left\{ {v}_{0}\right\} \) and the following conditions hold for all sentences \( \psi ,\chi \) of \( \mathcal{L} \):\n\n(i) if \( ... | Proof. There clearly is an elementary function \( f \) such that for any formula \( \psi \) and any \( m \in \omega, f\left( {{g}^{ + }\psi, m}\right) = {\mathcal{g}}^{ + }\psi \left( \mathbf{m}\right) \). Let \( \mathbf{O} \) be a binary operation symbol of \( \mathcal{L} \) which represents \( f \), in the sense of 1... | Yes |
Lemma 17.6. For each operation symbol \( \mathbf{O} \) of \( {\mathcal{L}}_{\mathrm{{el}}} \), either \( {}^{\# }\mathbf{O} \in \omega \) if \( \mathbf{O} \) is \( 0 \) -ary, or #O is an m-ary operation on \( \omega \) if \( \mathbf{O} \) is m-ary. In either case, for any \( {x}_{0},\ldots ,{x}_{m - 1}, y \in \omega \)... | The proof of this lemma is a straight-forward induction on \( \mathbf{O} \), following Definition 17.4. | No |
Lemma 17.7. If \( \mathbf{O} \) is an m-ary operation symbol of \( {\mathcal{L}}_{\mathrm{{el}}} \), then \( {}^{\# }{\mathbf{R}}_{\mathbf{o}} \) is an m-ary relation on \( \omega \) and for any \( {x}_{0},\ldots ,{x}_{m - 1} \in \omega \) , | \[ \left( {{x}_{0},\ldots ,{x}_{m - 1}}\right) \in {}^{\# }{\mathbf{R}}_{\mathbf{0}}\text{implies}\mathcal{P} \vDash {\mathbf{R}}_{\mathbf{0}}\left( {\Delta {x}_{0},\ldots ,\Delta {x}_{m - 1}}\right) \text{,} \] \[ \left( {{x}_{0},\ldots ,{x}_{m - 1}}\right) \notin {}^{\# }{\mathbf{R}}_{\mathbf{0}}\text{implies}\mathca... | Yes |
Lemma 17.8. If \( \Gamma \vDash \psi \), then \( \Gamma \vDash \Pr \Delta {g}^{ + }\psi \) . | Proof. Assume that \( \Gamma \vDash \psi \) . Thus there is a \( \Delta \) -proof \( \varphi \) with last member \( \psi \) .\n\nHence \( \left( {{g}^{ + }\psi ,{g}^{+ + }\varphi }\right) \in \operatorname{Prf} \), and \( \operatorname{Prf} = {\# }^{ \circ }\operatorname{Prf} \), so by 17.7\n\n\[ \mathcal{P} \vDash {}^... | Yes |
Lemma 17.9. \( \Gamma \vDash \Pr \Delta {g}^{ + }\left( {\chi \rightarrow \psi }\right) \rightarrow \left( {\Pr \Delta {g}^{ + }\chi \rightarrow \Pr \Delta {g}^{ + }\psi }\right) \) | Proof. From the choice of \( {}^{ \circ }{Prf} \), which is supposed to mimic the definition of \( \operatorname{Prf} \) (see Proposition 10.27), we have\n\n\[ \Gamma \vDash \operatorname{Prf}\left( {{v}_{0},{v}_{1}}\right) \leftrightarrow \forall {v}_{2}\left\{ {{v}_{2} \leq {}^{ \circ }1{v}_{1} \rightarrow \left\lbra... | Yes |
Proposition 18.5. Let \( {\mathcal{L}}^{\prime } \) be a rich expansion of \( \mathcal{L} \) by \( C \) . Let \( S \) be the set of all formally consistent sets \( \Gamma \) of sentences of \( {\mathcal{L}}^{\prime } \) such that \( \mid \{ \mathbf{c} \in C : \mathbf{c} \) occurs in some \( \varphi \in \Gamma \} \left|... | Proof. We need to check conditions (C0)-(C11) for an arbitrary \( \Gamma \in S \) . (C0) and (C1) are clear since \( \Gamma \) is consistent. (C2) follows from 18.2. Since \( \varphi \land \psi \in \Gamma \) implies that \( \Gamma \vDash \varphi \) and \( \Gamma \vDash \psi \) ,(C3) is clear. For (C4), assume that \( \... | Yes |
Proposition 18.6. Let \( {\mathcal{L}}^{\prime } \) be a rich expansion of \( \mathcal{L} \) by \( C \), where \( \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| = \) \( {\aleph }_{0} \) . Let \( S \) be the set of all sets \( \Gamma \) of sentences such that \( \Gamma \) has a model \( {\left( \mathfrak{A},{a}_{\mathrm{... | This proposition is clear. | No |
Lemma 18.7. If \( S \) satisfies conditions (C1)-(C9), then \( {S}^{\prime } = \{ \Gamma : \Gamma \subseteq \Delta \) for some \( \Delta \in S\} \) satisfies (C0)-(C9). | Proof. Assume that \( \Gamma \in {S}^{\prime } \), say \( \Gamma \subseteq \Delta \in S \) . Both (C0) and (C1) are clear for \( \Gamma \) . To check (C2), suppose \( \neg \varphi \in \Gamma \) . Thus \( \neg \varphi \in \Delta \), so by (C2) for \( S \) , \( \Delta \cup \{ \varphi \rightarrow \} \in S \) . Now \( \Gam... | No |
Let \( {\mathcal{L}}^{\prime } \) be a rich expansion of \( \mathcal{L} \) by \( C \), assume that \( \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| = {\aleph }_{0} \), and let \( S \) satisfy (C0)-(C9). Let \( \left\langle {{f}_{\alpha } : \alpha < \omega }\right\rangle \) be a family of admissible functions over \( S ... | Finally, using 18.7 we obtain the simplest form of the model existence theorem: | No |
Corollary 18.12. Let \( {\mathcal{L}}^{\prime } \) be a rich expansion of \( \mathcal{L} \) by \( C \), assume that \( \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \) \( = {\aleph }_{0} \), and let \( S \) satisfy (C1)-(C9). Then each \( \Gamma \in S \) has a model \( {\left( \mathfrak{A},{a}_{\mathrm{c}}\right) }_{\m... | Applying 18.9 to the consistency family given in 18.5 , we obtain the completeness theorem 11.19. Theorem 18.9 will be used for many of our later results, as we have mentioned. | No |
Proposition 18.17. Let \( F \) be a filter over \( I \) with \( 0 \notin F \) . Then the following conditions are equivalent:\n\n(i) \( F \) is an ultrafilter.\n\n(ii) for all \( a, b \subseteq I \), if \( a \cup b \in F \), then \( a \in F \) or \( b \in F \) .\n\n(iii) for any filter \( G \) on \( I \), if \( F \subs... | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Assume that \( a \cup b \in F \) while \( a \notin F \) . Then by \( \left( i\right), I \sim \) \( a \in F \) . Now \( \left( {I \sim a}\right) \cap \left( {a \cup b}\right) \subseteq b \), so \( b \in F \) . (ii) \( \Rightarrow \) (iii). Say \( a \in G \sim... | Yes |
Proposition 18.18. If \( F \) is a collection of subsets of \( I \neq 0 \) with the finite intersection property, then there is an ultrafilter \( G \) such that \( F \subseteq G \) . | Proof. Let \( \mathcal{A} \) be the collection of all filters \( G \) such that \( F \subseteq G \) and \( 0 \notin G \) . Then \( \mathcal{A} \) is nonempty; for, let \( H = \{ x \subseteq I \) : there exist \( m \in \omega \) and \( y \in {}^{m}F \) with \( \left. {{y}_{0} \cap \cdots \cap {y}_{m - 1} \subseteq x}\ri... | Yes |
Proposition 18.20. Under the assumptions of 18.19, \( \bar{F} \) is an equivalence relation on \( {\mathrm{P}}_{i \in I}{A}_{i} \) . Let \( \mathfrak{B} = {\mathrm{P}}_{i \in I}{\mathfrak{A}}_{i} \) . Then\n\n(i) if \( \mathbf{O} \) is an m-ary operation symbol, and if \( {x}_{t}\bar{F}{y}_{t} \) for all \( t < m \), t... | Proof. The proof is routine, and we just check \( \left( i\right) \) and the transitivity of \( \bar{F} \) as examples. Assume that \( x\bar{F}y\bar{F}z \) . Thus \( \left\{ {i \in I : {x}_{i} = {y}_{i}}\right\} \in F \) and \( \left\{ {i \in I : {y}_{i} = {z}_{i}}\right\} \) \( \in F \) . But\n\n\[ \left\{ {i \in I : ... | Yes |
Proposition 18.25. If \( I \) is a finite set, then every ultrafilter over \( I \) is principal. | Proof. Let \( F \) be an ultrafilter over \( I \) . Then \( \mathop{\bigcup }\limits_{{i \in I}}\{ i\} = I \in F \), so by \( {18.13}\left( {ii}\right) \) , since \( I \) is finite, there is an \( i \in I \) such that \( \{ i\} \in F \) . Then, in fact, \( F = \) \( \{ a \subseteq I : i \in a\} \) . For, if \( i \in a ... | Yes |
Proposition 18.26. If \( I \) is an infinite set, then there is a non-principal ultrafilter over I. | Proof. Let \( F = \{ I \sim \{ i\} : i \in I\} \) . Since \( I \) is infinite, it is clear that \( F \) has the finite intersection property. Let \( G \) be an ultrafilter containing \( F \), by 18.18. Clearly \( G \) is nonprincipal. | No |
Corollary 18.31. If a sentence \( \varphi \) holds in \( {\mathfrak{A}}_{i} \) for each \( i \in I \), then \( \varphi \) holds in \( {P}_{i \in I}{\mathfrak{A}}_{i}/\bar{F} \) | Assume that each finite subset \( \Delta \) of \( \Gamma \) has a model \( {\mathfrak{A}}_{\Delta } \) . Let \( I = \{ \Delta : \Delta \) is a finite subset of \( \Gamma \} \) . For each \( \Delta \in I \) let \( {G}_{\Delta } = \{ \Theta : \Delta \subseteq \Theta \in I\} \) . If \( {\Delta }_{0},\ldots \) , \( {\Delta... | Yes |
Proposition 18.32. If \( n \in \omega \) and \( \left\{ {i \in I : \left| {A}_{i}\right| = n}\right\} \in F \), then \( \left| {\mathop{\bigcap }\limits_{{i \in I}}{A}_{i}/\bar{F}}\right| = n \) . | Proof. Let \( \varphi \) be a sentence (involving equality only) which holds in a structure \( \mathfrak{B} \) iff \( \left| B\right| = n \) . Thus \( \left\{ {i \in I : \varphi }\right. \) holds in \( \left. {\mathfrak{A}}_{i}\right\} \in F \), so by \( {18.30},\varphi \) holds in \( {\mathrm{P}}_{i \in I}{\mathfrak{A... | No |
Proposition 18.35. If \( I \) is infinite, then \( \left| {\{ J : J \subseteq I, J\text{finite}\} }\right| = \left| I\right| \) . | Proof. For each \( m \in \omega \) and \( f \in {}^{m}I \) let \( {Ff} = \operatorname{Rng}f \) . Thus \( F \) maps \( \mathop{\bigcup }\limits_{{m \in \omega }}{}^{m}I \) onto \( \{ J \subseteq I : J \) finite \( \} \), so\n\n\[ \left| {\{ J \subseteq I : J\text{ finite }\} }\right| \leq \left| {\mathop{\bigcup }\limi... | Yes |
Proposition 18.36. If \( I \) is an infinite set, then there is a regular ultrafilter over I. | Proof. Let \( J = \{ X \subseteq I : X \) is finite \( \} \) . It clearly suffices to prove the proposition for \( J \) in place of \( I \), since \( \left| I\right| = \left| J\right| \) by 18.35. For each \( X \in J \) let \( {G}_{X} = \{ Y \in J : X \subseteq Y\} \) . If \( {X}_{0},\ldots ,{X}_{m - 1} \in J \), then\... | Yes |
Theorem 18.37. If \( A \) and \( I \) are infinite sets and \( F \) is a regular ultrafilter over \( I \), then \( \left| {{}^{I}A/\bar{F}}\right| \geq {2}^{\left| I\right| } \) . | Proof. By Definition 18.34 choose \( E \subseteq F \) so that \( \left| E\right| = \left| I\right| \) and \( \cap G = 0 \) whenever \( G \subseteq E \) and \( G \) is infinite. For each \( i \in I \) let \( {H}_{i} = \{ a \in E : i \in a\} \) . Thus by our choice of \( E,{H}_{i} \) cannot be infinite; hence we can choo... | Yes |
Proposition 19.3. If \( {\mathfrak{A}}_{i}{ \equiv }_{\mathrm{{ee}}}{\mathfrak{B}}_{i} \) for each \( i \in I \), and \( F \) is an ultrafilter over \( I \) , then \( \mathop{\bigcap }\limits_{{i \in I}}{\mathfrak{A}}_{i}/\bar{F}{ \equiv }_{\mathrm{{ee}}}\mathop{\bigcap }\limits_{{i \in I}}{\mathfrak{B}}_{i}/\bar{F} \)... | Proof. Let \( \varphi \) be a sentence which holds in \( \mathop{P}\limits_{{i \in I}}{\mathfrak{A}}_{i}/\bar{F} \) . Then by the basic theorem on ultraproducts, \( \left\{ {i \in I : {\mathfrak{A}}_{i} \vDash \varphi }\right\} \in F \) . But \( \left\{ {i \in I : {\mathfrak{A}}_{i} \vDash \varphi }\right\} = \left\{ {... | Yes |
Proposition 19.7. If \( f \) is an embedding of \( \mathfrak{A} \) into \( \mathfrak{B} \), then there is an \( \mathcal{L} \)-structure \( \mathfrak{C} \) and an isomorphism \( g \) of \( \mathfrak{C} \) onto \( \mathfrak{B} \) such that \( \mathfrak{A} \subseteq \mathfrak{C} \) and \( f \subseteq g \). | Proof. Let \( C = A \cup \{ \left( {A, x}\right) : x \in B \sim D\} \). Note that \( A \cap \{ \left( {A, x}\right) : x \in B \sim D\} = 0 \); for if \( \left( {A, x}\right) \in A \), then \( A \in \{ A\} \in \left( {A, x}\right) \in A \), contradicting the regularity axiom of set theory. Define \( g : C \rightarrow B ... | Yes |
Proposition 19.9. Let \( \mathfrak{A} \) and \( \mathfrak{B} \) be \( \mathcal{L} \) -structures with \( f \) an embedding of \( \mathfrak{A} \) into \( \mathfrak{B} \) . Let \( {\mathcal{L}}^{\prime } \) be an \( A \) -expansion of \( \mathcal{L} \) . Then \( {\left( \mathfrak{B}, fa\right) }_{a \in A} \) is a model o... | Proof. The proof is essentially trivial, and we only illustrate it by verifying that a member \( \varphi = \neg {\mathbf{{Rc}}}_{a0}\cdots {\mathbf{c}}_{a\left( {m - 1}\right) } \) of the diagram holds in \( {\left( \mathfrak{V}, fa\right) }_{a \in A} \) , where \( \mathbf{R} \) is an \( m \) -ary relation symbol of \(... | Yes |
Proposition 19.10. Let \( \mathfrak{A} \) be an \( \mathcal{L} \) -structure, \( {\mathcal{L}}^{\prime } \) an \( A \) -expansion of \( \mathcal{L} \), and \( {\left( \mathfrak{B},{l}_{a}\right) }_{a \in A} \) a model of the \( {\mathcal{L}}^{\prime } \) -diagram of \( \mathfrak{A} \) . Then \( l \) is an embedding of ... | Proof. Again the proof is almost trivial, and we will just check that \( l \) is one-one and that it preserves operations. If \( a, b \in A \) and \( a \neq b \), then \( \neg {\mathbf{c}}_{a} = {\mathbf{c}}_{b} \) is in the diagram of \( \mathfrak{A} \), so \( {\left( \mathfrak{B},{l}_{a}\right) }_{a \in A} \vDash \ne... | Yes |
Proposition 19.12. Let \( \mathfrak{A} \) be an \( \mathcal{L} \) -structure, \( 0 \neq X \subseteq A \), and let \( B \) be the subuniverse of \( \mathfrak{A} \) generated by \( X \) . Then \( \left| B\right| \leq \left| X\right| + \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \) . | Proof. Let \( {C}_{0} = X \), and for each \( m \in \omega \) let\n\n\( {C}_{m + 1} = {C}_{m} \cup \left\{ {{\mathbf{O}}^{\mathfrak{A}}x : \mathbf{O}}\right. \) is an \( n \) -ary operation symbol of \( \mathcal{L} \) and \( \left. {x \in {}^{n}{C}_{m}}\right\} \) .\n\nClearly \( B = \mathop{\bigcup }\limits_{{m \in \o... | Yes |
Theorem 19.13 (Henkin’s embedding theorem). Let \( \mathbf{K} \) be the class of all models of a set \( \Gamma \) of sentences and let \( \mathfrak{A} \) be an \( \mathcal{L} \) -structure. Suppose that every finitely generated substructure of \( \mathfrak{A} \) can be embedded in a member of \( \mathbf{K} \) . Then \(... | Proof. Let \( {\mathcal{L}}^{\prime } \) be an \( A \) -expansion of \( \mathcal{L} \), and let \( \Delta \) be the \( {\mathcal{L}}^{\prime } \) -diagram of \( \mathfrak{A} \) . Then every finite subset \( \Theta \) of \( \Gamma \cup \Delta \) has a model. In fact, let \( {\mathbf{c}}_{b0},\ldots ,{\mathbf{c}}_{b\left... | Yes |
Corollary 19.15. If \( \mathfrak{A} \preccurlyeq \mathfrak{B} \), then \( \mathfrak{A} \subseteq \mathfrak{B} \) . | Proof. Let \( \mathbf{O} \) be an \( m \) -ary operation symbol, and let \( {a}_{0},\ldots ,{a}_{m} \in A \) . Then\n\n\[ \n{\mathbf{O}}^{\mathfrak{A}}\left( {{a}_{0},\ldots ,{a}_{m - 1}}\right) = {a}_{m}\;\text{ iff }\mathfrak{A} \vDash \mathbf{O}{v}_{0}\cdots {v}_{m - 1} = {v}_{m}\left\lbrack {{a}_{0},\ldots ,{a}_{m}... | Yes |
Proposition 19.16. Let \( \mathfrak{A} \) and \( \mathfrak{B} \) be \( \mathcal{L} \) -structures, and assume that \( \mathfrak{A} \subseteq \mathfrak{B} \) . Then the following two conditions are equivalent: (i) \( \mathfrak{A} \preccurlyeq \mathfrak{B} \) ; (ii) for every formula \( \varphi \), every \( k \in \omega ... | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . By the hypothesis of \( \left( {ii}\right) \) and the meaning of \( \left( i\right) ,\mathfrak{A} \vDash \) \( \exists {v}_{k}\varphi \left\lbrack x\right\rbrack \) . Hence by the definition of satisfaction there is an \( a \in A \) such that \( \mathfrak{A}... | Yes |
Proposition 19.20. Let \( {\mathcal{L}}^{\prime } \) be a Skolem expansion of \( \mathcal{L} \), and let \( \mathfrak{B} \) be an \( {\mathcal{L}}^{\prime } \) structure which is a model of the Skolem set of \( {\mathcal{L}}^{\prime } \) over \( \mathcal{L} \). Then for any \( {\mathcal{L}}^{\prime } \) -structure \( \... | Proof. The implication \( \left( {ii}\right) \Rightarrow \left( i\right) \) is trivial. Now assume \( \left( i\right) \). We shall apply 19.16 in order to prove (ii). To this end, assume that \( x \in {}^{\omega }A \) and \( \mathfrak{B} \vDash \exists {v}_{i}\varphi \left\lbrack x\right\rbrack \). Then, since \( \math... | Yes |
Proposition 19.23. Let \( \mathfrak{A} \) be an \( \mathcal{L} \) -structure, \( I \) a non-empty set, and \( F \) an ultrafilter over \( I \) . For each \( a \in A \) set \( {fa} = {\left\lbrack \langle a : i \in I\rangle \right\rbrack }_{\bar{F}} \) . Then \( f \) is an elementary embedding of \( \mathfrak{A} \) into... | Proof. First, \( f \) is one-one. For, suppose \( {fa} = {fb} \) . Thus \( \langle a : i \in I\rangle \bar{F}\langle b : i \in I\rangle \) , i.e., \( \{ i \in I : a = b\} \in F \) . But \( \{ i : a = b\} \) is either empty or all of \( I \) depending upon whether \( a \neq b \) or \( a = b \) respectively. Since \( 0 \... | Yes |
Theorem 19.24 (Upward Löwenheim-Skolem theorem). Let \( \mathrm{m} \) be an infinite cardinal \( \geq \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \), and let \( \mathfrak{A} \) be an infinite \( \mathcal{L} \) -structure with \( \left| A\right| \leq \mathfrak{m} \) . Then \( \mathfrak{A} \) has an elementary extensio... | Proof. By 18.36 find 18.37 let \( I \) be a set and \( F \) an ultrafilter over \( I \) such that \( \left| {{}^{I}A/\bar{F}}\right| \geq \mathfrak{m} \) . By 19.23 we obtain an elementary embedding of \( \mathfrak{A} \) into \( {}^{I}\mathfrak{A}/\bar{F} \) , and by 19.22 we obtain an elementary extension (c) of \( \m... | Yes |
Corollary 19.25. If a set \( \Gamma \) of sentences has an infinite model, then \( \Gamma \) has a model of each cardinality \( \geq \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \) . | From the compactness theorem it is easy to see that if \( \Gamma \) has models of arbitrarily large finite cardinalities, then \( \Gamma \) has an infinite model, and hence by 19.25 has models of each cardinality \( \geq \left| {\mathrm{{Fmla}}}_{\mathcal{L}}\right| \) . | No |
Theorem 19.28. Let \( R \) be a Hanf system. Then there is a cardinal \( \mathfrak{m} \) such that for all \( \Gamma \in {\operatorname{pr}}_{0}^{ * }R \), condition (i) implies condition (ii): (i) there exist \( \mathfrak{A},\mathfrak{n} \) with \( \left( {\Gamma ,\mathfrak{A},\mathfrak{n}}\right) \in R \) and \( \mat... | Proof. For each \( \Gamma \in {\operatorname{pr}}_{0}^{ * }R \), let \[ {f}_{\Gamma } = 0\;\text{ if }\forall \mathfrak{p}\exists \mathfrak{A}\exists \mathfrak{n}\left\lbrack {\left( {\Gamma ,\mathfrak{A},\mathfrak{n}}\right) \in R\text{ and }\mathfrak{n} \geq \mathfrak{p}}\right\rbrack . \] \( {f}_{\Gamma } = \) least... | Yes |
Corollary 19.30. For any first-order language \( \mathcal{L} \), the Hanf number of \( {\mathrm{H}}_{\mathcal{L}} \) is \( {\aleph }_{0} \). | This corollary is just a restatement of the upward Löwenheim-Skolem theorem in a general framework. | No |
Proposition 19.32. Let \( \mathfrak{A} \) be an \( \mathcal{L} \) -structure, \( \mathfrak{B} \) an elementary extension of \( \mathfrak{A} \) , and \( {\mathcal{L}}^{\prime } \) an A-expansion of \( \mathcal{L} \) . Then \( {\left( \mathfrak{B}, a\right) }_{a \in A} \) is a model of the elementary \( {\mathcal{L}}^{\p... | Proof. Let the new constants of \( {\mathcal{L}}^{\prime } \) be \( {\mathbf{c}}_{a} \) for \( a \in A \) . The following is easily proved from the definition of satisfaction:\n\n(1)\n\nfor any \( x \in {}^{\omega }A \), any \( m \in \omega \), and any formula \( \varphi \) of \( \mathcal{L} \) such that\n\[ \text{Fv}\... | Yes |
Proposition 19.33. Let \( \mathfrak{A} \) be an \( \mathcal{L} \) -structure, \( {\mathcal{L}}^{\prime } \) an \( A \) -expansion of \( \mathcal{L} \), and \( {\left( \mathfrak{B}, la\right) }_{a \in A} \) a model of the elementary \( {\mathcal{L}}^{\prime } \) -diagram \( \Gamma \) of \( \mathfrak{A} \) . Then \( l \)... | Proof. Again, let the new constants of \( {\mathcal{L}}^{\prime } \) be \( {\mathbf{c}}_{a} \) for \( a \in A \) . By Proposition 19.10, \( l \) is an embedding of \( \mathfrak{A} \) into \( \mathfrak{B} \) . Now for any \( \mathcal{L} \) -formula \( \varphi \), say with Fv \( \varphi \subseteq \left\{ {{v}_{0},\ldots ... | Yes |
Theorem 19.36 (Tarski). If \( \mathbf{K} \) is a set of \( \mathcal{L} \) -structures directed by \( \preccurlyeq \), then \( \mathfrak{A} \preccurlyeq \bigcup \mathbf{K} \) for each \( \mathfrak{A} \in \mathbf{K} \) . | Proof. We proceed by induction on formulas to show that for every formula \( \varphi \), every \( \mathfrak{A} \in \mathbf{K} \), and every \( x \in {}^{\omega }A,\mathfrak{A} \vDash \varphi \left\lbrack x\right\rbrack \) iff \( \cup \mathbf{K} \vDash \varphi \left\lbrack x\right\rbrack \) . The case \( \varphi \) atom... | Yes |
Proposition 20.1. There is an infinite nonstandard natural number. | Proof. Define \( x \in {}^{\omega }\omega \) by \( {xi} = i \) for all \( i \in \omega \) . For each \( m \in \omega ,\{ i : m < {xi}\} = \) \( \omega \sim m \in F \) . Hence \( \left\lbrack {\langle m : i \in \omega \rangle }\right\rbrack < \left\lbrack x\right\rbrack \) in \( {}^{\omega }\langle \mathbb{R}, < \rangle... | Yes |
For any finite nonstandard real number \( r \) there is a unique standard real number \( s \) such that \( r * - s \) is infinitesimal. | Proof. We may assume that \( {r}^{ * } \geq 0 \) . Since \( r \) is finite, there is a natural number \( m \) such that \( {r}^{ * } \leq m \) . Let \( s \) be the inf of \( \{ t : t \) is a real number and \( r * \leq t\} \) . Thus \( s \) is a standard real number. If \( r * - s \) is not infinitesimal, choose a stan... | Yes |
Proposition 20.5. For any \( x \in {}^{\omega }\mathbb{R} \) and \( s \in \mathbb{R} \) the following conditions are equivalent:\n\n(i) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = s \) ;\n\n(ii) \( {}^{ * }{x}_{n} \simeq s \) for every infinite natural number \( n \) . | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Let \( \varepsilon > 0,\varepsilon \) standard. Then by \( \left( i\right) \) there is an \( m \in \omega \) such that \( \left| {{x}_{n} - s}\right| \leq \varepsilon \) for all \( n \geq m \) . Thus the following formula \( \varphi \) is satisfied in \( \ma... | Yes |
Corollary 20.6. Assume that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = s \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{y}_{n} = t \) . Then\n\n(i) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{x}_{n} + {y}_{n}}\right) = s + t \) ;\n\n(ii) \( \mathop{\lim }\limits_{{n \rightarro... | Proof. Let \( n \) be an infinite natural number. Then by 20.5,* \( {x}_{n} * - s \) and \( {}^{ * }{y}_{n}{}^{ * } - t \) are infinitesimal. Hence \( {}^{ * }{\left( x + y\right) }_{n}{}^{ * } - \left( {s + t}\right) = {}^{ * }{x}_{n}{}^{ * } + {}^{ * }{y}_{n}{}^{ * } - \) \( s * - t \) is also infinitesimal. So (i) h... | Yes |
Proposition 20.7 (Bolzano-Weierstrauss). A bounded infinite sequence has at least one limit point. | Proof. Let \( x \in {}^{\omega }\mathbb{R} \) be bounded. Thus there is an \( M \in \mathbb{R} \) such that \( \left| {x}_{n}\right| \leq M \) for all \( n \in \omega \) . Hence \( * \left| {t\left\lbrack x\right\rbrack }\right| * \leq M \) also, since \( \omega = \left\{ {n : \left| {x}_{n}\right| \leq M}\right\} \in ... | Yes |
Proposition 20.8. Let \( x \in {}^{\omega }\mathbb{R} \) . Then the following conditions are equivalent:\n\n(i) \( x \) is a bounded sequence;\n\n(ii) \( {}^{ * }{x}_{n} \) is finite for every infinite natural number \( n \) . | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Choose a positive real number \( M \) such that \( \left| {x}_{n}\right| \leq M \) for all \( n \) . Since * \( \mathbb{R} \) is an elementary extension of \( \mathbb{R} \) for any structures over \( \mathbb{R} \) , * \( \left| {*{x}_{n}}\right| * \leq M \) ... | Yes |
Proposition 20.9. For any \( f : \mathbb{R} \rightarrow \mathbb{R} \) the following conditions are equivalent:\n\n(i) \( f \) is continuous at \( a \) ;\n\n(ii) for all \( x \simeq a \) we have \( {}^{ * }{fx} \simeq {fa} \) . | Proof. First assume \( \left( i\right) \) . Thus, in \( \mathbb{R} \), \n\n(1) for every \( \varepsilon > 0 \) there is a \( \delta > 0 \) such that for all \( x \), if \( \left| {x - a}\right| < \delta \) then \( \left| {{fx} - {fa}}\right| < \varepsilon \) .\n\nNow assume that \( x \simeq a \) . Let \( \varepsilon \)... | Yes |
Proposition 20.10. Let \( f \) be a real-valued continuous function defined on a closed interval \( \left\lbrack {a, b}\right\rbrack \), such that \( {fa} < 0 \) and \( fb > 0 \) . Then there is a \( c \in \left\lbrack {a, b}\right\rbrack \) such that \( {fc} = 0 \) . | Proof. Let \( c = \inf \{ x \in \left\lbrack {a, b}\right\rbrack : {fx} \geq 0\} \) . Thus\n\n(1) \( \forall x\left( {x < c \Rightarrow {fx} < 0}\right) \) ,\n\n(2) \( \forall x\left\lbrack {c < x \Rightarrow \exists y\left( {c \leq y < x\text{and}{fy} \geq 0}\right) }\right\rbrack \) .\n\nLet \( i \) be a positive inf... | Yes |
Lemma 20.11. Let \( f \) be a real-valued continuous function defined on a closed interval \( \left\lbrack {a, b}\right\rbrack \) . Then \( f \) is bounded. Hence \( * {fc} \) is finite for each non-standard \( c \in \left\lbrack {a, b}\right\rbrack \) . | Proof. Suppose \( f \) is not bounded. Then it is easy to define a sequence \( {x}_{0},{x}_{1},\ldots \) of members of \( \left\lbrack {a, b}\right\rbrack \) such that \( \left| {f{x}_{i}}\right| \geq i \) for all \( i \in \omega \) . Let \( n \) be an infinite natural number. Since the sentence \( \forall i\left( {\le... | Yes |
Proposition 20.12. Let \( f \) be a real-valued continuous function defined on a closed interval \( \left\lbrack {a, b}\right\rbrack \) . Then fattains a maximum value on this interval. | Proof. For any two integers \( i, j \in \omega \) with \( j \neq 0 \), let \( {x}_{ij} = a + \left( {i/j}\right) \left( {b - a}\right) \) . Let \( m \) be an infinite natural number. Now\n\n(1) for any standard \( c \in \left\lbrack {a, b}\right\rbrack \) there is an \( {i}^{ * } \leq m \) such that \( c \simeq {}^{ * ... | Yes |
Corollary 21.2. Any categorical theory is complete. | Now we show that categorical theories are trivial: | No |
Theorem 21.3. For any consistent theory \( \Gamma \), the following conditions are equivalent:\n\n(i) \( \Gamma \) is categorical;\n\n(ii) there is a positive integer \( m \) such that \( \Gamma \) is \( m \) -categorical and every model of \( \Gamma \) has power \( m \) . | Proof. Obviously \( \left( {ii}\right) \Rightarrow \left( i\right) \) . Now assume \( \left( i\right) \) . Since any two models of \( \Gamma \) are isomorphic, they all have the same cardinality \( m \), and \( \Gamma \) is \( m \) -categorical. Suppose \( m \geq {\aleph }_{0} \) . Let \( \mathfrak{A} \) be a model of ... | Yes |
Theorem 21.4 (Łoś–Vaught test for completeness). If \( \Gamma \) is a theory with only infinite models and \( \Gamma \) is categorical in some power \( m \geq \left| {\mathrm{{Fmla}}}_{\mathfrak{B}}\right| \) , then \( \Gamma \) is complete. | Proof. We need to show that any two models \( \mathfrak{A},\mathfrak{B} \) of \( \Gamma \) are elementarily equivalent. By the Löwenheim-Skolem theorems there are \( \mathcal{L} \) -structures \( {\mathfrak{A}}^{\prime },{\mathfrak{B}}^{\prime } \) of power \( m \) with \( \mathfrak{A}{ \equiv }_{\mathrm{{ee}}}{\mathfr... | Yes |
Theorem 21.5 (Cantor). Let \( A \) and \( B \) be two denumerable densely linearly ordered sets, each without first or last elements. Then \( A \cong B \) . | Proof. Let \( < \) and \( { < }^{\prime } \) be the dense linear orderings of \( A \) and \( B \) respectively. Say \( A = \left\{ {{a}_{n} : n \in \omega }\right\} \) and \( B = \left\{ {{b}_{n} : n \in \omega }\right\} \), with \( a \) and \( b \) one-one. We define a sequence \( \left\langle {\left( {{x}_{n},{y}_{n}... | Yes |
Corollary 21.6. The theory of dense linear order without first or last elements is complete and decidable. | By Theorem 9.48 we know that any two denumerable atomless Boolean algebras are isomorphic. Hence: | No |
Theorem 21.7. The theory of nontrivial atomless Boolean algebras is complete and decidable. | It is well known that for any \( m > {\aleph }_{0} \), any two divisible torsion-free Abelian groups of power \( m \) are isomorphic. Hence: | No |
Proposition 21.12. Assume that \( m \in \omega ,\mathfrak{B} \subseteq \mathfrak{A} \), and that for any \( X \subseteq B \) with \( \left| X\right| < m \) and any \( a \in A \sim B \) there is an automorphism \( f \) of \( \mathfrak{A} \) such that \( f \upharpoonright X = \mathbf{I} \upharpoonright X \) and \( f \) a... | Proof. We proceed by induction on \( \varphi \) to show that \( {\varphi }^{\mathfrak{B}} = {\varphi }^{\mathfrak{A}} \cap {}^{\omega }B \) for any formula \( \varphi \) with at most \( m \) distinct variables. The only nontrivial step is the induction step to \( \varphi = \forall {v}_{i}\psi \) . Obviously \( {\varphi... | Yes |
Lemma 21.16. Let \( \mathfrak{A} \) be a model of \( {\Gamma }_{\text{equiv }} \), and let \( m \in \omega \sim 1 \) . Then there is a model \( \mathfrak{B} \) of \( {\Gamma }_{\text{equiv }} \) such that \( \mathfrak{B}{ \preccurlyeq }_{m}\mathfrak{A} \) and each \( \mathfrak{B} \) -equivalence class contains at most ... | Proof. Let \( f \) be a choice function for nonempty sets of subsets of \( A \) . For each \( X \in A/R \) we define\n\n\[ \n{gX} = X\;\text{ if }\left| X\right| \leq m \n\]\n\n\[ \n{gX} = f\{ Y : Y \subseteq X,\left| Y\right| = m\} \;\text{ if }\left| X\right| > m. \n\]\nThen let \( B = \mathop{\bigcup }\limits_{{X \i... | Yes |
Lemma 21.17. Let \( \mathfrak{A} \) be a model of \( {\Gamma }_{\text{equiv }}, m \in \omega \sim 1 \), and suppose that each \( \mathfrak{A} \) -equivalence class has at most \( m \) elements. Then there is an \( m \) -basic model \( \mathfrak{B} \) of \( {\Gamma }_{\text{equiv }} \) such that \( \mathfrak{B}{ \preccu... | Proof. Let \( \mathfrak{A} = \langle A, R\rangle \) . For \( i \in \{ 1,\ldots, m\} \) let \( {C}_{i} = \{ X : X \in A/R,\left| X\right| = i\} \) . Let \( f \) be a choice function for nonempty sets of subsets of \( \mathbf{S}A \) . For each \( i \in \{ 1,\ldots, m\} \) let\n\n\[ \n{gi} = {C}_{i}\;\text{ if }\left| {C}... | Yes |
Theorem 21.19. \( {\Gamma }_{\text{equiv }} \) is decidable. | Proof. We first claim\n\nIf \( \varphi \) is a formula with at most \( m \) distinct variables, then \( \varphi \in {\Gamma }_{\text{equiv }} \)\n\n(1)\n\niff \( \varphi \) holds in every \( m \) -basic model of \( {\Gamma }_{\text{equiv }} \) .\n\nIn fact, \( \Rightarrow \) is trivial, while \( \Leftarrow \) is an obv... | Yes |
Proposition 21.21. Let \( \Delta \) be the smallest set of formulas containing all atomic formulas and their negations and closed under the operations \( \mathbf{v},\mathbf{\Lambda } \) and \( \forall {v}_{i} \) for each \( i < \omega \) . Then for each \( \varphi \in \Delta \) there is a universal formula \( \psi \) w... | Proof. Let \( \Theta \) be the set of \( \varphi \) for which there is such a \( \psi \) . Obviously \( \Delta \subseteq \Theta \) since \( \Theta \) is trivially seen to satisfy the above conditions on \( \Delta \) . | No |
Lemma 21.24. Let \( \mathfrak{A} \) be an \( \mathcal{L} \) -structure, \( {\mathcal{L}}^{\prime } \) an \( A \) -expansion of \( \mathcal{L},\Delta \) the \( {\mathcal{L}}^{\prime } \) -diagram of \( \mathfrak{A} \), and \( \Gamma \cup \{ \varphi \} \subseteq {\operatorname{Sent}}_{\mathcal{L}} \) . Assume that \( \Ga... | Proof. By the deduction theorem for sentences, \( \Gamma \vDash \chi \rightarrow \varphi \) for \( \chi \) a conjunction of certain members of \( \Delta \) . Let \( {\chi }^{\prime } \) be obtained from \( \chi \) by replacing all of the new individual constants \( {\mathbf{c}}_{a0},\ldots ,{\mathbf{c}}_{a\left( {m - 1... | Yes |
Lemma 21.25. Let \( \Gamma \) be a model-complete set in a language \( \mathcal{L} \), and let \( {\mathcal{L}}^{\prime } \) be an expansion of \( \mathcal{L} \) by adjoining new individual constants. Then \( \Gamma \) is model-complete in \( {\mathcal{L}}^{\prime } \) . | Proof. Let \( {\mathfrak{A}}^{\prime } = {\left( \mathfrak{A},{l}_{k}\right) }_{k \in K} \) and \( {\mathfrak{B}}^{\prime } = {\left( \mathfrak{B},{s}_{k}\right) }_{k \in K} \) be \( {\mathcal{L}}^{\prime } \) -structures which are models of \( \Gamma \) such that \( {\mathfrak{A}}^{\prime } \subseteq {\mathfrak{B}}^{\... | Yes |
Theorem 21.27. Let \( \Gamma \) be a consistent set of sentences in a language \( \mathcal{L} \) . Then the following conditions are equivalent:\n\n(i) \( \Gamma \) is model-complete;\n\n(ii) for every model \( \mathfrak{A} \) of \( \Gamma \) and every \( A \) -expansion \( {\mathcal{L}}^{\prime } \) of \( \mathcal{L},... | Proof\n\n\( \left( i\right) \Rightarrow \left( {ii}\right) \) . Assume \( \left( i\right) \), let \( \mathfrak{A} \) be a model of \( \Gamma \), let \( {\mathcal{L}}^{\prime } \) be an \( A \) -expansion of \( \mathcal{L} \), and let \( \Delta \) be \( \Gamma \cup \left( {{\mathcal{L}}^{\prime }\text{-diagram of}\mathf... | Yes |
Proposition 21.29. If \( \Gamma \) is model complete and has a prime model, then \( \Gamma \) is complete. | Proof. Let \( \mathfrak{A} \) and \( \mathfrak{B} \) be any two models of \( \Gamma \) ; we show that \( \mathfrak{A} \equiv \mathfrak{B} \) . Let \( \mathfrak{C} \) be a prime model of \( \Gamma \) . Thus there are embeddings \( f \) and \( g \) of \( \mathfrak{C} \) into \( \mathfrak{A} \) and \( \mathfrak{B} \) resp... | Yes |
Theorem 21.32. Let \( \Gamma \) be a theory satisfying the following two conditions:\n\n(i) if \( \mathfrak{A} \) and \( \mathfrak{B} \) are any two models of \( \Gamma \), then every finitely generated substructure of \( \mathfrak{A} \) is embeddable in \( \mathfrak{B} \) ;\n\n(ii) if \( \mathfrak{A} \) and \( \mathfr... | Proof. First we show that \( \Gamma \) is model-complete; to prove this we shall apply 21.27(iii). Assume, then, that \( \mathfrak{A} \) and \( \mathfrak{B} \) are models of \( \Gamma ,\mathfrak{A} \subseteq \mathfrak{B}, x \in {}^{\omega }A \) , \( \varphi \) is a universal formula, and not \( \left( {\mathfrak{B} \vD... | Yes |
Theorem 22.2 (A. Robinson’s consistency theorem). Assume that \( {\mathcal{L}}_{0},{\mathcal{L}}_{1} \) , \( {\mathcal{L}}_{2},{\mathcal{L}}_{3},{\Gamma }_{0},{\Gamma }_{1},{\Gamma }_{2} \) are given satisfying the following conditions:\n\n(i) \( {\mathcal{L}}_{0},{\mathcal{L}}_{1},{\mathcal{L}}_{2},{\mathcal{L}}_{3} \... | Proof. Suppose that \( {\Gamma }_{1} \cup {\Gamma }_{2} \) is not consistent. Then, by the compactness theorem, there are finite subsets \( {\Delta }_{1} \) and \( {\Delta }_{2} \) of \( {\Gamma }_{1} \) and \( {\Gamma }_{2} \) respectively such\n\nthat \( {\Delta }_{1} \cup {\Delta }_{2} \) has no model; and we may as... | Yes |
Theorem 22.3 (Padoa’s method). Let \( \Gamma \) be a theory in a language \( \mathcal{L} \), and let \( \mathbf{\pi } \) be a nonlogical constant of \( \mathcal{L} \). Suppose that \( \mathfrak{A} \) and \( \mathfrak{B} \) are two models of \( \Gamma \) such that \( A = B,{\sigma }^{\mathfrak{A}} = {\sigma }^{\mathfrak... | Proof. We take the case of a relation symbol \( \mathbf{\pi } \) ; operation symbols are treated similarly. Suppose there is such a theory \( \Delta \), and let \( \varphi \) be a possible definition of \( \pi \), with \( \Gamma \vDash \forall {v}_{0}\cdots \forall {v}_{m - 1}\left( {\pi {v}_{0}\cdots {v}_{m - 1} \left... | Yes |
Lemma 22.6. If \( \Gamma \) is a theory in a countable language, then \( \Gamma \) is independently axiomatizable. | Proof. Write \( \Gamma = \left\{ {{\varphi }_{i} : i \in \omega }\right\} \) . For each \( i \in \omega \) let \( {\psi }_{i} \) be the sentence \( \mathop{\bigwedge }\limits_{{j < i}}{\varphi }_{j} \rightarrow {\varphi }_{i} \), where by convention \( {\psi }_{0} \) is \( {\varphi }_{0} \) . Let \( \Delta = \left\{ {{... | Yes |
Lemma 22.7. Suppose \( \Gamma ,\Delta \subseteq {\text{Sent}}_{\mathcal{L}} \) and:\n\n(i) \( \left| \Gamma \right| \leq \left| \Delta \right| \), and \( \Gamma \cap \Delta = 0 \)\n\n(ii) for every \( \varphi \in \Delta \), not \( \left( {\Gamma \cup \Delta \sim \{ \varphi \} \vDash \varphi }\right) \)\n\nThen there is... | Proof. Let \( \psi : \Gamma \rightarrowtail \Delta \) . Let \( \Theta = \left\{ {\varphi \land {\psi }_{\varphi } : \varphi \in \Gamma }\right\} \cup \left( {\Delta \sim \operatorname{Rng}\psi }\right) \) . Then obviously \( \Gamma \cup \Delta \subseteq \{ \varphi : \Theta \vDash \varphi \} \), and \( \Theta \subseteq ... | Yes |
Theorem 22.9. (Reznikoff). Every theory is independently axiomatizable. | Proof. Again, we let \( {C\varphi } \) denote the set of all nonlogical constants occurring in \( \varphi \) . By 22.6 we may assume that our language has infinitely many nonlogical constants. Let \( \Gamma \) be a theory. We define \( \left\langle {{\Delta }_{i} : i < \omega }\right\rangle \) by recursion:\n\n\[{\Delt... | Yes |
Lemma 23.2. If \( \left( {\varphi ,{\psi }_{0},\ldots ,{\psi }_{m}}\right) \) is an \( {\mathcal{L}}_{\mathrm{{fac}}},{\mathcal{L}}_{\mathrm{{ind}}} \) -sequence then one can effectively find a partitioning \( {\mathcal{L}}_{\text{tac }},{\mathcal{L}}_{\text{ind }} \) -sequence \( \left( {\chi ,{\theta }_{0},\ldots ,{\... | Proof. Let \( n = {2}^{m + 1} - 1 \), and let \( {r}_{0},\ldots ,{r}_{n} \) be a list of all subsets of \( \{ 0,\ldots, m\} \) . For each \( k \leq n \) let\n\n\[ \n{\theta }_{k} = \mathop{\bigwedge }\limits_{{j \in {rk}}}{\psi }_{j} \land \mathop{\bigwedge }\limits_{{j \in \left( {m + 1}\right) \sim {rk}}}\neg {\psi }... | Yes |
Theorem 23.4. If \( {\mathfrak{A}}_{i}{ \equiv }_{\mathrm{{ee}}}{\mathfrak{B}}_{i} \) for each \( i \in I \), then \( {\mathrm{P}}_{i \in I}{\mathfrak{A}}_{i}{ \equiv }_{\mathrm{{ee}}}{\mathrm{P}}_{i \in I}{\mathfrak{B}}_{i} \) . | Proof. Let \( \mathfrak{C} \) be the Boolean algebra of all subsets of \( I \) . As we saw following Definition 23.1, any sentence \( \chi \) of our given language can be considered as a sentence of \( {\mathcal{L}}_{\text{prod }} \) . Thus by 23.3, \( \mathop{\bigcap }\limits_{{i \in I}}{\mathfrak{A}}_{i} \vDash \chi ... | Yes |
Theorem 23.5. If \( \mathfrak{A}{ \equiv }_{\mathrm{{ee}}}\mathfrak{B} \) and \( I \) and \( J \) are index sets such that either \( \left| I\right| = \) \( \left| J\right| < {\aleph }_{0} \) or else \( \left| I\right| ,\left| J\right| \geq {\aleph }_{0} \), then \( {}^{I}\mathfrak{A}{ \equiv }_{\mathrm{{ee}}}{}^{I}\ma... | Proof. By Theorem 21.34, the hypothesis implies that the Boolean algebras \( \mathrm{S}I \) and \( \mathrm{S}J \) are elementarily equivalent. Now if \( \psi \) is a sentence of our language, then \( {K}_{\psi }^{\langle \mathfrak{A} : i \in I\rangle } \) is \( I \) or 0, and \( {K}^{\langle \mathfrak{B} : j \in J\rang... | Yes |
Theorem 23.8. If \( \mathfrak{A} \) has a decidable theory, then so does \( {}^{I}\mathfrak{A} \) . | Proof. From 21.34, \( \mathbf{S}I \) has a decidable theory. The decision procedure for \( \left\{ {\chi : {}^{I}\mathfrak{A} \vDash \chi }\right\} \) goes as follows. Given \( \chi \), determine \( \left( {\varphi ,{\psi }_{0},\ldots ,{\psi }_{m}}\right) \) by 23.3. Note that \( {K}_{\psi k}^{\mathfrak{A}} = I \) or 0... | Yes |
Theorem 23.9. If \( \{ \chi : \mathbf{K} \vDash \chi \} \) is decidable, then so is \( \{ \chi : \mathbf{{PK}} \vDash \chi \} \), where \( \mathbf{{PK}} \) is the class of all products of members of \( \mathbf{K} \) . | Proof. The decision procedure for \( \{ \chi : \mathbf{{PK}} \vDash \chi \} \) is as follows. Given \( \chi \), let \( \left( {\varphi ,{\psi }_{0},\ldots ,{\psi }_{m}}\right) \) be determined as in 23.3. Then let \( T = \left\{ {k \leq m : \mathbf{K} \vDash \neg {\psi }_{k}}\right\} \) , which can be effectively deter... | Yes |
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