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We show \( \{ \sim \left( {\exists x}\right) \left( { \sim p}\right) \} \vdash \left( {\forall x}\right) p \) for any element \( p \in P \) . (Recall that \( \left( {\exists x}\right) \) is an abbreviation for \( \sim \left( {\forall x}\right) \sim \) .) | \n\n\( \left( {\mathcal{A}}_{3}\right) \)\n\n(assumption)\n\n\[{p}_{3} = \left( {\forall x}\right) \left( { \sim \sim p}\right) ,\;\left( {{p}_{1} = {p}_{2} \Rightarrow {p}_{3}}\right)\]\n\n\[{p}_{4} = \left( {\forall x}\right) \left( { \sim \sim p\left( x\right) }\right) \Rightarrow \sim \sim p\left( y\right) ,\]\n\n\... | Yes |
Lemma 4.2. Let \( \left( {\alpha ,\beta }\right) : \left( {{P}_{1},{V}_{1}}\right) \rightarrow \left( {{P}_{2},{V}_{2}}\right) \) be a semi-homomorphism. Let \( p \in {P}_{1} \) and suppose \( x \in {V}_{1} - \operatorname{var}\left( p\right) \) . Then \( \beta \left( x\right) \notin \operatorname{var}\left( {\alpha \l... | Proof: We observe first that if \( x \neq y \), then \( \left( {\forall x}\right) p = \left( {\forall y}\right) p \) if and only if neither \( x \) nor \( y \) is in \( \operatorname{var}\left( \mathrm{p}\right) \) . \n\nSince \( \beta \left( {V}_{1}\right) \) is infinite, there is an element \( {y}^{\prime } \in \beta... | Yes |
Theorem 4.3. (The Substitution Theorem). Let \( \left( {\alpha ,\beta }\right) : \left( {{P}_{1},{V}_{1}}\right) \rightarrow \left( {{P}_{2},{V}_{2}}\right) \) be a semi-homomorphism. Let \( A \subseteq P, p \in {P}_{1} \). (a) If \( A \vdash p \), then \( \alpha \left( A\right) \vdash \alpha \left( p\right) \). (b) If... | Proof: (a) Let \( {p}_{1},\ldots ,{p}_{n} \) be a proof of \( p \) from \( A \). We use induction over \( n \) to show that \( \alpha \left( {p}_{1}\right) ,\ldots ,\alpha \left( {p}_{n}\right) \) is a proof of \( \alpha \left( p\right) \) from \( \alpha \left( A\right) \). If \( a = \left( {\left( {\forall x}\right) \... | Yes |
Theorem 4.7. (The Soundness Theorem). Let \( A \subseteq P\left( {V,\mathcal{R}}\right), p \in P\left( {V,\mathcal{R}}\right) \) . If \( A \vdash p \), then \( A \vDash p \) . | Proof: Let \( {p}_{1},\ldots ,{p}_{n} \) be a proof of \( p \) from \( A \) . Let \( \left( {U,\varphi ,\psi, v}\right) \) be an interpretation of \( P\left( {V,\mathcal{R}}\right) \) such that \( v\left( A\right) \subseteq \{ 1\} \) . We have to show that \( v\left( p\right) = 1 \) , and we shall use induction on \( n... | Yes |
Corollary 4.8. (The Consistency Theorem). \( F \) is not a theorem of \( \operatorname{Pred}\left( {V,\mathcal{R}}\right) \) . | Proof: Let \( U \) be any non-empty set, \( \varphi : V \rightarrow U \) any function, and \( \psi \) any function on \( \mathcal{R} \) such that if \( r \in {\mathcal{R}}_{n} \), then \( \psi \left( r\right) \) is an \( n \) -ary relation on \( U \) . Then there exists \( v : P\left( {V,\mathcal{R}}\right) \rightarrow... | Yes |
Theorem 4.9. (The Deduction Theorem). Let \( A \subseteq P = P\left( {V,\mathcal{R}}\right) \) and let \( p, q \in P \) . Then \( A \vdash p \Rightarrow q \) if and only if \( A \cup \{ p\} \vdash q \) . | Proof: If \( A \vdash p \Rightarrow q \), then it follows, as in the case of the Propositional Calculus, that \( A \cup \{ p\} \vdash q \) . Suppose \( A \cup \{ p\} \vdash q \) . We shall again use induction over the length of the proof. The argument used for the case of the Propositional Calculus again applies except... | Yes |
Example 4.10. As we did before, we use the techniques of the proof of the Deduction Theorem to convert the proof \( \sim p,\left( {\forall x}\right) \left( { \sim p}\right) , \sim \left( {\forall x}\right) \left( { \sim p}\right), F \) of \( F \) from \( \{ \left( {\exists x}\right) p, \sim p\} \left( {x \notin \operat... | <table><thead><tr><th>Given proof</th><th>Comment</th><th>Corresponding Steps of Constructed Proof</th></tr></thead><tr><td>\( \sim p \)</td><td>Assumption to be eliminated</td><td>\( \sim p \Rightarrow \left( {\left( { \sim p \Rightarrow \sim p}\right) \Rightarrow \sim p}\right) , \) \( \left( { \sim p \Rightarrow \le... | Yes |
Lemma 4.13. Let \( A \) be a consistent subset of \( P\left( {V,\mathcal{R}}\right) \). Suppose \( \left( {\exists x}\right) p\left( x\right) \in A \), and \( t \notin \operatorname{Var}\left( A\right) \). Then \( F \notin \operatorname{Ded}\left( {A\cup \{ p\left( t\right) \} }\right) \). | Proof: Suppose \( F \in \operatorname{Ded}\left( {A\cup \{ p\left( t\right) \} }\right) \). Then by the Deduction Theorem, \( \sim p\left( t\right) \in \operatorname{Ded}\left( A\right) \). Since \( t \notin \operatorname{Var}\left( A\right) \), we may apply Generalisation and obtain \( \left( {\forall x}\right) \left(... | Yes |
Lemma 4.14. Let \( A \) be a consistent subset of \( P\left( {V,\mathcal{R}}\right) \) . Then there exist \( {V}^{ * } \supseteq V \) and \( {A}^{ * } \supseteq A \), where \( {A}^{ * } \subseteq P\left( {{V}^{ * },\mathcal{R}}\right) \), such that\n\n(i) \( F \notin \operatorname{Ded}\left( {A}^{ * }\right) \), and\n\... | Proof: Put \( {V}_{0} = V,{A}_{0} = A,{P}_{0} = P\left( {V,\mathcal{R}}\right) \) . We construct inductively \( {V}_{i},{P}_{i} = P\left( {{V}_{i},\mathcal{R}}\right) ,{A}_{i}^{\prime } \) and \( {A}_{i} \) for \( i > 0 \) . Taking a new variable \( {t}_{p}^{\left( i\right) } \) for each \( p \in {A}_{i} \) of the form... | Yes |
Theorem 4.16. (The Adequacy Theorem). Let \( A \subseteq P\left( {V,\mathcal{R}}\right), p \in P\left( {V,\mathcal{R}}\right) \) . If \( A \vDash p \), then \( A \vdash p \) . | Proof: If \( F \notin \operatorname{Ded}\left( {A\cup \{ \sim p\} }\right) \), then by the Satisfiability Theorem, there exists an interpretation \( \left( {U,\varphi ,\psi, v}\right) \) of \( P\left( {V,\mathcal{R}}\right) \) such that \( v\left( {A\cup \{ \sim p\} }\right) \subseteq \{ 1\} \) , which contradicts the ... | Yes |
Theorem 1.7. (The Satisfiability Theorem) Suppose \( F \notin {\operatorname{Ded}}_{\mathcal{G}}\left( A\right) \) . Then there exists a proper interpretation of \( P\left( {V,\mathcal{R}}\right) \) with \( v\left( A\right) \subseteq \{ 1\} \) . | Proof: Since \( F \notin \operatorname{Ded}\left( {A \cup I}\right) \), there exists an interpretation \( \left( {U,\varphi ,\psi, v}\right) \) of \( P = P\left( {V,\mathcal{R}}\right) \) such that \( v\left( {A \cup I}\right) = \{ 1\} \) . The relation \( \psi \mathcal{I} \) is an equivalence relation on \( U \) . For... | Yes |
(i) \( {\operatorname{Con}}_{\mathcal{G}}\left( A\right) = \operatorname{Con}\left( {A \cup I}\right) \) | The soundness and consistency of \( {\operatorname{Pred}}_{\mathcal{G}}\left( {V,\mathcal{R}}\right) \) both follow immediately from the corresponding properties of \( \operatorname{Pred}\left( {V,\mathcal{R}}\right) \) . | No |
Theorem 3.4. The first-order theory \( \mathcal{T} \) is complete if and only if every \( p \in \mathcal{L}\left( \mathcal{T}\right) \) which is true in one model of \( \mathcal{T} \) is true in every model of \( \mathcal{T} \) . | Proof: The result is trivial if \( \mathcal{T} \) is inconsistent, so we suppose \( \mathcal{T} \) consistent. Suppose that \( \mathcal{T} \) is complete, and that \( p \in \mathcal{L}\left( \mathcal{T}\right) \) is true in the model \( M \) . Since \( \sim p \) is false for \( M \), it is not a theorem of \( \mathcal{... | Yes |
Theorem 3.5. Let \( \mathcal{T} = \left( {\mathcal{R}, A, C}\right) \) be a consistent theory. Then there exists \( {A}^{\prime } \subseteq \mathcal{L}\left( \mathcal{T}\right) \), with \( {A}^{\prime } \supseteq A \) and such that \( \left( {\mathcal{R},{A}^{\prime }, C}\right) \) is consistent and complete. | Proof: Since \( \mathcal{T} \) is consistent, it has a model \( M \) say. Put \( {A}^{\prime } = \{ p \in \mathcal{L}\left( \mathcal{T}\right) \mid p \) true in \( M\} \) . Then \( {A}^{\prime } \) has the required properties. | No |
Theorem 3.14. Suppose the theory \( \mathcal{T} \) has models of arbitrarily large finite cardinal. Then \( \mathcal{T} \) has an infinite model. | Proof: Let \( \mathcal{T} = \left( {\mathcal{R}, A, C}\right) \), and put \( {\mathcal{T}}^{\prime } = \left( {\mathcal{R},{A}^{\prime }, C}\right) \) where \( {A}^{\prime } = \) \( A \cup \left\{ {\operatorname{al}\left( n\right) \mid n \in {\mathbf{N}}^{ + }}\right\} \) . We show that \( {\mathcal{T}}^{\prime } \) is... | Yes |
Corollary 3.19. If the first-order theory \( \mathcal{T} \) has an infinite model, then \( \mathcal{T} \) is not categorical. | Proof obvious. | No |
Corollary 3.20. Suppose the theory \( \mathcal{T} \) has cardinal \( \chi \), and has only infinite models. Suppose also that \( \mathcal{T} \) is categorical in some infinite cardinal \( \aleph \geq \chi \) . Then \( \mathcal{T} \) is complete. | Proof: Let \( p \in \mathcal{L}\left( \mathcal{T}\right) \), and suppose that neither \( p \) nor \( \sim p \) is a theorem of \( \mathcal{T} = \left( {\mathcal{R}, A, C}\right) \) . Since \( \sim p \) is not a theorem, \( \mathcal{T} \) has a model (infinite) in which \( p \) is true, and so the theory \( {\mathcal{T}... | Yes |
Let \( \mathcal{E} = \left( {\{ \mathcal{I}\} ,\varnothing ,\varnothing }\right) \) be the theory of equality, with \( V = {V}_{0} = \left\{ {{x}_{n} \mid n \in \mathbf{N}}\right\} \) . We write \( \mathcal{I}\left( {x, y}\right) \) as \( x = y \), and \( \sim \mathcal{I}\left( {x, y}\right) \) as \( x \neq y \) . As t... | The following set of instructions constitutes the reduction process:\n\nStep 0. If \( p \) is quantifier-free, put \( q = p \) and stop. Otherwise, express \( p \) in prenex normal form (see Exercise 4.18 of Chapter IV) \( {Q}_{1}{Q}_{2}\cdots {Q}_{r}{p}_{1} \), where the \( {Q}_{i} \) are quantifiers and \( {p}_{1} \)... | No |
Theorem 4.3. The theory \( \mathcal{E} = \left( {\{ \mathcal{I}\} ,\varnothing ,\varnothing }\right) \) is decidable. | Proof. Let \( p \in \mathcal{L}\left( \mathcal{E}\right) \) . The reduction process described above gives an element \( q \), equivalent to \( p \), which is a propositional combination of elements of \( \Pi \) such that \( \operatorname{var}\left( q\right) \subseteq \operatorname{var}\left( p\right) \) . Since \( \ope... | Yes |
Lemma 2.4. Let \( S \) be a set of \( I \) . There exists a filter on \( I \) containing \( S \) if and only if \( S \) has the finite intersection property. | Proof: The necessity of the condition is immediate, so we prove its sufficiency. Suppose \( S \) has the finite intersection property, and put\n\n\[ T = \left\{ {U \subseteq I \mid U = {J}_{1} \cap \cdots \cap {J}_{n}\text{ for some }n\text{ and some }{J}_{1},\ldots ,{J}_{n} \in S}\right\} .\n\]\n\nLet\n\n\[ \mathcal{F... | Yes |
Lemma 2.5. Let \( \mathcal{F} \) be a filter on \( I \) . Then there exists an ultrafilter \( {\mathcal{F}}^{ * } \supseteq \mathcal{F} \) on \( I \) . | Proof: The set of filters containing \( \mathcal{F} \) is an inductive set. By Zorn’s Lemma, it has a maximal member \( {\mathcal{F}}^{ * } \) . | Yes |
Theorem 3.1. Let \( F \) be a field. Then there exists an algebraic closure of \( F \) . | Proof: Let \( \mathcal{T} \) be elementary field theory augmented by the addition of the elements of \( F \) to the set of constants, and of all the relations \( {a}_{1} + {a}_{2} = {a}_{3} \) , \( {b}_{1}{b}_{2} = {b}_{3} \) holding in \( F \) to the set of axioms. The models of \( \mathcal{T} \) are the extension fie... | No |
Lemma 3.2. Let \( {F}^{ * } \) be an extension of the field \( F \) such that every monic polynomial over \( F \) splits over \( {F}^{ * } \) . Let \( \bar{F} \) be the set of all elements of \( {F}^{ * } \) which are algebraic over \( F \) . Then \( \bar{F} \) is an algebraic closure of \( F \) . | Proof: Let \( f\left( x\right) \) be a monic polynomial over \( \bar{F} \) . Then \( f\left( x\right) = \left( {x - {a}_{1}}\right) \cdots \) \( \left( {x - {a}_{n}}\right) \) for some \( {a}_{1},\ldots ,{a}_{n} \) in the splitting field of \( f\left( x\right) \) considered as a polynomial over \( {F}^{ * } \) . But th... | Yes |
Lemma 4.4. Let \( \alpha \) be an infinite cardinal and let \( \mathcal{F} \) be an \( \alpha \) -incomplete ultrafilter on \( I \) . Then there exists a partition of \( I \) into \( \alpha \) disjoint subsets, none of which is in \( \mathcal{F} \) . | Proof: The cardinal \( \alpha \) is an ordinal, \( \alpha = \{ \beta \mid \beta \) ordinal, \( \beta < \alpha \} \) . Since \( \mathcal{F} \) is \( \alpha \) -incomplete, there exists \( \mathcal{G} \subseteq \mathcal{F} \) such that \( \left| \mathcal{G}\right| = \alpha \) and \( \cap \mathcal{G} \notin \mathcal{F} \)... | Yes |
Lemma 4.5. Let \( \mathcal{F} \) be an \( \alpha \) -complete ultrafilter on I. Then for every partition of \( I \) into a set \( \mathcal{G} \) of \( \alpha \) disjoint subsets, some member of \( \mathcal{G} \) is in \( \mathcal{F} \) . | Exercise 4.6. Prove Lemma 4.5. | No |
Theorem 4.7. Let \( \mathcal{F} \) be an ultrafilter on \( I \) and let \( \alpha = \left| M\right| \) . Then \( M = \) \( {M}^{I}/\mathcal{F} \) if and only if \( \mathcal{F} \) is \( \alpha \) -complete. | Proof: Suppose \( \mathcal{F} \) is \( \alpha \) -complete. An element of \( {M}^{I}/\mathcal{F} \) is \( f\mathcal{F} \) for some \( f : I \rightarrow M \) . For each \( m \in M \), put \( {J}_{m} = \{ i \in I \mid f\left( i\right) = m\} \) . Then \( \left\{ {{J}_{m} \mid m \in M}\right\} \) is a partition of \( I \) ... | Yes |
Theorem 5.3. Let \( \mathcal{T} \) be the theory of \( \mathbf{N} \). Let \( S \) be the set of all functions \( s : \mathbf{N} \rightarrow \mathbf{N} \) which are definable in \( \mathcal{T} \). Let \( \mathcal{F} \) be an ultrafilter on \( \mathbf{N} \). Then the subultraproduct \( S/\mathcal{F} \) is a countable mod... | Proof: To show that \( S/\mathcal{F} \) is a model of \( \mathcal{T} \), it remains for us to show that if \( a \in S \) and \( J = \left\{ {i \in \mathbf{N} \mid {a}_{i}}\right. \) satisfies \( \left. {\left( {\forall y}\right) q\left( {x, y}\right) }\right\} \notin \mathcal{F} \), then there exists \( b \in S \) such... | Yes |
Lemma 6.3. For any theory \( \mathcal{T} \), direct limits exist in \( \operatorname{Mod}\left( {\mathcal{T}}_{\varnothing }\right) \) . | Proof: Let \( \left\{ {{M}_{i},{f}_{j}^{i} \mid i, j \in I, i \leq j}\right\} \) be a direct family in \( \operatorname{Mod}\left( {\mathcal{T}}_{\varnothing }\right) \) . We construct a limit as a subultraproduct. Put \( {F}_{i} = \{ j \in I \mid j \geq i\} \), and let \( \mathcal{F} \) be any ultrafilter containing a... | Yes |
Lemma 6.4. The direct family \( \left\{ {{M}_{i},{f}_{j}^{i} \mid i, j \in I, i \leq j}\right\} \) has a limit in \( \operatorname{Mod}\left( \mathcal{T}\right) \) if and only if the subultraproduct \( S/\mathcal{F} \) constructed above is a model of \( \mathcal{T} \) . | Proof: If \( S/\mathcal{F} \) is a model of \( \mathcal{T} \), then it is clearly a limit in \( \operatorname{Mod}\left( \mathcal{T}\right) \) of the given family. If \( L \) is a limit in \( \operatorname{Mod}\left( \mathcal{T}\right) \) of the family, then \( L \) is also a limit in \( \operatorname{Mod}\left( {\math... | Yes |
Theorem 6.7. Let \( \mathcal{T} \) be an algebraic theory. Then direct limits exist in \( \operatorname{Mod}\left( \mathcal{T}\right) \) . | Proof: Let \( q = \left( {\forall x}\right) \left( {\exists y}\right) p\left( {x, y, c}\right) \) be an axiom of \( \mathcal{T} \), where \( p\left( {x, y, c}\right) \) is constructed from primitive elements of \( P\left( {V,\mathcal{R}}\right) \) using only \( \vee , \land .(x, y, c \) may denote \( n \) -tuples.) Let... | Yes |
Theorem 3.4. Let \( \mathcal{T} = \mathcal{T}\left( \mathrm{S}\right) \) and let \( \sigma \) be any set of concurrent relations of \( \mathrm{S} \) . Then \( {\mathcal{T}}^{\sigma } \) is consistent. | Proof. Suppose \( {\mathcal{T}}^{\sigma } \vdash F \) . Then \( {A}_{0}{ \vdash }_{\mathcal{T}}F \) for some finite subset \( {A}_{0} \) of \( {A}_{\sigma } \) . Let \( {A}_{0} = \left\{ {{\rho }_{j}\left( {{x}_{ij},{c}_{{\rho }_{j}}}\right) \mid i = 1,\ldots ,{r}_{j};j = 1,\ldots, n}\right\} \), where \( {x}_{ij} \in ... | Yes |
Theorem 3.10. Let \( \mathrm{S} \leq {\mathrm{S}}_{1} \), and let \( \sigma ,{\sigma }_{1} \) be sets of concurrent relations of \( \mathrm{S},{\mathrm{S}}_{1} \) such that \( \sigma \leq {\sigma }_{1} \) . Let \( {\mathrm{S}}_{1}^{{\sigma }_{1}} \) be an enlargement of \( {\mathrm{S}}_{1} \) with respect to \( {\sigma... | Proof. Put \( {\mathrm{S}}^{\sigma } = \left\{ {a \in {\mathrm{S}}_{1}^{{\sigma }_{1}} \mid p\left( a\right) }\right. \) is true in \( \left. {\mathrm{S}}_{1}^{{\sigma }_{1}}\right\} \), where \( p\left( x\right) \in P\left( {\mathrm{\;S}}_{1}\right) \) is such that \( \mathrm{S} = \left\{ {a \in {\mathrm{S}}_{1} \mid ... | Yes |
Theorem 4.4. Let \( * \mathrm{\;S} \) be a full enlargement of \( \mathrm{S} \) and let \( u \) be a definable subset of \( \mathrm{S} \). Then \( * u = u \) if and only if \( u \) is finite. | Proof. Suppose that \( u = \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) is finite. Then\n\n\[ u\left( x\right) = \left( {x = {u}_{1}}\right) \vee \left( {x = {u}_{2}}\right) \vee \cdots \vee \left( {x = {u}_{n}}\right) \]\n\nis a description of \( u \), and\n\n\[ {}^{ * }u = \left\{ {a \in {}^{ * }\mathbb{S}\left| {\le... | Yes |
Corollary 4.5. Suppose the enlargement \( {\mathcal{T}}^{\sigma } \) of \( \mathcal{T}\left( \mathbb{S}\right) \) is both full and trivial. Then \( \mathrm{S} \) is finite. | Proof. S is a definable subset with description \( p\left( x\right) = \sim F \) . | No |
Corollary 4.6. Let \( \rho \) be a definable n-ary relation on \( \mathrm{S} \) . Then \( * \rho = \rho \) if and only if \( \rho \) is finite. | Proof. If \( \rho \) is finite, we can give a description which lists its members and it follows that \( * \rho = \rho \) . If \( \rho \) is infinite, put\n\n\[ \n{u}_{i}\left( x\right) = \left( {\exists {x}_{1}}\right) \cdots \left( {\exists {x}_{i - 1}}\right) \left( {\exists {x}_{i + 1}}\right) \cdots \left( {\exist... | Yes |
Theorem 4.7. Let \( \rho \) be a definable relation on \( \mathrm{S} \) which defines a function \( f : D \rightarrow \mathbb{S} \) on some definable \( {}^{2} \) subset \( D \) of \( \mathbb{S} \) . Then \( {\rho }^{\sigma } \) defines a function \( {f}^{\sigma } : {D}^{\sigma } \rightarrow {S}^{\sigma } \) on the sub... | Proof. We have \( \mathcal{T} \vdash \left( {\forall x}\right) \left( {\left( {x \in D}\right) \Rightarrow \left( {\exists !y}\right) \rho \left( {x, y}\right) }\right) \) . Interpreting this in \( {\mathrm{S}}^{\sigma } \) gives the result. | Yes |
Lemma 5.2. Let \( u, v \) be internal n-ary relations on \( {\mathbb{S}}^{\sigma } \) . Then \( u \cap v, u \cup v \) , and the complement \( {u}^{ \sim } \) of \( u \) are also internal. | Proof. We have that\n\n\[ \mathcal{T}\left( {{\mathcal{O}}^{1}\left( \mathrm{\;S}\right) }\right) \vdash \left( {\forall x}\right) \left( {\forall y}\right) \left( {\left( {x \in {\mathcal{R}}_{n}^{\left( 1\right) }}\right) \land \left( {y \in {\mathcal{R}}_{n}^{\left( 1\right) }}\right) \Rightarrow }\right.\n\n\[ \lef... | Yes |
Lemma 6.3. There is no smallest infinite natural number. The set of infinite natural numbers is an external set. | Proof. If \( n \) is a natural number and \( n \neq 0 \), then \( n = m + 1 \) for some natural number \( m \), since this result is a theorem of \( \mathcal{T}\left( \mathbf{N}\right) \) . If \( n \) is the smallest infinite natural number, then \( m = n - 1 \) is also infinite, and \( m < n \), giving a contradiction... | Yes |
Lemma 6.4. Suppose \( n \in * \mathbf{N} \) . Then \( n \) is finite if and only if \( n \in \mathbf{N} \) . | Proof. If \( n \in \mathbf{N}, n \) is clearly finite. Suppose that \( n \) is finite. Then \( n < b \) for some standard real number \( b \), and \( b < m \) for some standard natural number \( m \) . Put \( u = \{ x \in \mathbf{N} \mid x < m\} \) . Then \( n \in * u = \{ x \in * \mathbf{N} \mid x < m\} \), and * \( u... | Yes |
Theorem 6.5. Each of \( \mathrm{N},\mathrm{R} \), the set of infinite real numbers, and the set of infinitesimal real numbers is an external set. | Proof.\n\n(a) By Lemma 6.3, the set of infinite natural numbers is an external set, and by Lemma 6.4, \( \mathrm{N} \) is its complement in the internal set \( {}^{ * }\mathrm{\;N} \) . Hence \( \mathrm{N} \) is external by Lemma 5.2.\n\n(b) If \( \mathbf{R} \) is internal, then so is \( \mathbf{N} = \mathbf{R} \cap * ... | Yes |
If \( a \) is a finite real number, then \( \mu \left( a\right) \) contains exactly one standard real number. If \( {R}_{0} \) is the set of finite real numbers and \( {R}_{1} \) the set of infinitesimal real numbers, then \( {R}_{0} \) is a ring, \( {R}_{1} \) is an ideal of \( {R}_{0} \) and \( {R}_{0}/{R}_{1} \) is ... | Proof. If \( r, s \in \mu \left( a\right) \) and \( r, s \) are standard, then \( \left| {r - s}\right| \) is an infinitesimal standard real number. Thus \( \left| {r - s}\right| = 0 \) and \( r = s \) . We have to show that there is a standard real number in \( \mu \left( a\right) \) . This is so if \( a \) is standar... | Yes |
Theorem 6.9. Let \( r \in \mathbb{R} \) and let \( s : \mathbb{N} \rightarrow \mathbb{R} \) be a sequence. Then \( \mathop{\operatorname{Lim}}\limits_{{n \rightarrow \infty }}s\left( n\right) = r \) if and only if \( {}^{ * }s\left( n\right) \in \mu \left( r\right) \) for all infinite natural numbers \( n \) . | Proof. Suppose that \( {\operatorname{Lim}}_{n \rightarrow \infty }s\left( n\right) = r \) . Then for every standard real number \( \varepsilon > 0 \), there exists \( {n}_{0} \in \mathbf{N} \) such that \( \left| {s\left( n\right) - r}\right| < \varepsilon \) for all \( n > {n}_{0} \) . For this \( \varepsilon \) and ... | Yes |
Theorem 6.14. Let \( U \) be a closed bounded interval \( \left\lbrack {p, q}\right\rbrack \) . If the real-valued function \( f \) is continuous on \( U \), then it is uniformly continuous on \( U \) . | Proof. Take any \( x \in * U.x \) is a finite real number, hence there is a unique \( r \in \mathbf{R} \) such that \( r \bumpeq x \) . If \( r < p \), then \( x = r + \left( {x - r}\right) < r + \left( {p - r}\right) = p \) , since \( x - r \) is infinitesimal and \( p - r \) is a standard positive real number. As \( ... | Yes |
Theorem 6.16. The sequence of functions \( {s}_{n}\left( x\right) \) converges uniformly on \( U \) to \( r\left( x\right) \) if and only if for all \( x \in * U \) and for all infinite \( n, * {s}_{n}\left( x\right) \bumpeq * r\left( x\right) \) . | Exercise 6.17. Prove Theorem 6.16 by suitably modifying the argument leading to Theorem 6.13. | No |
Theorem 6.18. Suppose the functions \( {s}_{n}\left( x\right) \) are continuous on \( U \), and converge uniformly on \( U \) to \( r\left( x\right) \) . Then \( r\left( x\right) \) is continuous on \( U \) . | Proof. Let \( a \in U \) . If \( x \in * U \) and \( x \bumpeq a \), and if \( n \) is infinite, then by Corollary 6.11 and Theorem 6.14,\n\n\[ * r\left( s\right) \bumpeq * {s}_{n}\left( x\right) \bumpeq * {s}_{n}\left( a\right) \bumpeq * r\left( a\right) ,\]\n\nshowing that \( r \) is continuous at \( a \) . | No |
The problem is to determine for all integers \( n, r \), whether or not there is a computation of \( {M}_{n} \) beginning with the state \( {q}_{0}\operatorname{code}\left( r\right) \) . I.e., the problem is to determine whether or not an arbitrary Turing machine \( {M}_{n} \), fed with an arbitrary integer \( r \) , s... | We show that this stopping problem is recursively insoluble. Put \( {f}_{n} = {\Psi }_{{M}_{n}}^{\left( 1,1\right) } \) . The problem is to determine those \( \left( {n, r}\right) \) for which \( {f}_{n}\left( r\right) \) exists, and we show that there is no Turing machine which computes for each \( n \) whether or not... | Yes |
Lemma 6.8. Let \( f \) be an initial state function (i.e., \( f\left( 0\right) = G\left( {q}_{0}\right) \) ) and let \( g\left( {n, r}\right) \) be the value at \( r \) of the state function after \( n \) steps of the computation by \( M \) starting at \( \left\lbrack f\right\rbrack \) . Then \( g \) is strongly defina... | Proof. \( f \) is strongly definable, and so we give a definition of \( g \) in terms of the definition of \( f \) . Put\n\n\[ \varphi \left( {x, y, z}\right) = \left( {\exists u}\right) \lbrack \left( {\forall v}\right) \left( {v \leq y + {2x} \Rightarrow \operatorname{seq}\left( {\operatorname{seq}\left( {u,0}\right)... | Yes |
Theorem 6.9. Let \( \mathcal{T} \supseteq {\mathcal{N}}_{0} \) be a theory with \( \mathbf{N} \) as model. Then every partial recursive function is strongly definable in \( \mathcal{T} \) . | Proof. The formulae given above, together with a description of the input function in terms of the arguments of \( {\Psi }_{M}^{\left( k,\ell \right) } \), can be adapted to give a definition of \( {\Psi }_{M}^{\left( k,\ell \right) } \) . The reader is asked to supply the details. | No |
Theorem 6.10. Let \( \mathcal{T} \supseteq {\mathcal{N}}_{0} \) be a theory which has \( \mathrm{N} \) as model. Then \( \mathcal{T} \) is undecidable. In particular, \( \mathcal{N} \) is undecidable. | Proof. A decision process for \( \mathcal{T} \) would provide a decision process for the family \( \{ \left( {\exists x}\right) \operatorname{comp}\left( {n, n, x}\right) \mid n \in \mathbb{N}\} \) . | Yes |
Theorem 6.11. Let \( \mathcal{T} \supseteq {\mathcal{N}}_{0} \) be an effectively axiomatised theory with \( \mathbf{N} \) as model. Then \( \mathcal{T} \) is incomplete. | Proof. By Exercise 5.4 and Theorem 6.10. However, it is of interest to construct an element \( q \in \mathcal{L}\left( \mathcal{T}\right) \) such that neither \( q \) nor \( \sim q \) is a theorem of \( \mathcal{T} \) . Let \( \mathcal{T} = \left( {\mathcal{R}, A, C}\right) \), and write \( P \) for \( P\left( {V,\math... | No |
Lemma 7.1. Let \( \mathcal{T},{\mathcal{T}}^{\prime } \) be theories, and let \( \varphi : \mathcal{L}\left( \mathcal{T}\right) \rightarrow \mathcal{L}\left( {\mathcal{T}}^{\prime }\right) \) be a recursive function such that for all \( p \in \mathcal{L}\left( \mathcal{T}\right) \), we have \( \mathcal{T} \vdash p \) i... | Proof: Clearly, to determine if \( p \) is a theorem, it suffices to calculate \( \varphi \left( p\right) \) and to apply the decision process for \( {\mathcal{T}}^{\prime } \) . | Yes |
Lemma 7.2. Let \( {\mathcal{N}}_{1} \) be the theory formed from the theory \( {\mathcal{N}}_{0} \) by omitting the constants and the axioms \( {e}_{n} \) which identify the constants. Then \( {\mathcal{N}}_{1} \) is undecidable. | Proof: Put, for each \( n \in \mathbb{N} \) , \[ {e}_{n}\left( x\right) = \left( {\exists {x}_{0}}\right) \left( {\exists {x}_{1}}\right) \cdots \left( {\exists {x}_{n - 1}}\right) \left( {\theta \left( {x}_{0}\right) \land s\left( {{x}_{1},{x}_{0}}\right) \land \cdots \land s\left( {x,{x}_{n - 1}}\right) }\right) . \]... | Yes |
Corollary 7.4. (Church’s Theorem) Let \( {\mathcal{R}}^{ * } = \left\{ {{r}_{ij} \mid i, j \in \mathbf{N}}\right\} \), with \( {r}_{ij} \) an i-ary relation symbol, be the universal relation alphabet. Then \( \operatorname{Pred}\left( {V,{\mathcal{R}}^{ * }}\right) \) is undecidable. | Proof: The inclusion \( P\left( {V,\mathcal{R}}\right) \rightarrow P\left( {V,{\mathcal{R}}^{ * }}\right) \) satisfies the conditions of Lemma 7.1. \( ▱ \) | No |
Theorem 7.5. Let \( r \) be a binary predicate symbol. Then \( \operatorname{Pred}\left( {V,\{ r\} }\right) \) is undecidable. | Before giving the formal proof, we note that the result implies that if \( \mathcal{R} \) contains at least one \( n \) -ary relation symbol with \( n \geq 2 \), then \( \operatorname{Pred}\left( {V,\mathcal{R}}\right) \) is undecidable. The theorem will be proved by constructing a function \( f : P\left( {V,\{ \rho \}... | "No" |
Lemma 7.6. Let \( \rho \) be a 4-ary relation on the non-empty set \( S \) . Put \( {S}^{\prime } = \{ K\} \cup {S}^{2} \cup {S}^{4} \) . For \( x \in S \), define \( \Delta \left( x\right) = \left( {x, x}\right) \in {S}^{\prime } \) . Let \( r \) be the binary relation on \( {S}^{\prime } \) consisting of those pairs ... | Proof: An element \( a \in {S}^{\prime } \) is in \( \Delta \left( S\right) \) if and only if \( \left( {a, a}\right) \in r \) . We claim that a quadruple \( \left( {x, y, z, t}\right) \) of elements of \( S \) is in \( \rho \) if and only if their images \( X, Y, Z, T \) under \( \Delta \) satisfy the condition that t... | Yes |
Example 1.6. The subset \( S \) of \( \mathbf{N} \), consisting of integers which are not powers of 2 , is diophantine, because | \[ x \in S\\text{iff}\\left( {\\exists y, z}\\right) \\left( {x - y\\left( {{2z} + 1}\\right) = 0}\\right) \\text{.} \] | Yes |
Lemma 1.13. Every diophantine function \( f \) is recursive. | Proof. Write\n\n\( y = f\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) iff \( \left( {\exists {t}_{1},\ldots ,{t}_{m}}\right) \left( {P\left( {{x}_{1},\ldots ,{x}_{n}, y,{t}_{1},\ldots ,{t}_{m}}\right) }\right. \n\n\[ \n= Q\left( {{x}_{1},\ldots ,{x}_{n}, y,{t}_{1},\ldots ,{t}_{m}}\right) ) \n\] \nwhere \( P, Q \) are poly... | Yes |
Corollary 1.15. The set of diophantine functions is closed under primitive recursion and minimalisation. | Proof of the Corollary. Suppose \( f, g \) are diophantine, and\n\n\[ h\left( {{x}_{1},\ldots ,{x}_{n},0}\right) = f\left( {{x}_{1},\ldots ,{x}_{n}}\right) \]\n\n\[ h\left( {{x}_{1},\ldots ,{x}_{n}, t + 1}\right) = g\left( {t, h\left( {{x}_{1},\ldots ,{x}_{n}, t}\right) ,{x}_{1},\ldots ,{x}_{n}}\right) . \]\n\nUsing th... | Yes |
Theorem 1.16. A function is recursive if and only if it is diophantine. | In chapter IX, we showed the existence of a subset \( E \) of \( \mathbf{N} \) which is recursively enumerable but not recursive. That is, \( E \) is the range of some recursive function, but the characteristic function of \( E \) is not a recursive function. Theorem 1.16 implies that a subset of \( \mathbf{N} \) is re... | Yes |
Lemma 2.8. Let \( w,{w}^{\prime } \) be special words. Suppose \( {w}^{\prime } \) is terminal. Then \( w \) , \( {w}^{\prime } \) are equivalent if and only if there is a path from \( w \) to \( {w}^{\prime } \) consisting only of forward steps. | Proof. Trivially, if such a path exists, then \( w \sim {w}^{\prime } \) . Suppose that \( w \sim {w}^{\prime } \) . Then there exists a path\n\n\[ w = {w}_{0} - {w}_{1} - \cdots - {w}_{n} = {w}^{\prime } \]\n\nfrom \( w \) to \( {w}^{\prime } \) . We may suppose the path is chosen so that the number \( n \) of steps i... | Yes |
Theorem 2.9. Let \( E \) be a recursively enumerable but non-recursive subset of \( \mathbf{N} \), and let \( M \) be a Turing machine which, when started in the state \( {q}_{0}{s}_{1}^{n} \) , stops with blank tape if \( n \in E \), and does not stop if \( n \notin E \) . Then the semigroup presentation associated wi... | Proof. By Lemma 2.8, the special word \( e{q}_{0}{s}_{1}^{n}e \) is equivalent to \( e{q}_{\infty }e \) if and only if there exists a forward path from \( e{q}_{0}{s}_{1}^{n}e \) to \( e{q}_{\infty }e \) . Such a path exists if and only if \( M \), started in the state \( {q}_{0}{s}_{1}^{n} \), stops with blank tape-i.... | Yes |
Proposition 1.5. If \( U \) belongs to \( \mathrm{U}\left( {2n}\right) \) then \( U \) belongs to \( \mathrm{{Sp}}\left( n\right) \) if and only if \( U \) commutes with \( J \) . | Proof. Fix some \( U \) in \( \mathrm{U}\left( {2n}\right) \) . Then for \( z \) and \( w \) in \( {\mathbb{C}}^{2n} \), we have, on the one hand,\n\n\[ \omega \left( {{Uz},{Uw}}\right) = \langle {JUz},{Uw}\rangle = \left\langle {{U}^{ * }{JUz}, w}\right\rangle = \left\langle {{U}^{-1}{JUz}, w}\right\rangle ,\]\n\nand,... | Yes |
Lemma 1.7. Suppose \( V \) is a complex subspace of \( {\mathbb{C}}^{2n} \) that is invariant under the conjugate-linear map \( J \) . Then the orthogonal complement \( {V}^{ \bot } \) of \( V \) (with respect to the inner product \( \langle \cdot , \cdot \rangle \) ) is also invariant under \( J \) . Furthermore, \( V... | Proof. If \( w \in {V}^{ \bot } \) then for all \( z \in V \), we have\n\n\[ \langle {Jw}, z\rangle = - \langle {Jz}, w\rangle = 0 \]\n\nbecause \( {Jz} \) is again in \( V \) . Thus, \( {V}^{ \bot } \) is invariant under \( J \) . Then if \( z \in V \) and \( w \in {V}^{ \bot } \) , we have\n\n\[ \omega \left( {z, w}\... | Yes |
Proposition 1.10. If \( G \) is a matrix Lie group, the identity component \( {G}_{0} \) of \( G \) is a normal subgroup of \( G \) . | Proof. If \( A \) and \( B \) are any two elements of \( {G}_{0} \), then there are continuous paths \( A\left( t\right) \) and \( B\left( t\right) \) connecting \( I \) to \( A \) and to \( B \) in \( G \) . Then the path \( A\left( t\right) B\left( t\right) \) is a continuous path connecting \( I \) to \( {AB} \) in ... | Yes |
Proposition 1.11. The group \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) is connected for all \( n \geq 1 \) . | Proof. We make use of the result that every matrix is similar to an upper triangular matrix (Theorem A.4). That is to say, we can express any \( A \in {M}_{n}\left( \mathbb{C}\right) \) in the form \( A = {CB}{C}^{-1} \), where\n\n\[ B = \left( \begin{matrix} {\lambda }_{1} & & * \\ & \ddots & \\ 0 & & {\lambda }_{n} \... | Yes |
Proposition 1.12. The group \( \mathrm{{SL}}\left( {n;\mathbb{C}}\right) \) is connected for all \( n \geq 1 \) . | Proof. The proof is almost the same as for \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \), except that we must make sure our path connecting \( A \in \mathrm{{SL}}\left( {n;\mathbb{C}}\right) \) to \( I \) lies entirely in \( \mathrm{{SL}}\left( {n;\mathbb{C}}\right) \) . We can ensure this by choosing \( {\lambda }_{... | Yes |
Proposition 1.13. The groups \( \mathrm{U}\left( n\right) \) and \( \mathrm{{SU}}\left( n\right) \) are connected, for all \( n \geq 1 \) . | Proof. By Theorem A.3, every unitary matrix has an orthonormal basis of eigenvectors, with eigenvalues having absolute value 1 . Thus, each \( U \in \mathrm{U}\left( n\right) \) can be written as \( {U}_{1}D{U}_{1}^{-1} \), where \( {U}_{1} \in \mathrm{U}\left( n\right) \) and \( D \) is diagonal with diagonal entries ... | Yes |
Proposition 1.15. The group \( \mathrm{{SU}}\left( 2\right) \) is simply connected. | Proof. Exercise 5 shows that \( \mathrm{{SU}}\left( 2\right) \) may be thought of (topologically) as the three-dimensional sphere \( {S}^{3} \) sitting inside \( {\mathbb{R}}^{4} \) . It is well known that \( {S}^{3} \) is simply connected; see, for example, Proposition 1.14 in [Hat]. | No |
Proposition 1.17. There is a continuous bijection between \( \mathrm{{SO}}\left( 3\right) \) and \( \mathbb{R}{P}^{3} \) . | Proof. If \( v \) is a unit vector in \( {\mathbb{R}}^{3} \), let \( {R}_{v,\theta } \) be the element of \( \mathrm{{SO}}\left( 3\right) \) consisting of a \ | No |
Proposition 1.19. The map \( U \mapsto {\Phi }_{U} \) is a 2-1 and onto map of \( \mathrm{{SU}}\left( 2\right) \) to \( \mathrm{{SO}}\left( 3\right) \) , with kernel equal to \( \{ I, - I\} \) . | Proof. Exercise 16 shows that the kernel of \( \Phi \) is precisely the set \( \{ I, - I\} \) . To see that \( \Phi \) maps onto \( \mathrm{{SO}}\left( 3\right) \), let \( R \) be a rotation of \( V \cong {\mathbb{R}}^{3} \) . By Exercise 14, there exists an \ | No |
Example 1.21. Let\n\n\\[ \nG = \\mathbb{R} \\times \\mathbb{R} \\times {S}^{1} = \\left\\{ {\\left( {x, y, u}\\right) \\mid x \\in \\mathbb{R}, y \\in \\mathbb{R}, u \\in {S}^{1} \\subset \\mathbb{C}}\\right\\} ,\n\\]\n\nequipped with the group product given by\n\n\\[ \n\\left( {{x}_{1},{y}_{1},{u}_{1}}\\right) \\cdot ... | Proof. It is easily checked that this operation is associative; the product of three elements with either grouping is\n\n\\[ \n\\left( {{x}_{1} + {x}_{2} + {x}_{3},{y}_{1} + {y}_{2} + {y}_{3},{e}^{i\\left( {{x}_{1}{y}_{2} + {x}_{1}{y}_{3} + {x}_{2}{y}_{3}}\\right) }{u}_{1}{u}_{2}{u}_{3}}\\right) .\n\\]\n\nThere is an i... | Yes |
Proposition 2.1. The series (2.1) converges for all \( X \in {M}_{n}\left( \mathbb{C}\right) \) and \( {e}^{X} \) is a continuous function of \( X \) . | Proof of Proposition 2.1. In light of (2.4), we see that\n\n\[ \begin{Vmatrix}{X}^{m}\end{Vmatrix} \leq \parallel X{\parallel }^{m} \]\n\nfor all \( m \geq 1 \), and, hence,\n\n\[ \mathop{\sum }\limits_{{m = 0}}^{\infty }\begin{Vmatrix}\frac{{X}^{m}}{m!}\end{Vmatrix} \leq \parallel I\parallel + \mathop{\sum }\limits_{{... | Yes |
Proposition 2.3. Let \( X \) and \( Y \) be arbitrary \( n \times n \) matrices. Then we have the following:\n\n1. \( {e}^{0} = I \) .\n\n2. \( {\left( {e}^{X}\right) }^{ * } = {e}^{{X}^{ * }} \) .\n\n3. \( {e}^{X} \) is invertible and \( {\left( {e}^{X}\right) }^{-1} = {e}^{-X} \) .\n\n4. \( {e}^{\left( {\alpha + \bet... | Proof. Point 1 is obvious and Point 2 follows from taking term-by-term adjoints of the series for \( {e}^{X} \) . Points 3 and 4 are special cases of Point 5 . To verify Point 5, we simply multiply the two power series term by term, which is permitted because both series converge absolutely. Multiplying out \( {e}^{X}{... | Yes |
Proposition 2.4. Let \( X \) be a \( n \times n \) complex matrix. Then \( {e}^{tX} \) is a smooth curve in \( {M}_{n}\left( \mathbb{C}\right) \) and\n\n\[ \n\frac{d}{dt}{e}^{tX} = X{e}^{tX} = {e}^{tX}X \n\]\n\nIn particular,\n\n\[ \n{\left. \frac{d}{dt}{e}^{tX}\right| }_{t = 0} = X \n\] | Proof. Differentiate the power series for \( {e}^{tX} \) term by term. This is permitted because, for each \( j \) and \( k,{\left( {e}^{tX}\right) }_{jk} \) is given by a convergent power series in \( t \), and one can differentiate a power series term by term inside its radius of convergence (e.g., Theorem 12 in Chap... | Yes |
Consider the matrices\n\n\[ \n{X}_{1} = \left( \begin{array}{rr} 0 & - a \\ a & 0 \end{array}\right) ;\;{X}_{2} = \left( \begin{array}{lll} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{array}\right) ;\;{X}_{3} = \left( \begin{array}{ll} a & b \\ 0 & a \end{array}\right) .\n\]\n\nThen\n\n\[ \n{e}^{{X}_{1}} = \left( \begin{a... | Proof. The eigenvectors of \( {X}_{1} \) are \( \left( {1, i}\right) \) and \( \left( {i,1}\right) \), with eigenvalues \( - {ia} \) and \( {ia} \) , respectively. Thus,\n\n\[ \n{e}^{{X}_{1}} = \left( \begin{array}{ll} 1 & i \\ i & 1 \end{array}\right) \left( \begin{matrix} {e}^{-{ia}} & 0 \\ 0 & {e}^{ia} \end{matrix}\... | Yes |
Lemma 2.6. The function\n\n\[ \log z = \mathop{\sum }\limits_{{m = 1}}^{\infty }{\left( -1\right) }^{m + 1}\frac{{\left( z - 1\right) }^{m}}{m} \]\n\n(2.7)\n\nis defined and holomorphic in a circle of radius 1 about \( z = 1 \) . | Proof. The usual logarithm for real, positive numbers satisfies\n\n\[ \frac{d}{dx}\log \left( {1 - x}\right) = \frac{-1}{1 - x} = - \left( {1 + x + {x}^{2} + \cdots }\right) \]\n\nfor \( \left| x\right| < 1 \) . Integrating term by term and noting that \( \log 1 = 0 \) gives\n\n\[ \log \left( {1 - x}\right) = - \left( ... | Yes |
Theorem 2.8. The function\n\n\\[ \log A = \\mathop{\\sum }\\limits_{{m = 1}}^{\\infty }{\\left( -1\\right) }^{m + 1}\\frac{{\\left( A - I\\right) }^{m}}{m} \\]\n\nis defined and continuous on the set of all \\( n \\times n \\) complex matrices \\( A \\) with \\( \\parallel A - I\\parallel < 1 \\) .\n\nFor all \\( A \\i... | Proof. Since \\( \\begin{Vmatrix}{\\left( A - I\\right) }^{m}\\end{Vmatrix} \\leq \\parallel \\left( {A - I}\\right) {\\parallel }^{m} \\) and since the series (2.7) has radius of convergence 1, the series (2.9) converges absolutely for all \\( A \\) with \\( \\parallel A - I\\parallel < 1 \\) . The proof of continuity... | Yes |
Proposition 2.9. There exists a constant \( c \) such that for all \( n \times n \) matrices \( B \) with \( \parallel B\parallel < 1/2 \), we have\n\n\[ \parallel \log \left( {I + B}\right) - B\parallel \leq c\parallel B{\parallel }^{2}. \] | Proof. Note that\n\n\[ \log \left( {I + B}\right) - B = \mathop{\sum }\limits_{{m = 2}}^{\infty }{\left( -1\right) }^{m + 1}\frac{{B}^{m}}{m} = {B}^{2}\mathop{\sum }\limits_{{m = 2}}^{\infty }{\left( -1\right) }^{m + 1}\frac{{B}^{m - 2}}{m} \]\n\nso that if \( \parallel B\parallel < 1/2 \), we have\n\n\[ \parallel \log... | Yes |
Theorem 2.11 (Lie Product Formula). For all \( X, Y \in {M}_{n}\left( \mathbb{C}\right) \), we have\n\n\[ \n{e}^{X + Y} = \mathop{\lim }\limits_{{m \rightarrow \infty }}{\left( {e}^{\frac{X}{m}}{e}^{\frac{Y}{m}}\right) }^{m}. \n\] | Proof. If we multiply the power series for \( {e}^{\frac{X}{m}} \) and \( {e}^{\frac{Y}{m}} \), all but three of the terms will involve \( 1/{m}^{2} \) or higher powers of \( 1/m \) . Thus,\n\n\[ \n{e}^{\frac{X}{m}}{e}^{\frac{Y}{m}} = I + \frac{X}{m} + \frac{Y}{m} + O\left( \frac{1}{{m}^{2}}\right) . \n\]\n\nNow, since... | Yes |
Theorem 2.12. For any \( X \in {M}_{n}\left( \mathbb{C}\right) \), we have\n\n\[ \det \left( {e}^{X}\right) = {e}^{\operatorname{trace}\left( X\right) } \] | Proof. If \( X \) is diagonalizable with eigenvalues \( {\lambda }_{1},\ldots ,{\lambda }_{n} \), then \( {e}^{X} \) is diagonalizable with eigenvalues \( {e}^{{\lambda }_{1}},\ldots ,{e}^{{\lambda }_{n}} \) . Thus, \( \operatorname{trace}\left( X\right) = \mathop{\sum }\limits_{j}{\lambda }_{j} \) and\n\n\[ \det \left... | No |
Lemma 2.15. Fix some \( \varepsilon \) with \( \varepsilon < \log 2 \) . Let \( {B}_{\varepsilon /2} \) be the ball of radius \( \varepsilon /2 \) around the origin in \( {M}_{n}\left( \mathbb{C}\right) \), and let \( U = \exp \left( {B}_{\varepsilon /2}\right) \) . Then every \( B \in U \) has a unique square root \( ... | Proof. It is evident that \( C \) is a square root of \( B \) and that \( C \) is in \( U \) . To establish uniqueness, suppose \( {C}^{\prime } \in U \) satisfies \( {\left( {C}^{\prime }\right) }^{2} = B \) . Let \( Y = \log {C}^{\prime } \) ; then \( \exp \left( Y\right) = \) \( {C}^{\prime } \) and\n\n\[ \exp \left... | Yes |
Proposition 2.16. The exponential map is an infinitely differentiable map of \( {M}_{n}\left( \mathbb{C}\right) \) into \( {M}_{n}\left( \mathbb{C}\right) \) . | Proof. Note that for each \( j \) and \( k \), the quantity \( {\left( {X}^{m}\right) }_{jk} \) is a homogeneous polynomial of degree \( m \) in the entries of \( X \) . Thus, the series for the function \( {\left( {X}^{m}\right) }_{jk} \) has the form of a multivariable power series on \( {M}_{n}\left( \mathbb{C}\righ... | Yes |
Lemma 2.18. If \( Q \) is a self-adjoint, positive matrix, then \( Q \) has a unique positive, self-adjoint square root. | Proof. Since \( Q \) has an orthonormal basis of eigenvectors, \( Q \) can be written as\n\n\[ Q = U\left( \begin{array}{lll} {\lambda }_{1} & & \\ & \ddots & \\ & & {\lambda }_{n} \end{array}\right) {U}^{-1} \]\n\nwith \( U \) unitary. Since \( Q \) is self-adjoint and positive, each \( {\lambda }_{j} \) is positive. ... | Yes |
Example 3.2. Let \( \mathfrak{g} = {\mathbb{R}}^{3} \) and let \( \left\lbrack {\cdot , \cdot }\right\rbrack : {\mathbb{R}}^{3} \times {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}^{3} \) be given by \[ \left\lbrack {x, y}\right\rbrack = x \times y \] where \( x \times y \) is the cross product (or vector product). Then \(... | Proof. Bilinearity and skew symmetry are standard properties of the cross product. To verify the Jacobi identity, it suffices (by bilinearity) to verify it when \( x = {e}_{j} \) , \( y = {e}_{k} \), and \( z = {e}_{l} \), where \( {e}_{1},{e}_{2} \), and \( {e}_{3} \) are the standard basis elements for \( {\mathbb{R}... | Yes |
Example 3.3. Let \( \mathcal{A} \) be an associative algebra and let \( \mathfrak{g} \) be a subspace of \( \mathcal{A} \) such that \( {XY} - {YX} \in \mathfrak{g} \) for all \( X, Y \in \mathfrak{g} \). Then \( \mathfrak{g} \) is a Lie algebra with bracket operation given by \[ \left\lbrack {X, Y}\right\rbrack = {XY}... | Proof. The bilinearity and skew symmetry of the bracket are evident. To verify the Jacobi identity, note that each double bracket generates four terms, for a total of 12 terms. It is left to the reader to verify that the product of \( X, Y \), and \( Z \) in each of the six possible orderings occurs twice, once with a ... | No |
Let \( \mathrm{{sl}}\left( {n;\mathbb{C}}\right) \) denote the space of all \( X \in {M}_{n}\left( \mathbb{C}\right) \) for which \( \operatorname{trace}\left( X\right) = 0 \) . Then \( \operatorname{sl}\left( {n;\mathbb{C}}\right) \) is a Lie algebra with bracket \( \left\lbrack {X, Y}\right\rbrack = {XY} - {YX} \) . | Proof. For any \( X \) and \( Y \) in \( {M}_{n}\left( \mathbb{C}\right) \), we have\n\n\[ \operatorname{trace}\left( {{XY} - {YX}}\right) = \operatorname{trace}\left( {XY}\right) - \operatorname{trace}\left( {YX}\right) = 0. \]\n\nThis holds, in particular, if \( X \) and \( Y \) have trace zero. Thus, Example 3.3 app... | Yes |
Proposition 3.8. If \( \mathfrak{g} \) is a Lie algebra, then\n\n\[{\operatorname{ad}}_{\left\lbrack X, Y\right\rbrack } = {\operatorname{ad}}_{X}{\operatorname{ad}}_{Y} - {\operatorname{ad}}_{Y}{\operatorname{ad}}_{X} = \left\lbrack {{\operatorname{ad}}_{X},{\operatorname{ad}}_{Y}}\right\rbrack\]\n\nthat is, ad: \( \m... | Proof. Observe that\n\n\[{\operatorname{ad}}_{\left\lbrack X, Y\right\rbrack }\left( Z\right) = \left\lbrack {\left\lbrack {X, Y}\right\rbrack, Z}\right\rbrack\]\n\nwhereas\n\n\[\left\lbrack {{\operatorname{ad}}_{X},{\operatorname{ad}}_{Y}}\right\rbrack \left( Z\right) = \left\lbrack {X,\left\lbrack {Y, Z}\right\rbrack... | Yes |
Proposition 3.12. The Lie algebra \( \mathrm{{sl}}\left( {2;\mathbb{C}}\right) \) is simple. | Proof. We use the following basis for \( \mathrm{{sl}}\left( {2;\mathbb{C}}\right) \) :\n\n\[ \nX = \left( \begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right) ;\;Y = \left( \begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right) ;\;H = \left( \begin{array}{rr} 1 & 0 \\ 0 & - 1 \end{array}\right) .\n\]\n\nDirect calculation... | Yes |
Proposition 3.16. If \( \mathfrak{g} \subset {M}_{3}\left( \mathbb{R}\right) \) denotes the space of \( 3 \times 3 \) upper triangular matrices with zeros on the diagonal, then \( \mathfrak{g} \) satisfies the assumptions of Example 3.3. The Lie algebra \( \mathfrak{g} \) is a nilpotent Lie algebra. | Proof. We will use the following basis for \( \mathfrak{g} \) ,\n\n\[ \nX = \left( \begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) ;\;Y = \left( \begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right) ;\;Z = \left( \begin{array}{lll} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{a... | Yes |
Proposition 3.17. If \( \mathfrak{g} \subset {M}_{2}\left( \mathbb{C}\right) \) denotes the space of \( 2 \times 2 \) matrices of the form\n\n\[ \left( \begin{array}{ll} a & b \\ 0 & c \end{array}\right) \]\n\nwith \( a, b \), and \( c \) in \( \mathbb{C} \), then \( \mathfrak{g} \) satisfies the assumptions of Example... | Proof. Direct calculation shows that\n\n\[ \left\lbrack {\left( \begin{array}{ll} a & b \\ 0 & c \end{array}\right) ,\left( \begin{array}{ll} d & e \\ 0 & f \end{array}\right) }\right\rbrack = \left( \begin{array}{ll} 0 & h \\ 0 & 0 \end{array}\right) \]\n\n(3.6)\n\nwhere \( h = {ae} + {bf} - {bd} - {ce} \), showing th... | Yes |
Proposition 3.19. Let \( G \) be a matrix Lie group, and \( X \) an element of its Lie algebra. Then \( {e}^{X} \) is an element of the identity component \( {G}_{0} \) of \( G \) . | Proof. By definition of the Lie algebra, \( {e}^{tX} \) lies in \( G \) for all real \( t \) . However, as \( t \) varies from 0 to \( 1,{e}^{tX} \) is a continuous path connecting the identity to \( {e}^{X} \) . | Yes |
Theorem 3.20. Let \( G \) be a matrix Lie group with Lie algebra \( \mathfrak{g} \). If \( X \) and \( Y \) are elements of \( \mathfrak{g} \), the following results hold.\n\n1. \( {AX}{A}^{-1} \in \mathfrak{g} \) for all \( A \in G \).\n\n2. \( {sX} \in \mathfrak{g} \) for all real numbers \( s \).\n\n3. \( X + Y \in ... | Proof. For Point 1, we observe that, by Proposition 2.3,\n\n\[ {e}^{t\left( {{AX}{A}^{-1}}\right) } = A{e}^{tX}{A}^{-1} \in G \]\n\nfor all \( t \), showing that \( {AX}{A}^{-1} \) is in \( \mathfrak{g} \). For Point 2, we observe that \( {e}^{t\left( {sX}\right) } = {e}^{\left( {ts}\right) X} \), which must be in \( G... | Yes |
Proposition 3.22. If \( G \) is commutative then \( \mathfrak{g} \) is commutative. | Proof. For any two matrices \( X, Y \in {M}_{n}\left( \mathbb{C}\right) \), the commutator of \( X \) and \( Y \) may be computed as\n\n\[\n\left\lbrack {X, Y}\right\rbrack = {\left. \frac{d}{dt}\left( {\left. \frac{d}{ds}{e}^{tX}{e}^{sY}{e}^{-{tX}}\right| }_{s = 0}\right) \right| }_{t = 0}.\n\]\n\n(3.7)\n\nIf \( G \) ... | Yes |
Proposition 3.23. The Lie algebra of \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) is the space \( {M}_{n}\left( \mathbb{C}\right) \) of all \( n \times n \) matrices with complex entries. Similarly, the Lie algebra of \( \mathrm{{GL}}\left( {n;\mathbb{R}}\right) \) is equal to \( {M}_{n}\left( \mathbb{R}}\right) \) ... | Proof. If \( X \in {M}_{n}\left( \mathbb{C}\right) \), then \( {e}^{tX} \) is invertible, so that \( X \) belongs to the Lie algebra of \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \) . If \( X \in {M}_{n}\left( \mathbb{R}}\right) \), then \( {e}^{tX} \) is invertible and real, so that \( X \) is in the Lie algebra of ... | Yes |
The Lie algebra of \( \mathrm{U}\left( n\right) \) consists of all complex matrices satisfying \( {X}^{ * } = - X \) and the Lie algebra of \( \mathrm{{SU}}\left( n\right) \) consists of all complex matrices satisfying \( {X}^{ * } = - X \) and \( \operatorname{trace}\left( X\right) = 0 \) . The Lie algebra of the orth... | Proof. A matrix \( U \) is unitary if and only if \( {U}^{ * } = {U}^{-1} \) . Thus, \( {e}^{tX} \) is unitary if and only if\n\n\[{\left( {e}^{tX}\right) }^{ * } = {\left( {e}^{tX}\right) }^{-1} = {e}^{-{tX}}.\]\n\n(3.8)\n\nBy Point 2 of Proposition 2.3, \( {\left( {e}^{tX}\right) }^{ * } = {e}^{t{X}^{ * }} \), and so... | Yes |
Proposition 3.25. If \( g \) is the matrix in Exercise 1 of Chapter 1, then the Lie algebra of \( \mathrm{O}\left( {n;k}\right) \) consists precisely of those real matrices \( X \) such that\n\n\[ g{X}^{tr}g = - X \]\n\nand the Lie algebra of \( \mathrm{{SO}}\left( {n;k}\right) \) is the same as that of \( \mathrm{O}\l... | The verification of Proposition 3.25 is similar to our previous computations and is omitted. | No |
Proposition 3.26. The Lie algebra of the Heisenberg group \( H \) in Sect. 1.2.6 is the space of all matrices of the form\n\n\[ \nX = \left( \begin{array}{lll} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{array}\right)\n\]\n\n(3.10)\n\nwith \( a, b, c \in \mathbb{R} \) . | Proof. If \( X \) is strictly upper triangular, it is easy to verify that \( {X}^{m} \) will be strictly upper triangular for all positive integers \( m \) . Thus, for \( X \) as in (3.10), we will have \( {e}^{tX} = I + B \) with \( B \) strictly upper triangular, showing that \( {e}^{tX} \in H \) . Conversely, if \( ... | Yes |
The following elements form a basis for the Lie algebra \( \mathrm{su}(2) \) : | \[ E_1 = \frac{1}{2}\left( \begin{array}{rr} i & 0 \\ 0 & -i \end{array}\right) ;\; E_2 = \frac{1}{2}\left( \begin{array}{ll} 0 & i \\ i & 0 \end{array}\right) ;\; E_3 = \frac{1}{2}\left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right) . \]\n\nThese elements satisfy the commutation relations \( \left\lbrack E_1, E... | Yes |
Theorem 3.28. Let \( G \) and \( H \) be matrix Lie groups, with Lie algebras \( \mathfrak{g} \) and \( \mathfrak{h} \) , respectively. Suppose that \( \Phi : G \rightarrow H \) is a Lie group homomorphism. Then there exists a unique real-linear map \( \phi : \mathfrak{g} \rightarrow \mathfrak{h} \) such that\n\n\[ \Ph... | Proof. The proof is similar to the proof of Theorem 3.20. Since \( \Phi \) is a continuous group homomorphism, \( \Phi \left( {e}^{tX}\right) \) will be a one-parameter subgroup of \( H \), for each \( X \in \mathfrak{g} \) . Thus, by Theorem 2.14, there is a unique matrix \( Z \) such that\n\n\[ \Phi \left( {e}^{tX}\r... | Yes |
Let \( \Phi : \mathrm{{SU}}\left( 2\right) \rightarrow \mathrm{{SO}}\left( 3\right) \) be the homomorphism in Proposition 1.19. Then the associated Lie algebra homomorphism \( \phi : \mathrm{{su}}\left( 2\right) \rightarrow \mathrm{{so}}\left( 3\right) \) satisfies\n\n\[ \phi \left( {E}_{j}\right) = {F}_{j},\;j = 1,2,3... | Proof. If \( X \) is in \( \mathrm{{su}}\left( 2\right) \) and \( Y \) is in the space \( V \) in (1.14), then\n\n\[ {\left. \frac{d}{dt}\Phi \left( {e}^{tX}\right) Y\right| }_{t = 0} = {\left. \frac{d}{dt}{e}^{tX}Y{e}^{-{tX}}\right| }_{t = 0} = \left\lbrack {X, Y}\right\rbrack . \]\n\nThus, \( \phi \left( X\right) \) ... | No |
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