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Problem 25.1. | Solution: Suppose \( A \) is a connected subspace of \( {\mathbb{R}}_{\ell } \) containing a given pair of points \( x \) and \( y \) where \( x < y \) . It follows that \( A = \left( {\left( {\infty, y}\right) \cap A}\right) \cup \left( {\lbrack y,\infty }\right) \cap A) \) , where the two rays are open in \( A \) and... | Yes |
Problem 25.3. | Solution: Let \( U \) be any neighborhood of a given \( x \in {I}_{0}^{2} \) . Since \( {I}_{0}^{2} \) is a linear continuum, it is connected, as is any interval contained in it. There is some basis element of \( {I}_{0}^{2} \) containing \( x \) that is contained in \( U \), which must be an interval of \( {I}_{0}^{2}... | Yes |
Problem 25.9. | Solution: Suppose \( C \) is a component of \( G \) containing the identity element \( e \) . If \( x \in G \) , then \( x \cdot e = x \) is in \( {xC} \) and \( e \cdot x = x \) is in \( {Cx} \) . Since the operator \( x \cdot y \) for any \( y \) is a continuous map, it follows that \( {xC} \) and \( {Cx} \) are conn... | Yes |
Problem 26.1. | Solution: Part (a) Suppose \( X \) is compact in \( {\mathcal{T}}^{\prime } \). If \( \mathcal{A} \) is an open covering of \( X \) in \( \mathcal{T} \), then every element of \( \mathcal{A} \) is also open in \( {\mathcal{T}}^{\prime } \), so \( X \) is also an open covering in that topology. Since \( X \) is compact ... | Yes |
Problem 26.7. | Solution: I found this one very tricky. Let \( A \) be closed in \( X \times Y \) and \( B = {\pi }_{1}\left( A\right) \) . We will show that \( B \) is equal to its closure, so it is closed. Designate \( Z = \left( {X \times Y}\right) \smallsetminus A \), which must be open. If \( {x}_{0} \notin B \), then \( \left\{ ... | Yes |
Suppose \( X \) is an ordered set where every closed set is compact. Let \( A \) be a non-empty subset of \( X \) that is bounded above by \( m \in X \). If \( A \) contains the largest element of \( X \), then \( m \in X \) and \( m \) is the supremum of \( A \). Otherwise, let \( {a}_{0} \) be an arbitrary element of... | \[ D = \mathop{\bigcap }\limits_{{a \in {A}^{\prime }}}{D}_{a} \] Since each \( {D}_{a} \) is closed, \( D \) is closed and, by hypothesis, compact. First we will show that \( D \) is non-empty. Let \( {\mathcal{D}}^{\prime } = \) \( {\left\{ {D}_{{a}_{j}}\right\} }_{j = 1}^{N} \) be a finite subcollection of \( D \). ... | Yes |
Problem 28.1. | Solution: Designate \( S = \{ \mathbf{s}\left( j\right) : j \in \mathbb{N}\} \) where \( \mathbf{s}\left( j\right) \) is an omega-tuple whose elements are:\n\n\[ \n{\left( \mathbf{s}\left( j\right) \right) }_{k} = \left\{ \begin{array}{lll} 1 & \text{ for } & j = k, \\ 0 & \text{ for } & j \neq k. \end{array}\right. \]... | Yes |
Problem 29.1. | Solution: Given \( x \in \mathbb{Q} \) as a subspace of \( \mathbb{R} \) , suppose there is a compact subspace \( C \) of \( \mathbb{Q} \) that contains a neighborhood of \( x \) in \( \mathbb{Q} \) . It follows that there is some \( \left\lbrack {a, b}\right\rbrack \cap \mathbb{Q} \) contained in \( C \) where \( a, b... | Yes |
Problem 29.4. | Solution: Let \( \mathbf{x} \in {\left\lbrack 0,1\right\rbrack }^{\omega } \) . If \( {\left\lbrack 0,1\right\rbrack }^{\omega } \) is locally compact, there is a compact subspace \( C \) that contains a neighborhood \( U \) of \( \mathbf{x} \) . Accordingly, some ball \( B = {B}_{\bar{\rho }}\left( {\mathbf{x},\epsilo... | Yes |
Problem 29.7. | Solution: Let \( R = {S}_{\Omega } \cup \{ \infty \} \) where \( \infty \) is some element not contained in \( {S}_{\omega } \) . The set \( R \) is Hausdorff. Given \( a, b \in R \) where \( a < b \) , if \( b \neq \infty \) , then if \( b \) is the immediate successor of \( a \) , then \( \left( {-\infty, b}\right) \... | Yes |
Problem 30.1. | Solution: Part (a) Suppose \( X \) is a finite-countable \( {T}_{1} \) space. Let \( \{ x\} \) be a one-point set in \( X \) , which must be closed. Let \( \mathcal{B} = \left\{ {B}_{n}\right\} \) be a collection of neighborhoods of \( x \) such that every neighborhood of \( x \) contains at least one \( {B}_{n} \) . C... | Yes |
Problem 30.3. | Solution: Let \( X \) be a space with a countable basis \( \{ {B}_{n}\} \) . Given an uncountable subset \( A \) of \( X \) , there must be some basis element \( {B}_{k} \) that contains an uncountable number of points in \( A \) (otherwise, each \( {B}_{n} \) would contain a countable number of points of \( A \) , so ... | Yes |
Problem 30.4. | The solution is analogous to that for exercise \( {30.5}\left( \mathrm{\;b}\right) \) . | No |
Problem 30.5. Part (a) This is an interesting problem with an analog to the density of rational numbers in \( \mathbb{R} \) under the standard topology. Let \( X \) be a metrizable space with a countably dense subset \( A = {\left\{ {a}_{n}\right\} }_{n} \). Designate \( \mathcal{B} = \left\{ {{B}_{d}\left( {{a}_{n}, m... | The topology generated by \( \mathcal{B} \) is identical to the metric topology. Let \( {B}_{0} = {B}_{d}\left( {x,\epsilon }\right) \) where \( x \in X \) and \( \epsilon > 0 \). Given \( y \in {B}_{0} \), let \( \delta < 1/4\left( {\epsilon - d\left( {x, y}\right) }\right) \), from which it follows that \( {B}_{d}\le... | Yes |
Problem 30.10. | Solution: Let \( f : X \rightarrow Y \) be continous. Suppose \( X \) is Lindelöf. Designate \( \mathcal{C} \) as a covering of \( f\left( X\right) \) of open sets in \( X \) . The collection \( {\mathcal{C}}^{\prime } = \{ {f}^{-1}\left( C\right) : C \in \mathcal{C}\} \) is an open covering of \( X \) , of which there... | Yes |
Problem 30.12. | Solution: Let \( f : X \rightarrow Y \) be a continuous open map. Suppose \( X \) satisfies the first countability axiom. Let \( {U}^{\prime } \) be a neighborhood of some \( y \in f\left( X\right) \) in \( X \) ; therefore \( U = {U}^{\prime } \cap f\left( X\right) \) is open in the subspace \( f\left( X\right) \) . C... | Yes |
Problem 31.1. | Solution: Given distinct points \( x \) and \( y \) in the regular space \( X \) , there are disjoint neighborhoods \( {U}_{x} \) and \( {U}_{y} \) of \( x \) and \( y \) , respectively, because \( X \) is Hausdorff. By Lemma 31.1(a), there are neighborhoods \( {V}_{x} \) and \( {V}_{y} \) of \( x \) and \( y \), respe... | Yes |
Theorem 1. Suppose \( f : X \rightarrow Y \) is surjective. If \( A \) is a subset of \( X \), then \( Y \smallsetminus f\left( {X \smallsetminus A}\right) \) is contained in \( f\left( A\right) \) . | Proof. Since \( f \) is surjective, \( f\left( A\right) \cup f\left( {X \smallsetminus A}\right) = Y \) . If \( y \in Y \smallsetminus f\left( {X \smallsetminus A}\right) \), then \( y \notin f\left( {X \smallsetminus A}\right) \) . Accordingly, \( y \in f\left( A\right) \) . | Yes |
Theorem 2. If \( A \) is an open set contained in closed set \( B \) in space \( X \) , then \( \overline{A} \) is contained in \( B \) . | Proof. If \( x \in \overline{A} \) , then every neighborhood \( U \) of \( x \) intersects \( A \) . Therefore \( U \) also intersects \( B \) , so \( x \in \overline{B} = B \) . We conclude that \( \bar{A} \) is a subset of \( B \) . | Yes |
Theorem 3. Let \( f : X \rightarrow Y \) be a closed surjective map. If \( U \) is open in \( X \) and contains \( {f}^{-1}\left( {\{ y\} }\right) \) for some \( y \in Y \), there is a neighborhood \( V \) of \( y \) in \( Y \) that is contained in \( f\left( U\right) \) . | Proof. Since \( U \) is open, \( X \smallsetminus U \) is closed and does not contain \( {f}^{-1}\left( {\{ y\} }\right) \), so \( f\left( {X \smallsetminus U}\right) \) is closed and does not contain \( y \) . The set \( V = Y \smallsetminus f\left( {X \smallsetminus U}\right) \) is open and contains \( y \) . It foll... | Yes |
Problem 32.3. | Solution: Given a locally compact Hausdorff space \( X \) , if \( X \) is compact then by Theorem 32.3 \( X \) is normal and therefore regular. Otherwise, by Theorem 29.1 there is a one-point compactification \( Y \) of \( X \) which, by Theorem 32.3, is normal and thus regular. Since \( X \) is a subspace of \( Y \), ... | Yes |
Suppose \( Y \) be a subspace of \( X \) and \( {C}^{\prime } \) and \( {D}^{\prime } \) be disjoint closed sets in \( Y \). | By Theorem 17.2., \( {C}^{\prime } = C \cap Y \) and \( {D}^{\prime } = D \cap Y \) where \( C \) and \( D \) are closed in \( X \). If \( x \in {C}^{\prime } \cap {D}^{\prime } \), then \( x \in {D}^{\prime } \) and every neighborhood \( W \) of \( x \) intersects \( {C}^{\prime } \) (since \( x \in C \), which is clo... | Yes |
Problem 32.7. | Solution: Part (a) Yes. Given a subspace \( {Y}_{a} \) of a completely normal space \( {X}_{a} \), a subspace \( {Z}_{a} \) of \( {Y}_{a} \) is also a subspace of \( {X}_{a} \) by exercise 16.1. Therefore \( {Z}_{a} \) is normal, so \( {Y}_{a} \) is completely normal. | No |
Problem 33.1. | Solution: Observe that for the function \( f : X \rightarrow \left\lbrack {a, b}\right\rbrack \) constructed in the proof of Urysohn’s Lemma, any real number in \( \left\lbrack {a, b}\right\rbrack \) may be in \( f\left( X\right) \) since the value of any \( f\left( x\right) \) is the infimum of \( \mathbb{Q}\left( x\r... | Yes |
Proposition 1. Every countable space is Lindelöf. | Proof. Let \( X \) be a countable space. Let \( \mathcal{C} = \left\{ {U}_{\alpha }\right\} \) be an open covering of \( X \) . For each \( x \in X \), choose some \( {U}_{\alpha } \) that contains \( x \) and designate \( {V}_{x} = {U}_{\alpha } \) . Since there are countably many \( x \), it follows that \( {\left\{ ... | Yes |
Proposition 2. Let \( A \) be a nonempty subset of metric space \( X \) with metric \( d \) . The function \( d\left( {x, A}\right) = \inf \{ d\left( {x, a}\right) : a \in A\} \) with \( x \in X \) and fixed \( A \) is continuous. | Proof. Given \( x, y \in X \), by the definition of \( d\left( {x, A}\right) \) and the triangle inequality, for any \( a \in A \) :\n\n\[ d\left( {x, A}\right) \leq d\left( {x, a}\right) \leq d\left( {x, y}\right) + d\left( {y, a}\right) . \]\n\nSince \( d\left( {x, A}\right) - d\left( {x, y}\right) \leq d\left( {y, a... | Yes |
Problem 33.6. Solution: Part (a) Suppose \( X \) is a metrizable space, so it is normal by Theorem 32.2. Let \( A \) be a closed subset of \( X \). If \( A \) is empty or equal to \( X \), it is trivially a \( {G}_{\delta } \) set: In either case, it is equal to the open set \( \varnothing \) or \( X \), respectively. ... | where: \[ f\left( x\right) = \frac{d\left( {x, A}\right) }{d\left( {x, A}\right) + d\left( {x, B}\right) }.\]\n\nThus \( f\left( A\right) = \{ 0\} \) and \( f\left( B\right) = \{ 1\} \). If \( x \) is contained in neither \( A \) nor \( B \), then \( d\left( {x, A}\right) > 0 \) and \( d\left( {x, B}\right) > 0 \). Thi... | Yes |
Proposition 3. Let \( f, g : X \rightarrow Y \) be continuous. The function \( g\left( x\right) = \max \{ f\left( x\right), g\left( x\right) \} \) is continuous. | Proof. We have \( f\left( x\right) > g\left( x\right) \) if and only if \( - f\left( x\right) < - g\left( x\right) \) . Therefore \( \min \{ - f\left( x\right) , - g\left( x\right) \} = \max \{ f\left( x\right), g\left( x\right) \} \) . Since the \( \min \) function is continuous, the max function is continuous by exer... | No |
Problem 34.1. | Solution: One example is \( {I}_{0}^{2} \) . By exercise 30.6, \( {I}_{0}^{2} \) is not metrizable; however, it has a countable basis ( \( \mathbb{R} \) has a countable basis, so as a subspace of \( \mathbb{R} \) it follows that \( {I}_{0}^{2} \) has a countable basis). | No |
Problem 34.2. | Solution: An example is \( {\mathbb{R}}_{\ell } \) . By exercise 30.6, \( {\mathbb{R}}_{\ell } \) is not metrizable. By exercise 32.7(f), \( {\mathbb{R}}_{\ell } \) is completely normal, and by exercise 30.3 it is Lindelöf, countably dense, and first countable. | No |
Problem 34.3. | Solution: Let \( X \) be a compact Hausdorff space. From Theorem 32.3, \( X \) is normal and thus regular. If \( X \) has a countable basis, we infer from the Urysohn metrization theorem that \( X \) is metrizable. Conversely, if \( X \) is metrizble, then by exercise \( {30.4}\mathrm{X} \) has a countable basis. | No |
Proposition 4. An uncountable set in the discrete topology is metrizable. | Proof. Let \( Y \) be an uncountable set in the discrete toplogy. The collection \( {\mathcal{B}}_{0} \) of all one-point sets in \( Y \) is a basis for \( Y \) . Define \( d : Y \times Y \rightarrow {\mathbb{R}}_{ \geq 0} \) as follows. For any \( a \in Y \), let \( d\left( {a, a}\right) = 0 \) . For any \( b \in Y \)... | Yes |
Proposition 5. Let \( Y \) be a compact space, \( X \) a subspace of \( Y \) , and \( A \) a non-empty subset of \( Y \) . If \( A \) is a closed subset of \( Y \), then \( A \) is compact in \( X \) . | Proof. Since \( A \) is closed in \( Y \), it is compact subspace of \( Y \) . Let \( \left\{ {U}_{\alpha }^{\prime }\right\} \) be a covering of \( A \) by sets open in \( X \) . Each \( {U}_{\alpha }^{\prime } = {U}_{\alpha } \cap X \) for some open set \( {U}_{\alpha } \) in \( Y \) . It follows that \( \left\{ {U}_... | Yes |
Proposition 6. Let \( Y \) be a compact space and \( X \) a subspace of \( Y \). If \( A \) a compact subspace of \( X \), then \( A \) is compact in \( Y \). | Proof. Let \( \left\{ {U}_{\alpha }\right\} \) be a covering of \( A \) by sets open in \( Y \). Since \( \left\{ {{U}_{\alpha } \cap X}\right\} \) is a covering of \( A \) by sets open in \( X \), there is a finite subcollection \( {\left\{ {U}_{{\alpha }_{k}} \cap X\right\} }_{k = 1}^{N} \) that covers \( A \). Since... | Yes |
Problem 35.1. | Solution: Define \( f : A \rightarrow \left\lbrack {\alpha ,\beta }\right\rbrack \) where \( f\left( a\right) = \alpha \) for \( a \in A \) and \( g : B \rightarrow \left\lbrack {\alpha ,\beta }\right\rbrack \) where \( f\left( b\right) = \beta \) for \( b \in B \) . Each is continuous as a constant function. By the Pa... | Yes |
Problem 36.1. | Solution: Suppose \( X \) is an m-manifold, so it has a countable basis and is Hausdorff. Given a point \( x \in X \), there is a neighborhood \( {U}_{x} \) of \( x \) that is homeomorphic to an open set \( {V}_{x} \) of \( {\mathbb{R}}^{m} \) by the homeomorphism \( f : x \rightarrow {V}_{x} \) . Since \( {\mathbb{R}}... | No |
Problem 36.3 | Solution: The solution will rely on the following generalized result of exercise 34.9. | No |
Proposition 8. If a compact Hausdorff space \( X \) is the union of \( N \) closed spaces \( {\left\{ {X}_{n}\right\} }_{n = 1}^{N} \) , each of which is metrizable, then \( X \) is metrizable. | Proof. We will prove the result by mathematical induction. If \( X \) equals a single closed metrizable space \( {X}_{1} \) , then trivially \( X \) is metrizable. Assume the inductive hypothesis holds for some \( n \in \mathbb{N} \), and suppose \( X = {X}_{1} \cup \cdots \cup {X}_{n} \cup {X}_{n + 1} = \) \( \left( {... | No |
The circle group \( {\mathbb{S}}^{1} = \mathrm{{SO}}\left( 2\right) \). | Viewed as a set of points in \( \mathbb{C} \) or \( {\mathbb{R}}^{2} \), the unit circle is a closed set because its complement (the set of points not on the circle) is clearly open. Figure 8.1 shows a typical point \( P \) not on the circle and an \( \varepsilon \) -neighborhood of \( P \) that lies in the complement ... | No |
The groups \( \mathrm{O}\left( n\right) \) and \( \mathrm{{SO}}\left( n\right) \). | We view \( \mathrm{O}\left( n\right) \) as a subset of the space \( {\mathbb{R}}^{{n}^{2}} \) of \( n \times n \) real matrices, which we also call \( {M}_{n}\left( \mathbb{R}\right) \) . The complement of \( \mathrm{O}\left( n\right) \) is \[ {M}_{n}\left( \mathbb{R}\right) - \mathrm{O}\left( n\right) = \left\{ {A \in... | Yes |
Example 3. The group \( \operatorname{Aff}\left( 1\right) \) . We view \( \operatorname{Aff}\left( 1\right) \) as in Section 4.6, namely, as the group of real matrices of the form \( A = \left( \begin{array}{ll} a & b \\ 0 & 1 \end{array}\right) \), where \( a, b \in \mathbb{R} \) and \( a > 0 \) . It is now easy to se... | However, \( \operatorname{Aff}\left( 1\right) \) is closed in the \ | No |
The general linear group \( \mathrm{{GL}}\left( {n,\mathbb{C}}\right) \) is the set of all invertible \( n \times n \) complex matrices. This set is a group because it is closed under products (since \( {A}^{-1}{B}^{-1} = \) \( {\left( BA\right) }^{-1} \) ) and under inverses (obviously). | It follows that every group of real or complex matrices is a subgroup of some \( \mathrm{{GL}}\left( {n,\mathbb{C}}\right) ,{}^{8} \) which is why we bring it up now. We are about to define what a \ | No |
Lemma 2.4. If \( \left( {W, < }\right) \) is a well-ordered set and \( f : W \rightarrow W \) is an increasing function, then \( f\left( x\right) \geq x \) for each \( x \in W \) . | Proof. Assume that the set \( X = \{ x \in W : f\left( x\right) < x\} \) is nonempty and let \( z \) be the least element of \( X \) . If \( w = f\left( z\right) \), then \( f\left( w\right) < w \), a contradiction. | Yes |
Corollary 2.5. The only automorphism of a well-ordered set is the identity. | Proof. By Lemma 2.4, \( f\left( x\right) \geq x \) for all \( x \), and \( {f}^{-1}\left( x\right) \geq x \) for all \( x \) . | No |
Lemma 2.7. No well-ordered set is isomorphic to an initial segment of itself. | Proof. If \( \operatorname{ran}\left( f\right) = \{ x : x < u\} \), then \( f\left( u\right) < u \), contrary to Lemma 2.4. | Yes |
Theorem 2.8. If \( {W}_{1} \) and \( {W}_{2} \) are well-ordered sets, then exactly one of the following three cases holds:\n\n(i) \( {W}_{1} \) is isomorphic to \( {W}_{2} \) ;\n\n(ii) \( {W}_{1} \) is isomorphic to an initial segment of \( {W}_{2} \) ;\n\n(iii) \( {W}_{2} \) is isomorphic to an initial segment of \( ... | Proof. For \( u \in {W}_{i},\left( {i = 1,2}\right) \), let \( {W}_{i}\left( u\right) \) denote the initial segment of \( {W}_{i} \) given by \( u \) . Let\n\n\[ f = \left\{ {\left( {x, y}\right) \in {W}_{1} \times {W}_{2} : {W}_{1}\left( x\right) \text{ is isomorphic to }{W}_{2}\left( y\right) }\right\} . \]\n\nUsing ... | Yes |
(i) \( 0 = \varnothing \) is an ordinal. | Proof. (i), (ii) by definition. | No |
Theorem 2.12. Every well-ordered set is isomorphic to a unique ordinal number. | Proof. The uniqueness follows from Lemma 2.7. Given a well-ordered set \( W \) , we find an isomorphic ordinal as follows: Define \( F\left( x\right) = \alpha \) if \( \alpha \) is isomorphic to the initial segment of \( W \) given by \( x \) . If such an \( \alpha \) exists, then it is unique. By the Replacement Axiom... | Yes |
Theorem 2.14 (Transfinite Induction). Let \( C \) be a class of ordinals and assume that:\n\n(i) \( 0 \in C \) ;\n\n(ii) if \( \alpha \in C \), then \( \alpha + 1 \in C \) ;\n\n(iii) if \( \alpha \) is a nonzero limit ordinal and \( \beta \in C \) for all \( \beta < \alpha \), then \( \alpha \in C \) .\n\nThen \( C \) ... | Proof. Otherwise, let \( \alpha \) be the least ordinal \( \alpha \notin C \) and apply (i),(ii), or (iii). | Yes |
Corollary 2.16. Let \( X \) be a set and \( \theta \) an ordinal number. For every function \( G \) on the set of all transfinite sequences in \( X \) of length \( < \theta \) such that \( \operatorname{ran}\left( G\right) \subset X \) there exists a unique \( \theta \) -sequence \( \left\langle {{a}_{\alpha } : \alpha... | Proof. Let\n\n(2.6)\n\n\( F\left( \alpha \right) = x \leftrightarrow \) there is a sequence \( \left\langle {{a}_{\xi } : \xi < \alpha }\right\rangle \) such that:\n\n(i) \( \left( {\forall \xi < \alpha }\right) {a}_{\xi } = G\left( \left\langle {{a}_{\eta } : \eta < \xi }\right\rangle \right) \) ;\n\n(ii) \( x = G\lef... | Yes |
Lemma 2.21. For all ordinals \( \alpha ,\beta \) and \( \gamma \) , (i) \( \alpha + \left( {\beta + \gamma }\right) = \left( {\alpha + \beta }\right) + \gamma \) , (ii) \( \alpha \cdot \left( {\beta \cdot \gamma }\right) = \left( {\alpha \cdot \beta }\right) \cdot \gamma \) . | Proof. By induction on \( \gamma \) . | No |
Lemma 2.24. For all ordinals \( \alpha \) and \( \beta ,\alpha + \beta \) and \( \alpha \cdot \beta \) are, respectively, isomorphic to the sum and to the product of \( \alpha \) and \( \beta \) . | Proof. By induction on \( \beta \) . | No |
Theorem 2.26 (Cantor’s Normal Form Theorem). Every ordinal \( \alpha > \) 0 can be represented uniquely in the form\n\n\[ \alpha = {\omega }^{{\beta }_{1}} \cdot {k}_{1} + \ldots + {\omega }^{{\beta }_{n}} \cdot {k}_{n} \]\n\nwhere \( n \geq 1,\alpha \geq {\beta }_{1} > \ldots > {\beta }_{n} \), and \( {k}_{1},\ldots ,... | Proof. By induction on \( \alpha \) . For \( \alpha = 1 \) we have \( 1 = {\omega }^{0} \cdot 1 \) ; for arbitrary \( \alpha > 0 \) let \( \beta \) be the greatest ordinal such that \( {\omega }^{\beta } \leq \alpha \) . By Lemma 2.25(iv) there exists a unique \( \delta \) and a unique \( \rho < {\omega }^{\beta } \) s... | Yes |
Theorem 2.27. If \( E \) is a well-founded relation on \( P \), then there exists a unique function \( \rho \) from \( P \) into the ordinals such that for all \( x \in P \), (2.7) \[ \rho \left( x\right) = \sup \{ \rho \left( y\right) + 1 : {yEx}\} . | Proof. We shall define a function \( \rho \) satisfying (2.7) and then prove its uniqueness. By induction, let \[ {P}_{0} = \varnothing ,\;{P}_{\alpha + 1} = \left\{ {x \in P : \forall y\left( {{yEx} \rightarrow y \in {P}_{\alpha }}\right) }\right\} , \] \[ {P}_{\alpha } = \mathop{\bigcup }\limits_{{\xi < \alpha }}{P}_... | Yes |
Theorem 3.1 (Cantor). For every set \( X,\left| X\right| < \left| {P\left( X\right) }\right| \) . | Proof. Let \( f \) be a function from \( X \) into \( P\left( X\right) \) . The set\n\n\[ Y = \{ x \in X : x \notin f\left( x\right) \} \]\n\nis not in the range of \( f \) : If \( z \in X \) were such that \( f\left( z\right) = Y \), then \( z \in Y \) if and only if \( z \notin Y \), a contradiction. Thus \( f \) is ... | Yes |
Theorem 3.2 (Cantor-Bernstein). If \( \\left| A\\right| \\leq \\left| B\\right| \) and \( \\left| B\\right| \\leq \\left| A\\right| \), then \( \\left| A\\right| = \\left| B\\right| \) . | Proof. If \( {f}_{1} : A \\rightarrow B \) and \( {f}_{2} : B \\rightarrow A \) are one-to-one, then if we let \( {B}^{\\prime } = \) \( {f}_{2}\\left( B\\right) \) and \( {A}_{1} = {f}_{2}\\left( {{f}_{1}\\left( A\\right) }\\right) \), we have \( {A}_{1} \\subset {B}^{\\prime } \\subset A \) and \( \\left| {A}_{1}\\ri... | Yes |
Lemma 3.3. If \( \left| A\right| = \kappa \), then \( \left| {P\left( A\right) }\right| = {2}^{\kappa } \) . | Proof. For every \( X \subset A \), let \( {\chi }_{X} \) be the function\n\n\[ \n{\chi }_{X}\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }x \in X \\ 0 & \text{ if }x \in A - X \end{array}\right.\n\]\n\nThe mapping \( f : X \rightarrow {\chi }_{X} \) is a one-to-one correspondence between \( P\left( A\righ... | No |
Lemma 3.6. \( \operatorname{cf}\left( {\operatorname{cf}\alpha }\right) = \operatorname{cf}\alpha \) . | Proof. If \( \left\langle {{\alpha }_{\xi } : \xi < \beta }\right\rangle \) is cofinal in \( \alpha \) and \( \langle \xi \left( \nu \right) : \nu < \gamma \rangle \) is cofinal in \( \beta \), then \( \left\langle {{\alpha }_{\xi \left( \nu \right) } : \nu < \gamma }\right\rangle \) is cofinal in \( \alpha \) . | Yes |
Lemma 3.7. Let \( \alpha > 0 \) be a limit ordinal.\n\n(i) If \( A \subset \alpha \) and \( \sup A = \alpha \), then the order-type of \( A \) is at least \( \operatorname{cf}\alpha \) .\n\n(ii) If \( {\beta }_{0} \leq {\beta }_{1} \leq \ldots \leq {\beta }_{\xi } \leq \ldots ,\xi < \gamma \), is a nondecreasing \( \ga... | Proof. (i) The order-type of \( A \) is the length of the increasing enumeration of \( A \) which is an increasing sequence with limit \( \alpha \) .\n\n(ii) If \( \gamma = \mathop{\lim }\limits_{{\nu \rightarrow \operatorname{cf}\gamma }}\xi \left( \nu \right) \), then \( \alpha = \mathop{\lim }\limits_{{\nu \rightarr... | Yes |
Lemma 3.8. For every limit ordinal \( \alpha , \) cf \( \alpha \) is a regular cardinal. | Proof. It is easy to see that if \( \alpha \) is not a cardinal, then using a mapping of \( \left| \alpha \right| \) onto \( \alpha \), one can construct a cofinal sequence in \( \alpha \) of length \( \leq \left| \alpha \right| \), and therefore cf \( \alpha < \alpha \). Since \( \operatorname{cf}\left( {\operatorname... | Yes |
Lemma 3.9. Let \( \kappa \) be an aleph.\n\n(i) If \( X \subset \kappa \) and \( \left| X\right| < \operatorname{cf}\kappa \) then \( X \) is bounded.\n\n(ii) If \( \lambda < \operatorname{cf}\kappa \) and \( f : \lambda \rightarrow \kappa \) then the range of \( f \) is bounded. | Proof. (i) Lemma 3.7(i).\n\n(ii) If \( X = \operatorname{ran}\left( f\right) \) then \( \left| X\right| \leq \lambda \), and use (i). | No |
Lemma 3.10. An infinite cardinal \( \kappa \) is singular if and only if there exists a cardinal \( \lambda < \kappa \) and a family \( \left\{ {{S}_{\xi } : \xi < \lambda }\right\} \) of subsets of \( \kappa \) such that \( \left| {S}_{\xi }\right| < \kappa \) for each \( \xi < \lambda \), and \( \kappa = \mathop{\big... | Proof. If \( \kappa \) is singular, then there is an increasing sequence \( \left\langle {{\alpha }_{\xi } : \xi < \operatorname{cf}\kappa }\right\rangle \) with \( \mathop{\lim }\limits_{\xi }{\alpha }_{\xi } = \kappa \) . Let \( \lambda = \operatorname{cf}\kappa \), and \( {S}_{\xi } = {\alpha }_{\xi } \) for all \( ... | Yes |
Theorem 3.11. If \( \kappa \) is an infinite cardinal, then \( \kappa < {\kappa }^{\text{cf }\kappa } \) . | Proof. Let \( F \) be a collection of \( \kappa \) functions from cf \( \kappa \) to \( \kappa : F = \left\{ {{f}_{\alpha } : \alpha < \kappa }\right\} \) . It is enough to find \( f : \) cf \( \kappa \rightarrow \kappa \) that is different from all the \( {f}_{\alpha } \) . Let \( \kappa = \) \( \mathop{\lim }\limits_... | Yes |
Theorem 4.1 (Cantor). The set of all real numbers is uncountable. | Proof. Let us assume that the set \( \mathbf{R} \) of all reals is countable, and let \( {c}_{0} \) , \( {c}_{1},\ldots ,{c}_{n},\ldots, n \in \mathbf{N} \), be an enumeration of \( \mathbf{R} \) . We shall find a real number different from each \( {c}_{n} \) .\n\nLet \( {a}_{0} = {c}_{0} \) and \( {b}_{0} = {c}_{{k}_{... | Yes |
Theorem 4.4. Let \( \left( {P, < }\right) \) be a dense unbounded linearly ordered set. Then there is a complete unbounded linearly ordered set \( \left( {C, \prec }\right) \) such that:\n\n(i) \( P \subset C \), and \( < \) and \( \prec \) agree on \( P \) ;\n\n(ii) \( P \) is dense in \( C \) . | Proof. A Dedekind cut in \( P \) is a pair \( \left( {A, B}\right) \) of disjoint nonempty subsets of \( P \) such that\n\n(i) \( A \cup B = P \) ;\n\n(ii) \( a < b \) for any \( a \in A \) and \( b \in B \) ;\n\n(iii) \( A \) does not have a greatest element.\n\nLet \( C \) be the set of all Dedekind cuts in \( P \) a... | Yes |
Theorem 4.5. Every perfect set has cardinality \( \mathfrak{c} \) . | Proof. Given a perfect set \( P \), we want to find a one-to-one function \( F \) from \( \{ 0,1{\} }^{\omega } \) into \( P \) . Let \( S \) be the set of all finite sequences of 0 ’s and 1’s. By induction on the length of \( s \in S \) one can find closed intervals \( {I}_{s} \) such that for each \( n \) and all \( ... | Yes |
If \( F \) is a closed set, then either \( \left| F\right| \leq {\aleph }_{0} \) or \( \left| F\right| = {2}^{{\aleph }_{0}} \). | For every \( A \subset \mathbf{R} \), let\n\n\[ {A}^{\prime } = \text{the set of all limit points of}A \]\n\nIt is easy to see that \( {A}^{\prime } \) is closed, and if \( A \) is closed then \( {A}^{\prime } \subset A \) . Thus we let\n\n\[ {F}_{0} = F,\;{F}_{\alpha + 1} = {F}_{\alpha }^{\prime }, \]\n\n\[ {F}_{\alph... | Yes |
Theorem 4.8 (The Baire Category Theorem). If \( {D}_{0},{D}_{1},\ldots ,{D}_{n},\ldots \) , \( n \in \mathbf{N} \), are dense open sets of reals, then the intersection \( D = \mathop{\bigcap }\limits_{{n = 0}}^{\infty }{D}_{n} \) is dense in \( \mathbf{R} \) . | Proof. We show that \( D \) intersects every nonempty open interval \( I \) . First note that for each \( n,{D}_{0} \cap \ldots \cap {D}_{n} \) is dense and open. Let \( \left\langle {{J}_{k} : k \in \mathbf{N}}\right\rangle \) be an enumeration of rational intervals. Let \( {I}_{0} = I \), and let, for each \( n \) , ... | Yes |
Lemma 4.11. A closed set \( F \subset \mathcal{N} \) is perfect if and only if the tree \( {T}_{F} \) is a perfect tree. | The Cantor-Bendixson analysis for closed sets in the Baire space is carried out as follows: For each tree \( T \subset {Seq} \), we let\n\n(4.7)\n\n\[ \n{T}^{\prime } = \left\{ {t \in T : }\right. \text{there exist incomparable}\left. {{s}_{1} \supset t\text{and}{s}_{2} \supset t\text{in}T}\right\} \text{.} \]\n\n(Thus... | Yes |
Lemma 5.2. \( \left| {\bigcup S}\right| \leq \left| S\right| \cdot \sup \{ \left| X\right| : X \in S\} \) . | Proof. Let \( \kappa = \left| S\right| \) and \( \lambda = \sup \{ \left| X\right| : X \in S\} \) . We have \( S = \left\{ {{X}_{\alpha } : \alpha < \kappa }\right\} \) and for each \( \alpha < \kappa \), we choose an enumeration \( {X}_{\alpha } = \left\{ {{a}_{\alpha ,\beta } : \beta < {\lambda }_{\alpha }}\right\} \... | Yes |
Corollary 5.3. Every \( {\aleph }_{\alpha + 1} \) is a regular cardinal. | Proof. This is because otherwise \( {\omega }_{\alpha + 1} \) would be the union of at most \( {\aleph }_{\alpha } \) sets of cardinality at most \( {\aleph }_{\alpha } \) . | No |
Theorem 5.4 (Zorn’s Lemma). If \( \left( {P, < }\right) \) is a nonempty partially ordered set such that every chain in \( P \) has an upper bound, then \( P \) has a maximal element. | Proof. We construct (using a choice function for nonempty subsets of \( P \) ), a chain in \( P \) that leads to a maximal element of \( P \) . We let, by induction,\n\n\( {a}_{\alpha } = \) an element of \( P \) such that \( {a}_{\alpha } > {a}_{\xi } \) for every \( \xi < \alpha \) if there is one.\n\nClearly, if \( ... | Yes |
Lemma 5.6. If \( 2 \leq \kappa \leq \lambda \) and \( \lambda \) is infinite, then \( {\kappa }^{\lambda } = {2}^{\lambda } \) . | Proof.\n\n(5.6)\n\n\[ \n{2}^{\lambda } \leq {\kappa }^{\lambda } \leq {\left( {2}^{\kappa }\right) }^{\lambda } = {2}^{\kappa \cdot \lambda } = {2}^{\lambda } \n\] | Yes |
Lemma 5.7. If \( \left| A\right| = \kappa \geq \lambda \), then the set \( {\left\lbrack A\right\rbrack }^{\lambda } \) has cardinality \( {\kappa }^{\lambda } \) . | Proof. On the one hand, every \( f : \lambda \rightarrow A \) is a subset of \( \lambda \times A \), and \( \left| f\right| = \lambda \) . Thus \( {\kappa }^{\lambda } \leq {\left| \left\lbrack \lambda \times A\right\rbrack \right| }^{\lambda } = \left| {\left\lbrack A\right\rbrack }^{\lambda }\right| \) . On the other... | Yes |
Lemma 5.8. If \( \lambda \) is an infinite cardinal and \( {\kappa }_{i} > 0 \) for each \( i < \lambda \), then\n\n\[ \mathop{\sum }\limits_{{i < \lambda }}{\kappa }_{i} = \lambda \cdot \mathop{\sup }\limits_{{i < \lambda }}{\kappa }_{i} \] | Proof. Let \( \kappa = \mathop{\sup }\limits_{{i < \lambda }}{\kappa }_{i} \) and \( \sigma = \mathop{\sum }\limits_{{i < \lambda }}{\kappa }_{i} \) . On the one hand, since \( {\kappa }_{i} \leq \kappa \) for all \( i \), we have \( \mathop{\sum }\limits_{{i < \lambda }}\kappa \leq \lambda \cdot \kappa \) . On the oth... | Yes |
Lemma 5.9. If \( \lambda \) is an infinite cardinal and \( \left\langle {{\kappa }_{i} : i < \lambda }\right\rangle \) is a nondecreasing sequence of nonzero cardinals, then\n\n\[ \mathop{\prod }\limits_{{i < \lambda }}{\kappa }_{i} = {\left( \mathop{\sup }\limits_{i}{\kappa }_{i}\right) }^{\lambda } \] | Proof. Let \( \kappa = \mathop{\sup }\limits_{i}{\kappa }_{i} \) . Since \( {\kappa }_{i} \leq \kappa \) for each \( i < \lambda \), we have\n\n\[ \mathop{\prod }\limits_{{i < \lambda }}{\kappa }_{i} \leq \mathop{\prod }\limits_{{i < \lambda }}\kappa = {\kappa }^{\lambda } \]\n\nTo prove that \( {\kappa }^{\lambda } \l... | Yes |
Theorem 5.10 (König). If \( {\kappa }_{i} < {\lambda }_{i} \) for every \( i \in I \), then\n\n\[ \mathop{\sum }\limits_{{i \in I}}{\kappa }_{i} < \mathop{\prod }\limits_{{i \in I}}{\lambda }_{i} \]\n | Proof. We shall show that \( \mathop{\sum }\limits_{i}{\kappa }_{i} \ngeqslant \mathop{\prod }\limits_{i}{\lambda }_{i} \) . Let \( {T}_{i}, i \in I \), be such that \( \left| {T}_{i}\right| = {\lambda }_{i} \) for each \( i \in I \) . It suffices to show that if \( {Z}_{i}, i \in I \), are subsets of \( T = \mathop{\p... | Yes |
Corollary 5.11. \( \kappa < {2}^{\kappa } \) for every \( \kappa \) . | Proof. \( \underset{\kappa \text{ times }}{\underbrace{1 + 1 + \ldots }} < \underset{\kappa \text{ times }}{\underbrace{2 \cdot 2 \cdot \ldots }}. \) | No |
Corollary 5.12. \( \operatorname{cf}\left( {2}^{{\aleph }_{\alpha }}\right) > {\aleph }_{\alpha } \) . | Proof. It suffices to show that if \( {\kappa }_{i} < {2}^{{\aleph }_{\alpha }} \) for \( i < {\omega }_{\alpha } \), then \( \mathop{\sum }\limits_{{i < {\omega }_{\alpha }}}{\kappa }_{i} < {2}^{{\aleph }_{\alpha }} \) . Let \( {\lambda }_{i} = {2}^{{\aleph }_{\alpha }} \) .\n\n\[\n\mathop{\sum }\limits_{{i < {\omega ... | Yes |
Corollary 5.13. \( \operatorname{cf}\left( {\aleph }_{\alpha }^{{\aleph }_{\beta }}\right) > {\aleph }_{\beta } \) . | Proof. We show that if \( {\kappa }_{i} < {\aleph }_{\alpha }^{{\aleph }_{\beta }} \) for \( i < {\omega }_{\beta } \), then \( \mathop{\sum }\limits_{{i < {\omega }_{\beta }}}{\kappa }_{i} < {\aleph }_{\alpha }^{{\aleph }_{\beta }} \) . Let \( {\lambda }_{i} = {\aleph }_{\alpha }^{{\aleph }_{\beta }} \n\n\[ \n\mathop{... | Yes |
Corollary 5.14. \( {\kappa }^{\text{cf }\kappa } > \kappa \) for every infinite cardinal \( \kappa \) . | Proof. Let \( {\kappa }_{i} < \kappa, i < \) cf \( \kappa \), be such that \( \kappa = \mathop{\sum }\limits_{{i < \operatorname{cf}\kappa }}{\kappa }_{i} \) . Then\n\n\[ \kappa = \mathop{\sum }\limits_{{i < \operatorname{cf}\kappa }}{\kappa }_{i} < \mathop{\prod }\limits_{{i < \operatorname{cf}\kappa }}\kappa = {\kapp... | Yes |
Theorem 5.15. If \( \mathrm{{GCH}} \) holds and \( \kappa \) and \( \lambda \) are infinite cardinals then:\n\n(i) If \( \kappa \leq \lambda \), then \( {\kappa }^{\lambda } = {\lambda }^{ + } \).\n\n(ii) If cf \( \kappa \leq \lambda < \kappa \), then \( {\kappa }^{\lambda } = {\kappa }^{ + } \).\n\n(iii) If \( \lambda... | Proof. (i) Lemma 5.6.\n\n(ii) This follows from (5.7) and (5.8).\n\n(iii) By Lemma 3.9(ii), the set \( {\kappa }^{\lambda } \) is the union of the sets \( {\alpha }^{\lambda },\alpha < \kappa \), and \( \left| {\alpha }^{\lambda }\right| \leq {2}^{\left| \alpha \right| \cdot \lambda } = {\left( \left| \alpha \right| \c... | No |
(ii) cf \( {2}^{\kappa } > \kappa \) . | By Corollary 5.12, | No |
Corollary 5.17. If \( \kappa \) is a singular cardinal and if the continuum function is eventually constant below \( \kappa \), with value \( \lambda \), then \( {2}^{\kappa } = \lambda \) . | Proof. If \( \kappa \) is a singular cardinal that satisfies the assumption of the theorem, then there is \( \mu \) such that cf \( \kappa \leq \mu < \kappa \) and that \( {2}^{ < \kappa } = \lambda = {2}^{\mu } \) . Thus\n\n\[ \n{2}^{\kappa } = {\left( {2}^{ < \kappa }\right) }^{\operatorname{cf}\kappa } = {\left( {2}... | Yes |
Lemma 5.19. If \( \kappa \) is a limit cardinal, and \( \lambda \geq \operatorname{cf}\kappa \), then\n\n\[{\kappa }^{\lambda } = {\left( \mathop{\lim }\limits_{{\alpha \rightarrow \kappa }}{\alpha }^{\lambda }\right) }^{\operatorname{cf}\kappa }\] | Proof. Let \( \kappa = \mathop{\sum }\limits_{{i < \operatorname{cf}\kappa }}{\kappa }_{i} \), where \( {\kappa }_{i} < \kappa \) for each \( i \) . We have \( {\kappa }^{\lambda } \leq \) \( {\left( \mathop{\prod }\limits_{{i < \operatorname{cf}\kappa }}{\kappa }_{i}\right) }^{\lambda } = \mathop{\prod }\limits_{i}{\k... | Yes |
Theorem 5.20. Let \( \lambda \) be an infinite cardinal. Then for all infinite cardinals \( \kappa \), the value of \( {\kappa }^{\lambda } \) is computed as follows, by induction on \( \kappa \) :\n\n(i) If \( \kappa \leq \lambda \) then \( {\kappa }^{\lambda } = {2}^{\lambda } \).\n\n(ii) If there exists some \( \mu ... | Proof. (i) Lemma 5.6\n\n(ii) \( {\mu }^{\lambda } \leq {\kappa }^{\lambda } \leq {\left( {\mu }^{\lambda }\right) }^{\lambda } = {\mu }^{\lambda } \).\n\n(iii) If \( \kappa \) is a successor cardinal, we use the Hausdorff formula. If \( \kappa \) is a limit cardinal, we have \( \mathop{\lim }\limits_{{\alpha \rightarro... | Yes |
For every \( \kappa \) and \( \lambda \), the value of \( {\kappa }^{\lambda } \) is either \( {2}^{\lambda } \), or \( \kappa \), or \( \gimel \left( \mu \right) \) for some \( \mu \) such that \( \operatorname{cf}\mu \leq \lambda < \mu \) . | If \( {\kappa }^{\lambda } > {2}^{\lambda } \cdot \kappa \), let \( \mu \) be the least cardinal such that \( {\mu }^{\lambda } = {\kappa }^{\lambda } \), and by Theorem 5.20 (for \( \mu \) and \( \lambda \) ), \( {\mu }^{\lambda } = {\mu }^{\text{cf }\mu } \) . | No |
Theorem 5.22. Assume that SCH holds.\n\n(i) If \( \kappa \) is a singular cardinal then\n\n(a) \( {2}^{\kappa } = {2}^{ < \kappa } \) if the continuum function is eventually constant below \( \kappa \) ,\n\n(b) \( {2}^{\kappa } = {\left( {2}^{ < \kappa }\right) }^{ + } \) otherwise.\n\n(ii) If \( \kappa \) and \( \lamb... | Proof. (i) If \( \kappa \) is a singular cardinal, then by Theorem 5.16, \( {2}^{\kappa } \) is either \( \lambda \) or \( {\lambda }^{\text{cf }\kappa } \) where \( \lambda = {2}^{ < \kappa } \) . The latter occurs if \( {2}^{\alpha } \) is not eventually constant below \( \kappa \) . Then cf \( \lambda = \operatornam... | Yes |
Lemma 6.1. For every set \( S \) there exists a transitive set \( T \supset S \) . | Proof. We define by induction\n\n\[ \n{S}_{0} = S,\;{S}_{n + 1} = \bigcup {S}_{n} \n\]\n\nand\n\n(6.1)\n\n\[ \nT = \mathop{\bigcup }\limits_{{n = 0}}^{\infty }{S}_{n} \n\]\n\nClearly, \( T \) is transitive and \( T \supset S \) . | Yes |
Lemma 6.2. Every nonempty class \( C \) has an \( \in \) -minimal element. | Proof. Let \( S \in C \) be arbitrary. If \( S \cap C = \varnothing \), then \( S \) is a minimal element of \( C \) ; if \( S \cap C \neq \varnothing \), we let \( X = T \cap C \) where \( T = \operatorname{TC}\left( S\right) .X \) is a nonempty set and by the Axiom of Regularity, there is \( x \in X \) such that \( x... | Yes |
Lemma 6.3. For every \( x \) there is \( \alpha \) such that \( x \in {V}_{\alpha } \): | Proof. Let \( C \) be the class of all \( x \) that are not in any \( {V}_{\alpha } \) . If \( C \) is nonempty, then \( C \) has an \( \in \) -minimal element \( x \) . That is, \( x \in C \), and \( z \in \mathop{\bigcup }\limits_{\alpha }{V}_{\alpha } \) for every \( z \in x \) . Hence \( x \subset \mathop{\bigcup }... | Yes |
Theorem 6.4 ( \( \in \) -Induction). Let \( T \) be a transitive class, let \( \Phi \) be a property. Assume that\n\n(i) \( \Phi \left( \varnothing \right) \) ;\n\n(ii) if \( x \in T \) and \( \Phi \left( z\right) \) holds for every \( z \in x \), then \( \Phi \left( x\right) \) .\n\nThen every \( x \in T \) has proper... | Proof. Let \( C \) be the class of all \( x \in T \) that do not have the property \( \Phi \) . If \( C \) is nonempty, then it has an \( \in \) -minimal element \( x \) ; apply (i) or (ii). | No |
Theorem 6.5 ( \( \\in \) -Recursion). Let \( T \) be a transitive class and let \( G \) be a function (defined for all \( x \) ). Then there is a function \( F \) on \( T \) such that\n\n(6.6)\n\n\[ F\\left( x\\right) = G\\left( {F \\upharpoonright x}\\right) \]\n\nfor every \( x \\in T \) .\n\nMoreover, \( F \) is the... | Proof. We let, for every \( x \\in T \) ,\n\n\( F\\left( x\\right) = y \\leftrightarrow \) there exists a function \( f \) such that\n\n\( \\operatorname{dom}\\left( f\\right) \) is a transitive subset of \( T \) and:\n\n(i) \( \\left( {\\forall z \\in \\operatorname{dom}\\left( f\\right) }\\right) f\\left( z\\right) =... | Yes |
Corollary 6.6. Let \( A \) be a class. There is a unique class \( B \) such that\n\n\[ B = \{ x \in A : x \subset B\} . \] | Proof. Let\n\n\[ F\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }x \in A\text{ and }F\left( z\right) = 1\text{ for all }z \in x, \\ 0 & \text{ otherwise. } \end{array}\right. \]\n\nLet \( B = \{ x : F\left( x\right) = 1\} \) . The uniqueness of \( B \) is proved by \( \in \) -induction. | No |
Theorem 6.7. Let \( {T}_{1},{T}_{2} \) be transitive classes and let \( \pi \) be an \( \in \) -isomorphism of \( {T}_{1} \) onto \( {T}_{2} \) ; i.e., \( \pi \) is one-to-one and\n\n(6.8)\n\n\[ u \in v \leftrightarrow {\pi u} \in {\pi v} \]\n\nThen \( {T}_{1} = {T}_{2} \) and \( {\pi u} = u \) for every \( u \in {T}_{... | Proof. We show, by \( \in \) -induction, that \( {\pi x} = x \) for every \( x \in {T}_{1} \) . Assume that \( {\pi z} = z \) for each \( z \in x \) and let \( y = {\pi x} \) .\n\nWe have \( x \subset y \) because if \( z \in x \), then \( z = {\pi z} \in {\pi x} = y \) .\n\nWe also have \( y \subset x \) : Let \( t \i... | Yes |
Lemma 6.9. If \( E \) is a well-founded relation on \( P \), then every nonempty class \( C \subset P \) has an \( E \) -minimal element. | Proof. We follow the proof of Lemma 6.2; we are looking for \( x \in C \) such that \( {\operatorname{ext}}_{E}\left( x\right) \cap C = \varnothing \) . Let \( S \in C \) be arbitrary and assume that \( {\operatorname{ext}}_{E}\left( S\right) \cap C \neq \varnothing \) . We let \( X = T \cap C \) where\n\n\[ T = \matho... | No |
Theorem 6.10 (Well-Founded Induction). Let \( E \) be a well-founded relation on \( P \) . Let \( \Phi \) be a property. Assume that:\n\n(i) every \( E \) -minimal element \( x \) has property \( \Phi \) ;\n\n(ii) if \( x \in P \) and if \( \Phi \left( z\right) \) holds for every \( z \) such that \( {zEx} \), then \( ... | Proof. A modification of the proof of Theorem 6.4. | No |
Theorem 6.11 (Well-Founded Recursion). Let \( E \) be a well-founded relation on \( P \) . Let \( G \) be a function \( \left( {\text{on}V \times V}\right) \) . Then there is a unique function \( F \) on \( P \) such that\n\n\[ F\left( x\right) = G\left( {x, F \upharpoonright {\operatorname{ext}}_{E}\left( x\right) }\r... | Proof. A modification of the proof of Theorem 6.5.\n\n(Note that if \( F\left( x\right) = G\left( {F \upharpoonright \operatorname{ext}\left( x\right) }\right) \) for some \( G \), then \( F\left( x\right) = F\left( y\right) \) whenever ext \( \left( x\right) = \operatorname{ext}\left( y\right) \) ; in particular, \( F... | No |
## Theorem 6.15 (Mostowski’s Collapsing Theorem).\n\n(i) If \( E \) is a well-founded and extensional relation on a class \( P \), then there is a transitive class \( M \) and an isomorphism \( \pi \) between \( \left( {P, E}\right) \) and \( \left( {M, \in }\right) \) . The transitive class \( M \) and the isomorphism... | Proof. Since (ii) is a special case of (i) \( \left( {E = \in \text{in case (ii)), we shall prove}}\right) \) the existence of an isomorphism in the general case.\n\nSince \( E \) is a well-founded relation, we can define \( \pi \) by well-founded induction (Theorem 6.11), i.e., \( \pi \left( x\right) \) can be defined... | Yes |
Lemma 7.4. A filter \( F \) on \( S \) is an ultrafilter if and only if it is maximal. | Proof. (a) An ultrafilter \( U \) is clearly a maximal filter: Assume that \( U \subset F \) and \( X \in F - U \) . Then \( S - X \in U \), and so both \( S - X \in F \) and \( X \in F \) , a contradiction.\n\n(b) Let \( F \) be a filter that is not an ultrafilter. We will show that \( F \) is not maximal. Let \( Y \s... | Yes |
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