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Theorem 14.4.5 (Biró theorem [31]). Let \( K \) be an uncountable Kronecker set in \( \mathbb{T} \). Then the subgroup \( \langle K\rangle \) is not Polishable. In particular, \( \langle K\rangle \) cannot be characterized. | Since \( \langle K\rangle \) is obviously \( {F}_{\sigma } \), this provides an example of a noncharacterized \( {F}_{\sigma } \) -subgroup of \( \mathbb{T} \), thereby answering a question of the second named author (see also [91], where some special classes of \( {F}_{\sigma } \) -subgroups were shown to be character... | Yes |
Theorem 14.4.7 (Borel theorem [36]). Every countable subgroup of \( \mathbb{T} \) is characterized. | Three proofs of Borel theorem 14.4.7 were given in [32]. | No |
Lemma 15.1.1. (a) Closed subspaces and continuous images of countably compact spaces are countably compact. | Proof. (a) The first assertion is obvious. To prove the second, assume that \( X \) is a countably compact space and \( f : X \rightarrow Y \) is a continuous surjective function. Let \( \mathcal{U} = \left\{ {{U}_{n} : n \in }\right. \) \( \mathbb{N}\} \) be an open cover of \( Y \), so \( \mathcal{V} = \left\{ {{f}^{... | Yes |
Theorem 15.1.2. For a Tichonov space \( X \), the following are equivalent:\n\n(a) \( X \) is pseudocompact;\n\n(b) every locally finite family of nonempty open sets of \( X \) is finite;\n\n(c) \( \mathop{\bigcap }\limits_{{n \in \mathbb{N}}}{V}_{n} \neq \varnothing \) for every chain \( \left\{ {{V}_{n} : n \in \math... | Proof. (a) \( \Rightarrow \) (b) Assume that \( \left\{ {{V}_{n} : n \in {\mathbb{N}}_{ + }}\right\} \) is an infinite locally finite family of nonempty open sets of \( X \) . For every \( n \in {\mathbb{N}}_{ + } \), fix a point \( {x}_{n} \in {V}_{n} \) ; since \( X \) is Tichonov, there exists a continuous function ... | Yes |
Proposition 15.1.3. If \( X \) is a dense pseudocompact subspace of a Tichonov space \( Y \) , then \( X \) is \( {G}_{\delta } \) -dense in \( Y \) . | Proof. Let \( O \) be a nonempty \( {G}_{\delta } \) -set of \( Y \) . Then there exist \( y \in O \) and open sets \( {U}_{n} \) of \( Y \) such that \( O = \mathop{\bigcap }\limits_{{n \in \mathbb{N}}}{U}_{n} \) . By the regularity of \( Y \), we can find, for each \( n \in \mathbb{N} \), an open set \( {V}_{n} \) of... | Yes |
Corollary 15.1.4. A Tichonov space \( X \) is pseudocompact if and only if \( X \) is \( {G}_{\delta } \) -dense in \( {\beta X} \) . | Proof. Assume that \( X \) is not pseudocompact and arrange for an unbounded continuous function \( f : X \rightarrow \mathbb{R} \) with \( f\left( x\right) \geq 1 \) for all \( x \in X \) . Then \( g \mathrel{\text{:=}} 1/f : X \rightarrow (0,1\rbrack \) has \( \inf \{ g\left( x\right) :x \in X\} = 0 \), so its contin... | Yes |
Theorem 15.1.5. Every pseudocompact Tichonov space \( X \) is a Baire space. | Proof. Any compactification \( Y \) of \( X \) is a Baire space, by Theorem B.5.20. Since \( X \) is \( {G}_{\delta } \) -dense in \( Y \) by Proposition 15.1.3, Exercise 15.4.1 applies. | No |
Corollary 15.2.2. If \( G \) is a \( {G}_{\delta } \) -dense subgroup of a pseudocompact group \( H \), then \( G \) is pseudocompact. | Proof. Since \( {G}_{\delta } \) -density is transitive, \( G \) is \( {G}_{\delta } \) -dense in the completion of \( H \) . So, Comfort-Ross criterion 15.2.1 applies. | Yes |
Corollary 15.2.3. Direct products of pseudocompact groups are pseudocompact. | Proof. If \( \left\{ {{G}_{i} : i \in I}\right\} \) is a family of pseudocompact groups, then their completions \( \left\{ {{K}_{i} : i \in }\right. \) \( I\} \) are compact and the completion of \( G = \mathop{\prod }\limits_{{i \in I}}{G}_{i} \) is \( K = \mathop{\prod }\limits_{{i \in I}}{K}_{i} \) . Then \( K \) is... | Yes |
Corollary 15.2.4. Every countably compact group is precompact. In particular, a countably compact group is compact precisely when it is complete. | Proof. This follows from Comfort-Ross criterion 15.2.1, since every countably compact group is pseudocompact by Lemma 15.1.1(c) and Theorem 10.2.6. | No |
Lemma 15.2.9. Let \( \left( {G,\tau }\right) \) be a topological group. Then \( \Lambda \left( G\right) \) is a neighborhood base at \( {e}_{G} \) of \( \left( {G,{\tau }_{\delta }}\right) \) . If \( G \) is precompact, then \( {\Lambda }_{ \trianglelefteq }\left( G\right) \) is a neighborhood base at \( {e}_{G} \) of ... | Proof. It is clear that every \( H \in \Lambda \left( G\right) \) is a neighborhood of \( {e}_{G} \) in \( {\tau }_{\delta } \) . Conversely, let \( O = \) \( \mathop{\bigcap }\limits_{{n \in \mathbb{N}}}{U}_{n} \), where each \( {U}_{n} \) is an open neighborhood of \( {e}_{G} \) in \( \tau \) . We can build inductive... | Yes |
Lemma 15.2.10. Let \( \left( {G,\tau }\right) \) be a topological group and \( H \) a subgroup of \( G \) . Then:\n\n(a) if \( H \in \Lambda \left( G\right) \) and \( L \in \Lambda \left( H\right) \), then \( L \in \Lambda \left( G\right) \) ; | Proof. (a) The subgroup \( L \) of \( G \) is \( \tau \) -closed and \( L \in {\left( \tau { \upharpoonright }_{H}\right) }_{\delta } \subseteq {\tau }_{\delta } \), since \( H \in {\tau }_{\delta } \) . | Yes |
Lemma 15.3.1. Let \( X \) be a topological space and \( Y \) a pseudocompact \( {G}_{\delta } \) -dense subset of \( X \) . Then \( Y \) is \( C \) -embedded in \( X \) if and only if \( Y \) is \( {C}^{ * } \) -embedded in \( X \) . | Proof. By the above remark, it is sufficient to prove that \( Y \) is \( C \) -embedded if it is \( {C}^{ * } \) -embedded. Pick a continuous function \( f : Y \rightarrow \mathbb{R} \) . Since \( Y \) is pseudocompact, \( f \) is bounded, and so we can assume without loss of generality that \( f\left( Y\right) \subset... | Yes |
Theorem 15.3.3. If a dense subspace \( Y \) of a Tichonov space \( X \) is \( C \) -embedded in \( X \), then \( Y \) is \( {G}_{\delta } \) -dense in \( X \) . | Proof. Assume that \( Y \) is dense but not \( {G}_{\delta } \) -dense in \( X \) . We build a continuous real-valued function on \( Y \) that cannot be extended to \( X \) .\n\nLet \( O \) be a nonempty \( {G}_{\delta } \) -set contained in \( X \smallsetminus Y \) and let \( x \in O \) . By Claim 15.3.2, there exists... | Yes |
Lemma 15.3.4. In a compact group \( K \), for every nonempty open set \( U \) of \( K \), there exists \( N \in {\Lambda }_{ \trianglelefteq }\left( K\right) \) such that \( \bar{U} = \bar{U}N \) . | Proof. According to [251, Theorem 1.6], \( \bar{U} \) is a Baire set, hence [164, Theorem 64 G] applies.\n\nIn other words, if \( {e}_{K} \in U \), then \( \bar{U} \) not only contains a subgroup \( N \in {\Lambda }_{ \trianglelefteq }\left( K\right) \), but it is actually a union of cosets of \( N \) . | Yes |
Corollary 15.3.7. If \( G \) is a \( {G}_{\delta } \) -dense subgroup of a compact group \( K \), then \( K = {\beta G} \) and \( G \) is pseudocompact. | Proof. In view of Lemma 15.3.6, \( G \) is \( C \) -embedded in \( K \), and hence \( K = {\beta G} \) . Since \( G \) is \( {G}_{\delta } \) -dense in \( K = {\beta G}, G \) is pseudocompact by Corollary 15.1.4. | Yes |
Corollary 15.3.8. If \( G \) is a pseudocompact group, then \( {bG} = \widetilde{G} = {\beta G} \) . | Proof. Since \( G \) is pseudocompact, \( G \) is \( {G}_{\delta } \) -dense in \( \widetilde{G} \) by Theorem 15.2.1. By Corollary 15.3.7, \( \widetilde{G} = {\beta G} \) . That \( {bG} = \widetilde{G} \) follows from Theorem 10.2.15, since \( G \) is precompact by Theorem 15.2.1. | Yes |
Lemma 15.3.10. If \( X \) is a pseudocompact Tichonov space, then every singleton \( \{ x\} \) in \( X \) that is a \( {G}_{\delta } \) -set in \( X \) is also a \( {G}_{\delta } \) -set in \( {\beta X} \) . | Proof. According to Claim 15.3.2, if a singleton \( \{ x\} \) in a Tichonov space \( X \) is a \( {G}_{\delta } \) -set, then \( \{ x\} \) is functionally closed, i. e., there exists a continuous function \( f : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( {f}^{-1}\left( 0\right) = \{ x\} \) . Let us s... | Yes |
Corollary 15.3.11. A pseudocompact submetrizable group \( G \) is compact. | Proof. By Theorem 15.2.1, \( G \) is precompact. Let \( K \) be the compact completion of \( G \). If \( G \) admits a coarser metrizable group topology, then the singleton \( \left\{ {e}_{G}\right\} \) is a \( {G}_{\delta } \)-set of \( G \). Since \( K = {\beta G} \) by Corollary 15.3.7, Lemma 15.3.10 yields that \( ... | Yes |
Corollary 15.3.12. A metrizable compact group admits no strictly finer pseudocompact group topology. | Proof. If \( \left( {G,\tau }\right) \) is a metrizable compact group and \( {\tau }^{\prime } \geq \tau \) is a finer pseudocompact group topology on \( G \), then \( {\tau }^{\prime } \) is submetrizable, hence compact by Corollary 15.3.11. Since compact groups are minimal, this entails \( {\tau }^{\prime } = \tau \)... | Yes |
Theorem 16.1.3. Let \( A \) be a ring and \( \mathcal{V}\left( {0}_{A}\right) \) the filter of all neighborhoods of \( {0}_{A} \) in some ring topology \( \tau \) on A. Then (gt1),(gt2), and the following conditions hold:\n\n(rt1) for every \( U \in \mathcal{V}\left( {0}_{A}\right) \) and \( a \in A \), there is \( V \... | Proof. Since \( \left( {A,+,\tau }\right) \) is a topological group,(gt1) and (gt2) hold by Theorem 2.1.10. To prove (rt2), it suffices to apply the definition of continuity of the multiplication \( m \) at \( \left( {{0}_{A},{0}_{A}}\right) \in A \times A \) . Analogously, for (rt1) use the continuity of the multiplic... | Yes |
Theorem 16.1.12. If \( K \) is a locally retrobounded field, then:\n\n(a) ([216, Theorems 2 and 6]) every finite dimensional K-vector space admits a unique topological K-vector space topology, namely, the product topology;\n\n(b) ([287, Theorem 13.9]) the completion \( \widetilde{K} \) is a locally retrobounded field. | Since locally compact fields are locally retrobounded (see [287]), item (a) of this theorem applies for locally compact fields. Item (b) provides examples of topological fields that are not locally retrobounded (see Exercise 16.3.6). | No |
Take the compact ring \( k\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \), where \( k \) is a finite field, and its field of fractions \( K = k\left( \left( x\right) \right) \), consisting of formal Laurent power series of the form \( \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \) , \( {n}_{... | The topology in the above example is generated by an appropriate absolute value of \( K \) (e. g., letting \( \left| f\right| = {2}^{-{n}_{0}} \) for a nonzero \( f = \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{a}_{n}{x}^{n} \in K \) with \( {a}_{{n}_{0}} \neq 0 \), and \( \left| 0\right| = 0 \) ) and the same appl... | Yes |
Theorem 16.2.9 ([96, Theorem 5.17], [99, §3.4]). Let \( R \) be a discrete ring. The unique continuous functorial duality on \( {\mathcal{L}}_{R} \) is the Pontryagin-van Kampen duality if and only if \( \operatorname{Pic}\left( R\right) = \{ 0\} \) . | Prodanov proved in [235] (see also [99,§3.4]) that every functorial duality on \( \mathcal{L} = \) \( {\mathcal{L}}_{\mathbb{Z}} \) is continuous. In view of \( \operatorname{Pic}\left( \mathbb{Z}\right) = \{ 0\} \), this result, combined with Theorem 16.2.9, gives another proof of Roeder theorem 13.4.19. | No |
It turns out that \( {\mathcal{L}}_{R}^{0} \) has much more functorial dualities than the category \( {\mathcal{L}}_{R} \) . Actually, every pair \( \left( {T,\kappa }\right) \) where \( T \) is a torus \( T \) and \( \kappa : T \rightarrow T \) an involution give rise to a functorial duality \( {}^{\# } \) on \( {\mat... | see [96,10.2] \( - \) by means of \( \kappa \), one builds a concrete involutive equivalence \( \mu : {\mathcal{C}}_{R} \rightarrow {\mathcal{C}}_{R} \) which extends to \( {\mathcal{L}}_{R}^{0} \) and defines \( {X}^{\# } = {\Delta }_{T}\left( {\mu \left( X\right) }\right) \), as above | Yes |
A full subcategory of \( {\mathcal{L}}_{\mathbb{Q}} \) containing \( {\mathcal{L}}_{\mathbb{Q}}^{0} \) is finitely closed if it is closed with respect to isomorphisms, taking quotients (with respect to closed submodules), finite products, and closed submodules. | Using the fact that the Adele ring \( {\mathbf{A}}_{\mathbb{Q}} \) has no ring automorphisms beyond the identity (see [117]), it was proved in [94] that all dualities of the finitely closed subcategories of \( {\mathcal{L}}_{\mathbb{Q}} \) are continuous (actually, among all finitely closed subcategories \( \mathcal{L}... | No |
Proposition 16.2.18. Let \( M \) be a linearly topologized \( A \) -module.\n\n(a) If \( M \) is locally linearly compact, then \( M \) is complete.\n\n(b) If \( N \) is a locally linearly compact submodule of \( M \), then \( N \) is closed.\n\n(c) If \( N \) is a closed linear submodule of \( M \), then \( M \) is lo... | Proof. (a) is a consequence of Proposition 16.2.14(h) and implies (b), since a complete topological subgroup of a Hausdorff group is closed by Proposition 7.1.22.\n\n(c) If \( M \) is locally linearly compact, then \( N \) and \( M/N \) are locally linearly compact as well. Conversely, assume that \( N \) and \( M/N \)... | Yes |
Theorem 16.2.19. Every \( V \in {\mathrm{{LLC}}}_{K} \) is topologically isomorphic to \( {V}_{c} \times {V}_{d} \), where \( {V}_{c} \) and \( {V}_{d} \) are linear subspaces of \( V \), with \( {V}_{c} \) linearly compact open and \( {V}_{d} \) discrete. | Proof. Let \( {V}_{c} \) be an open linear subspace of \( V \) that is linearly compact. There exists a linear subspace \( {V}_{d} \) of \( V \) such that \( V = {V}_{c} \oplus {V}_{d} \), where \( {V}_{d} \) is discrete, being topologically isomorphic to \( V/{V}_{c} \) . It is straightforward to prove that the isomor... | Yes |
Theorem 16.2.21 (Lefschetz duality theorem [195, Theorem 29.1]). The biduality functor \( {}^{\land \land } : {\mathrm{{LLC}}}_{K} \rightarrow {\mathrm{{LLC}}}_{K} \) and \( {1}_{{\mathrm{{LLC}}}_{K}} : {\mathrm{{LLC}}}_{K} \rightarrow {\mathrm{{LLC}}}_{K} \) are naturally isomorphic. This induces a duality between the... | As in Pontryagin-van Kampen duality theorem 13.4.17, one proves that, for every locally linearly compact vector space \( V \), the evaluation map \( {\omega }_{V} : V \rightarrow {V}^{\land \land }, v \mapsto {\omega }_{V}\left( v\right) \) , with \( {\omega }_{V}\left( v\right) \left( \chi \right) = \chi \left( v\righ... | Yes |
Corollary 16.2.22. Let \( V \) be a linearly compact vector space over a discrete field \( K \) . Then \( V \) is compact if and only if \( K \) is finite. In particular, \( V \) is a hereditarily disconnected locally compact abelian group whenever \( K \) is finite. | Proof. By Exercise 16.3.16, \( V = \mathop{\prod }\limits_{{i \in I}}{K}_{i} \) with \( {K}_{i} = K \) for all \( i \in I \) . If \( V \) is compact, then each \( {K}_{i} \) is compact as well, hence \( K \) is finite, being compact and discrete. Conversely, if \( K \) is finite, then each \( {K}_{i} \) is compact, so ... | No |
The Pontryagin-van Kampen duality functor, when restricted to \( p \) -torsion discrete abelian groups, gives a duality between \( p \) -torsion discrete abelian groups and abelian pro- \( p \) -groups. | Both categories consist of \( {\mathrm{J}}_{p} \) -modules; moreover, \( \widehat{G} = \operatorname{Hom}\left( {G,\mathbb{T}}\right) = {\operatorname{Hom}}_{{\mathbb{J}}_{p}}\left( {G,\mathbb{Z}\left( {p}^{\infty }\right) }\right) \) for a discrete \( p \) -torsion abelian group \( G \), and \( \widehat{K} = \operator... | Yes |
Lemma 3.1. Two equivalence classes \( E \) and \( {E}^{\prime } \) are either disjoint or equal. | Proof. Let \( E \) be the equivalence class determined by \( x \), and let \( {E}^{\prime } \) be the equivalence class determined by \( {x}^{\prime } \) . Suppose that \( E \cap {E}^{\prime } \) is not empty; let \( \mathrm{y} \) be a point of \( E \cap {E}^{\prime } \) . See Figure 3.1. We show that \( E = {E}^{\prim... | Yes |
Theorem 4.1 (Well-ordering property). Every nonempty subset of \( {\mathbb{Z}}_{ + } \) has a smallest element. | Proof. We first prove that, for each \( n \in {\mathbb{Z}}_{ + } \), the following statement holds: Every nonempty subset of \( \{ 1,\ldots, n\} \) has a smallest element.\n\nLet \( A \) be the set of all positive integers \( \mathrm{n} \) for which this statement holds. Then \( A \) contains 1, since if \( n = 1 \), t... | Yes |
Theorem 4.2 (Strong induction principle). Let \( A \) be a set of positive integers. Suppose that for each positive integer \( n \), the statement \( {S}_{n} \subset A \) implies the statement \( n \in A \) . Then \( A = {\mathbb{Z}}_{ + } \) . | Proof. If \( A \) does not equal all of \( {\mathbb{Z}}_{ + } \), let \( n \) be the smallest positive integer that is not in \( A \) . Then every positive integer less than \( n \) is in \( A \), so that \( {S}_{n} \subset A \) . Our hypothesis implies that \( n \in A \), contrary to assumption. | Yes |
Lemma 6.1. Let \( n \) be a positive integer. Let \( A \) be a set; let \( {a}_{0} \) be an element of \( A \) . Then there exists a bijective correspondence \( f \) of the set \( A \) with the set \( \{ 1,\ldots, n + 1\} \) if and only if there exists a bijective correspondence \( g \) of the set \( A - \left\{ {a}_{0... | Proof. There are two implications to be proved. Let us first assume that there is a bijective correspondence\n\n\[ g : A - \left\{ {a}_{0}\right\} \rightarrow \{ 1,\ldots, n\} .\n\]\n\nWe then define a function \( f : A \rightarrow \{ 1,\ldots, n + 1\} \) by setting\n\n\[ f\left( x\right) = g\left( x\right) \;\text{ fo... | Yes |
Theorem 6.2. Let \( A \) be a set; suppose that there exists a bijection \( f : A \rightarrow \{ 1,\ldots, n\} \) for some \( n \in {\mathbb{Z}}_{ + } \) . Let \( B \) be a proper subset of \( A \) . Then there exists no bijection \( g : B \rightarrow \{ 1,\ldots, n\} \) ; but (provided \( B \neq \varnothing \) ) there... | Proof. The case in which \( B = \varnothing \) is trivial, for there cannot exist a bijection of the empty set \( B \) with the nonempty set \( \{ 1,\ldots, n\} \) .\n\nWe prove the theorem \ | No |
Corollary 6.3. If \( A \) is finite, there is no bijection of \( A \) with a proper subset of itself. | Proof. Assume that \( B \) is a proper subset of \( A \) and that \( f : A \rightarrow B \) is a bijection. By assumption, there is a bijection \( g : A \rightarrow \{ 1,\ldots, n\} \) for some \( n \) . The composite \( g \circ {f}^{-1} \) is then a bijection of \( B \) with \( \{ 1,\ldots, n\} \) . This contradicts t... | Yes |
Corollary 6.4. \( {\mathbb{Z}}_{ + } \) is not finite. | Proof. The function \( f : {\mathbb{Z}}_{ + } \rightarrow {\mathbb{Z}}_{ + } - \{ 1\} \) defined by \( f\left( n\right) = n + 1 \) is a bijection of \( {\mathbb{Z}}_{ + } \) with a proper subset of itself. | Yes |
Corollary 6.5. The cardinality of a finite set \( A \) is uniquely determined by \( A \) . | Proof. Let \( m < n \) . Suppose there are bijections\n\n\[ f : A \rightarrow \{ 1,\ldots, n\} \]\n\n\[ g : A \rightarrow \{ 1,\ldots, m\} . \]\n\nThen the composite\n\n\[ g \circ {f}^{-1} : \{ 1,\ldots, n\} \rightarrow \{ 1,\ldots, m\} \]\n\nis a bijection of the finite set \( \{ 1,\ldots, n\} \) with a proper subset ... | Yes |
Corollary 6.7. Let \( B \) be a nonempty set. Then the following are equivalent:\n\n(1) \( B \) is finite.\n\n(2) There is a surjective function from a section of the positive integers onto \( B \) .\n\n(3) There is an injective function from \( B \) into a section of the positive integers. | Proof. (1) \( \Rightarrow \) (2). Since \( B \) is nonempty, there is, for some \( n \), a bijective function \( f : \{ 1,\ldots, n\} \rightarrow B \) .\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) . If \( f : \{ 1,\ldots, n\} \rightarrow B \) is surjective, define \( g : B \rightarrow \{ 1,\ldots, n\} \) by th... | Yes |
Theorem 7.1. Let \( B \) be a nonempty set. Then the following are equivalent:\n\n(1) \( B \) is countable.\n\n(2) There is a surjective function \( f : {\mathbb{Z}}_{ + } \rightarrow B \) .\n\n(3) There is an injective function \( g : B \rightarrow {\mathbb{Z}}_{ + } \) . | Proof. (1) \( \Rightarrow \) (2). Suppose that \( B \) is countable. If \( B \) is countably infinite, there is a bijection \( f : {\mathbb{Z}}_{ + } \rightarrow B \) by definition, and we are through. If \( B \) is finite, there is a\nbijection \( h : \{ 1,\ldots, n\} \rightarrow B \) for some \( n \geq 1 \) . (Recall... | Yes |
Lemma 7.2. If \( C \) is an infinite subset of \( {\mathbb{Z}}_{ + } \), then \( C \) is countably infinite. | Proof. We define a bijection \( h : {\mathbb{Z}}_{ + } \rightarrow C \) . We proceed by induction. Define \( h\left( 1\right) \) to be the smallest element of \( C \) ; it exists because every nonempty subset \( C \) of \( {\mathbb{Z}}_{ + } \) has a smallest element. Then assuming that \( h\left( 1\right) ,\ldots, h\l... | Yes |
Corollary 7.3. A subset of a countable set is countable. | Proof. Suppose \( A \subset B \), where \( B \) is countable. There is an injection \( f \) of \( B \) into \( {\mathbb{Z}}_{ + } \) ; the restriction of \( f \) to \( A \) is an injection of \( A \) into \( {\mathbb{Z}}_{ + } \). | Yes |
Corollary 7.4. The set \( {\mathbb{Z}}_{ + } \times {\mathbb{Z}}_{ + } \) is countably infinite. | Proof. In view of Theorem 7.1, it suffices to construct an injective map \( f : {\mathbb{Z}}_{ + } \times \) \( {\mathbb{Z}}_{ + } \rightarrow {\mathbb{Z}}_{ + } \) . We define \( f \) by the equation\n\n\[ f\left( {n, m}\right) = {2}^{n}{3}^{m}\text{.} \]\n\nIt is easy to check that \( f \) is injective. For suppose t... | Yes |
Theorem 7.5. A countable union of countable sets is countable. | Proof. Let \( {\left\{ {A}_{n}\right\} }_{n \in J} \) be an indexed family of countable sets, where the index set \( J \) is either \( \{ 1,\ldots, N\} \) or \( {\mathbb{Z}}_{ + } \) . Assume that each set \( {A}_{n} \) is nonempty, for convenience; this assumption does not change anything.\n\nBecause each \( {A}_{n} \... | Yes |
Theorem 7.7. Let \( X \) denote the two element set \( \{ 0,1\} \) . Then the set \( {X}^{\omega } \) is uncountable. | Proof. We show that, given any function\n\n\[ g : {\mathbb{Z}}_{ + } \rightarrow {X}^{\omega } \]\n\n\( g \) is not surjective. For this purpose, let us denote \( g\left( n\right) \) as follows :\n\n\[ g\left( n\right) = \left( {{x}_{n1},{x}_{n2},{x}_{n3},\ldots {x}_{nm},\ldots }\right) ,\]\n\nwhere each \( {x}_{ij} \)... | Yes |
Theorem 7.8. Let \( A \) be a set. There is no injective map \( f : \mathcal{P}\left( A\right) \rightarrow A \), and there is no surjective map \( g : A \rightarrow \mathcal{P}\left( A\right) \) . | Proof. In general, if \( B \) is a nonempty set, the existence of an injective map \( f : B \rightarrow \) \( C \) implies the existence of a surjective map \( g : C \rightarrow B \) ; one defines \( g\left( c\right) = {f}^{-1}\left( c\right) \) for each \( c \) in the image set of \( f \), and defines \( g \) arbitrar... | Yes |
Lemma 8.1. Given \( n \in {\mathbb{Z}}_{ + } \), there exists a function\n\n\[ f : \{ 1,\ldots, n\} \rightarrow C \]\n\nthat satisfies \( \left( *\right) \) for all \( i \) in its domain. | Proof. The point of this lemma is that it is a statement that depends on \( n \) ; therefore, it is capable of being proved by induction. Let \( A \) be the set of all \( n \) for which the lemma holds. We show that \( A \) is inductive. It then follows that \( A = {\mathbb{Z}}_{ + } \).\n\nThe lemma is true for \( n =... | Yes |
Lemma 8.2. Suppose that \( f : \{ 1,\ldots, n\} \rightarrow C \) and \( g : \{ 1,\ldots, m\} \rightarrow C \) both satisfy \( \left( *\right) \) for all \( i \) in their respective domains. Then \( f\left( i\right) = g\left( i\right) \) for all \( i \) in both domains. | Proof. Suppose not. Let \( i \) be the smallest integer for which \( f\left( i\right) \neq g\left( i\right) \) . The integer \( i \) is not 1, because\n\n\[ f\left( 1\right) = \text{ smallest element of }C = g\left( 1\right) ,\]\n\nby \( \left( *\right) \) . Now for all \( j < i \), we have \( f\left( j\right) = g\left... | Yes |
There exists a unique function \( h : {\mathbb{Z}}_{ + } \rightarrow C \) satisfying \( \left( *\right) \) for all \( i \in {\mathbb{Z}}_{ + } \) . | Proof. By Lemma 8.1, there exists for each \( n \) a function that maps \( \{ 1,\ldots, n\} \) into \( C \) and satisfies \( \left( *\right) \) for all \( i \) in its domain. Given \( n \), Lemma 8.2 shows that this function is unique; two such functions having the same domain must be equal. Let \( {f}_{n} \) : \( \{ 1... | Yes |
Theorem 8.4 (Principle of recursive definition). Let \( A \) be a set; let \( {a}_{0} \) be an element of \( A \) . Suppose \( \rho \) is a function that assigns, to each function \( f \) mapping a nonempty section of the positive integers into \( A \), an element of \( A \) . Then there exists a unique function\n\n\[ ... | EXAMPLE 1. Let us show that Theorem 8.3 is a special case of this theorem. Given the infinite subset \( C \) of \( {\mathbb{Z}}_{ + } \), let \( {a}_{0} \) be the smallest element of \( C \), and define \( \rho \) by the equation\n\n\[ \rho \left( f\right) = \text{ smallest element of }\left\lbrack {C - \left( {\text{ ... | No |
Theorem 9.1. Let \( A \) be a set. The following statements about \( A \) are equivalent:\n\n(1) There exists an injective function \( f : {\mathbb{Z}}_{ + } \rightarrow A \) .\n\n(2) There exists a bijection of \( A \) with a proper subset of itself.\n\n(3) \( A \) is infinite. | Proof. We prove the implications \( \left( 1\right) \Rightarrow \left( 2\right) \Rightarrow \left( 3\right) \Rightarrow \left( 1\right) \) . To prove that \( \left( 1\right) \Rightarrow \left( 2\right) \) , we assume there is an injective function \( f : {\mathbb{Z}}_{ + } \rightarrow A \) . Let the image set \( f\left... | Yes |
Lemma 9.2 (Existence of a choice function). Given a collection \( \mathcal{B} \) of nonempty sets (not necessarily disjoint), there exists a function\n\n\[ c : \mathcal{B} \rightarrow \mathop{\bigcup }\limits_{{B \in \mathcal{B}}}B \]\n\n such that \( c\left( B\right) \) is an element of \( B \), for each \( B \in \mat... | Proof of the lemma. Given an element \( B \) of \( \mathcal{B} \), we define a set \( {B}^{\prime } \) as follows:\n\n\[ {B}^{\prime } = \{ \left( {B, x}\right) \mid x \in B\} . \]\n\nThat is, \( {B}^{\prime } \) is the collection of all ordered pairs, where the first coordinate of the ordered pair is the set \( B \), ... | Yes |
Theorem 10.1. Every nonempty finite ordered set has the order type of a section \( \{ 1,\ldots, n\} \) of \( {\mathbb{Z}}_{ + } \), so it is well-ordered. | Proof. This was given as an exercise in \( §6 \) ; we prove it here. First, we show that every finite ordered set \( A \) has a largest element. If \( A \) has one element, this is trivial. Supposing it true for sets having \( n - 1 \) elements, let \( A \) have \( n \) elements and let \( {a}_{0} \in A \) . Then \( A ... | Yes |
Lemma 10.2. There exists a well-ordered set \( A \) having a largest element \( \Omega \), such that the section \( {S}_{\Omega } \) of \( A \) by \( \Omega \) is uncountable but every other section of \( A \) is countable. | Proof. We begin with an uncountable well-ordered set \( B \) . Let \( C \) be the well-ordered set \( \{ 1,2\} \times B \) in the dictionary order; then some section of \( C \) is uncountable. (Indeed, the section of \( C \) by any element of the form \( 2 \times b \) is uncountable.) Let \( \Omega \) be the smallest e... | Yes |
Theorem 10.3. If \( A \) is a countable subset of \( {S}_{\Omega } \), then \( A \) has an upper bound in \( {S}_{\Omega } \) . | Proof. Let \( A \) be a countable subset of \( {S}_{\Omega } \) . For each \( a \in A \), the section \( {S}_{a} \) is countable. Therefore, the union \( B = \mathop{\bigcup }\limits_{{a \in A}}{S}_{a} \) is also countable. Since \( {S}_{\Omega } \) is uncountable, the set \( B \) is not all of \( {S}_{\Omega } \) ; le... | Yes |
Lemma 13.1. Let \( X \) be a set; let \( \mathcal{B} \) be a basis for a topology \( \mathcal{T} \) on \( X \) . Then \( \mathcal{T} \) equals the collection of all unions of elements of \( \mathcal{B} \) . | Proof. Given a collection of elements of \( \mathcal{B} \), they are also elements of \( \mathcal{T} \) . Because \( \mathcal{T} \) is a topology, their union is in \( \mathcal{T} \) . Conversely, given \( U \in \mathcal{T} \), choose for each \( x \in U \) an element \( {B}_{x} \) of \( \mathcal{B} \) such that \( x \... | Yes |
Lemma 13.2. Let \( X \) be a topological space. Suppose that \( \mathcal{C} \) is a collection of open sets of \( X \) such that for each open set \( U \) of \( X \) and each \( x \) in \( U \), there is an element \( C \) of \( \mathcal{C} \) such that \( x \in C \subset U \). Then \( \mathcal{C} \) is a basis for the... | Proof. We must show that \( \mathcal{C} \) is a basis. The first condition for a basis is easy: Given \( x \in X \), since \( X \) is itself an open set, there is by hypothesis an element \( C \) of \( \mathcal{C} \) such that \( x \in C \subset X \). To check the second condition, let \( x \) belong to \( {C}_{1} \cap... | Yes |
Lemma 13.3. Let \( \mathcal{B} \) and \( {\mathcal{B}}^{\prime } \) be bases for the topologies \( \mathcal{T} \) and \( {\mathcal{T}}^{\prime } \), respectively, on \( X \) . Then the following are equivalent:\n\n(1) \( {\mathcal{T}}^{\prime } \) is finer than \( \mathcal{T} \) .\n\n(2) For each \( x \in X \) and each... | Proof. \( \;\left( 2\right) \Rightarrow \left( 1\right) \) . Given an element \( U \) of \( \mathcal{T} \), we wish to show that \( U \in {\mathcal{T}}^{\prime } \) . Let \( x \in U \) . Since \( \mathcal{B} \) generates \( \mathcal{T} \), there is an element \( B \in \mathcal{B} \) such that \( x \in B \subset U \) . ... | Yes |
Lemma 13.4. The topologies of \( {\mathbb{R}}_{\ell } \) and \( {\mathbb{R}}_{K} \) are strictly finer than the standard topology on \( \mathbb{R} \), but are not comparable with one another. | Proof. Let \( \mathcal{T},{\mathcal{T}}^{\prime } \), and \( {\mathcal{T}}^{\prime \prime } \) be the topologies of \( \mathbb{R},{\mathbb{R}}_{\ell } \), and \( {\mathbb{R}}_{K} \), respectively. Given a basis element \( \left( {a, b}\right) \) for \( \mathcal{T} \) and a point \( x \) of \( \left( {a, b}\right) \), t... | Yes |
Theorem 15.1. If \( \mathcal{B} \) is a basis for the topology of \( X \) and \( \mathcal{C} \) is a basis for the topology of \( Y \), then the collection\n\n\[ \mathcal{D} = \{ B \times C \mid B \in \mathcal{B}\text{ and }C \in \mathcal{C}\}\} \n\nis a basis for the topology of \( X \times Y \) . | Proof. We apply Lemma 13.2. Given an open set \( W \) of \( X \times Y \) and a point \( x \times y \) of \( W \), by definition of the product topology there is a basis element \( U \times V \) such that \( x \times y \in U \times V \subset W \) . Because \( \mathcal{B} \) and \( \mathcal{C} \) are bases for \( X \) a... | Yes |
Theorem 15.2. The collection\n\n\[ S = \left\{ {{\pi }_{1}^{-1}\left( U\right) \mid U\text{ open in }X}\right\} \cup \left\{ {{\pi }_{2}^{-1}\left( V\right) \mid V\text{ open in }Y}\right\} \]\n\nis a subbasis for the product topology on \( X \times Y \) . | Proof. Let \( \mathcal{T} \) denote the product topology on \( X \times Y \) ; let \( {\mathcal{T}}^{\prime } \) be the topology generated by \( \mathcal{S} \) . Because every element of \( \mathcal{S} \) belongs to \( \mathcal{T} \), so do arbitrary unions of finite intersections of elements of \( \mathcal{S} \) . Thu... | Yes |
Lemma 16.1. If \( \mathcal{B} \) is a basis for the topology of \( X \) then the collection\n\n\[{\mathcal{B}}_{Y} = \{ B \cap Y \mid B \in \mathcal{B}\}\]\n\nis a basis for the subspace topology on \( Y \) . | Proof. Given \( U \) open in \( X \) and given \( y \in U \cap Y \), we can choose an element \( B \) of \( \mathcal{B} \) such that \( y \in B \subset U \) . Then \( y \in B \cap Y \subset U \cap Y \) . It follows from Lemma 13.2 that \( {\mathcal{B}}_{Y} \) is a basis for the subspace topology on \( Y \) . | No |
Lemma 16.2. Let \( Y \) be a subspace of \( X \) . If \( U \) is open in \( Y \) and \( Y \) is open in \( X \), then \( U \) is open in \( X \) . | Proof. Since \( U \) is open in \( Y, U = Y \cap V \) for some set \( V \) open in \( X \) . Since \( Y \) and \( V \) are both open in \( X \), so is \( Y \cap V \) . | Yes |
Theorem 16.3. If \( A \) is a subspace of \( X \) and \( B \) is a subspace of \( Y \), then the product topology on \( A \times B \) is the same as the topology \( A \times B \) inherits as a subspace of \( X \times Y \) . | Proof. The set \( U \times V \) is the general basis element for \( X \times Y \), where \( U \) is open in \( X \) and \( V \) is open in \( Y \) . Therefore, \( \left( {U \times V}\right) \cap \left( {A \times B}\right) \) is the general basis element for the subspace topology on \( A \times B \) . Now\n\n\[ \left( {... | Yes |
Theorem 16.4. Let \( X \) be an ordered set in the order topology; let \( Y \) be a subset of \( X \) that is convex in \( X \) . Then the order topology on \( Y \) is the same as the topology \( Y \) inherits as a subspace of \( X \) . | Proof. Consider the ray \( \left( {a, + \infty }\right) \) in \( X \) . What is its intersection with \( Y \) ? If \( a \in Y \) , then\n\n\[ \left( {a, + \infty }\right) \cap Y = \{ x \mid x \in Y\text{ and }x > a\} \]\n\nthis is an open ray of the ordered set \( Y \) . If \( a \notin Y \), then \( a \) is either a lo... | Yes |
Theorem 17.1. Let \( X \) be a topological space. Then the following conditions hold:\n\n(1) \( \varnothing \) and \( X \) are closed.\n\n(2) Arbitrary intersections of closed sets are closed.\n\n(3) Finite unions of closed sets are closed. | Proof. (1) \( \varnothing \) and \( X \) are closed because they are the complements of the open sets \( X \) and \( \varnothing \), respectively.\n\n(2) Given a collection of closed sets \( {\left\{ {A}_{\alpha }\right\} }_{\alpha \in J} \), we apply DeMorgan’s law,\n\n\[ X - \mathop{\bigcap }\limits_{{\alpha \in J}}{... | Yes |
Theorem 17.2. Let \( Y \) be a subspace of \( X \) . Then a set \( A \) is closed in \( Y \) if and only if it equals the intersection of a closed set of \( X \) with \( Y \) . | Proof. Assume that \( A = C \cap Y \), where \( C \) is closed in \( X \) . (See Figure 17.1.) Then \( X - C \) is open in \( X \), so that \( \left( {X - C}\right) \cap Y \) is open in \( Y \), by definition of the subspace topology. But \( \left( {X - C}\right) \cap Y = Y - A \) . Hence \( Y - A \) is open in \( Y \)... | Yes |
Theorem 17.4. Let \( Y \) be a subspace of \( X \) ; let \( A \) be a subset of \( Y \) ; let \( \bar{A} \) denote the closure of \( A \) in \( X \) . Then the closure of \( A \) in \( Y \) equals \( \bar{A} \cap Y \) . | Proof. Let \( B \) denote the closure of \( A \) in \( Y \) . The set \( \bar{A} \) is closed in \( X \), so \( \bar{A} \cap Y \) is closed in \( Y \) by Theorem 17.2. Since \( \bar{A} \cap Y \) contains \( A \), and since by definition \( B \) equals the intersection of all closed subsets of \( Y \) containing \( A \)... | Yes |
Theorem 17.5. Let \( A \) be a subset of the topological space \( X \) . (a) Then \( x \in \bar{A} \) if and only if every open set \( U \) containing \( x \) intersects \( A \) . | Proof. Consider the statement in (a). It is a statement of the form \( P \Leftrightarrow Q \) . Let us transform each implication to its contrapositive, thereby obtaining the logically equivalent statement (not \( P \) ) \( \Leftrightarrow \) (not \( Q \) ). Written out, it is the following: \( x \notin \bar{A} \Leftri... | Yes |
Theorem 17.6. Let \( A \) be a subset of the topological space \( X \) ; let \( {A}^{\prime } \) be the set of all limit points of \( A \) . Then\n\n\[ \bar{A} = A \cup {A}^{\prime } \] | Proof. If \( x \) is in \( {A}^{\prime } \), every neighborhood of \( x \) intersects \( A \) (in a point different from \( x \) ). Therefore, by Theorem 17.5, \( x \) belongs to \( \bar{A} \) . Hence \( {A}^{\prime } \subset \bar{A} \) . Since by definition \( A \subset \bar{A} \), it follows that \( A \cup {A}^{\prim... | Yes |
Corollary 17.7. A subset of a topological space is closed if and only if it contains all its limit points. | Proof. The set \( A \) is closed if and only if \( A = \bar{A} \), and the latter holds if and only if \( {A}^{\prime } \subset A \) . | Yes |
Theorem 17.9. Let \( X \) be a space satisfying the \( {T}_{1} \) axiom; let \( A \) be a subset of \( X \) . Then the point \( x \) is a limit point of \( A \) if and only if every neighborhood of \( x \) contains infinitely many points of \( A \) . | Proof. If every neighborhood of \( x \) intersects \( A \) in infinitely many points, it certainly intersects \( A \) in some point other than \( x \) itself, so that \( x \) is a limit point of \( A \) .\n\nConversely, suppose that \( x \) is a limit point of \( A \), and suppose some neighborhood \( U \) of \( x \) i... | Yes |
Theorem 17.10. If \( X \) is a Hausdorff space, then a sequence of points of \( X \) converges to at most one point of \( X \) . | Proof. Suppose that \( {x}_{n} \) is a sequence of points of \( X \) that converges to \( x \) . If \( y \neq x \) , let \( U \) and \( V \) be disjoint neighborhoods of \( x \) and \( y \), respectively. Since \( U \) contains \( {x}_{n} \) for all but finitely many values of \( n \), the set \( V \) cannot. Therefore... | Yes |
Theorem 18.1. Let \( X \) and \( Y \) be topological spaces; let \( f : X \rightarrow Y \) . Then the following are equivalent:\n\n(1) \( f \) is continuous.\n\n(2) For every subset \( A \) of \( X \), one has \( f\left( \bar{A}\right) \subset \overline{f\left( A\right) } \).\n\n(3) For every closed set \( B \) of \( Y... | Proof. We show that \( \left( 1\right) \Rightarrow \left( 2\right) \Rightarrow \left( 3\right) \Rightarrow \left( 1\right) \) and that \( \left( 1\right) \Rightarrow \left( 4\right) \Rightarrow \left( 1\right) \) .\n\n\( \left( 1\right) \Rightarrow \left( 2\right) \) . Assume that \( f \) is continuous. Let \( A \) be ... | Yes |
Theorem 18.2 (Rules for constructing continuous functions). Let \( X, Y \), and \( Z \) be topological spaces.\n\n(a) (Constant function) If \( f : X \rightarrow Y \) maps all of \( X \) into the single point \( {y}_{0} \) of \( Y \) , then \( f \) is continuous.\n\n(b) (Inclusion) If \( A \) is a subspace of \( X \), ... | Proof. (a) Let \( f\left( x\right) = {y}_{0} \) for every \( x \) in \( X \) . Let \( V \) be open in \( Y \) . The set \( {f}^{-1}\left( V\right) \) equals \( X \) or \( \varnothing \), depending on whether \( V \) contains \( {y}_{0} \) or not. In either case, it is open.\n\n(b) If \( U \) is open in \( X \), then \(... | Yes |
Theorem 18.3 (The pasting lemma). Let \( X = A \cup B \), where \( A \) and \( B \) are closed in \( X \). Let \( f : A \rightarrow Y \) and \( g : B \rightarrow Y \) be continuous. If \( f\left( x\right) = g\left( x\right) \) for every \( x \in A \cap B \), then \( f \) and \( g \) combine to give a continuous functio... | Proof. Let \( C \) be a closed subset of \( Y \). Now\n\n\[ \n{h}^{-1}\left( C\right) = {f}^{-1}\left( C\right) \cup {g}^{-1}\left( C\right) ,\n\]\n\nby elementary set theory. Since \( f \) is continuous, \( {f}^{-1}\left( C\right) \) is closed in \( A \) and, therefore, closed in \( X \). Similarly, \( {g}^{-1}\left( ... | Yes |
Theorem 18.4 (Maps into products). Let \( f : A \rightarrow X \times Y \) be given by the equation\n\n\[ f\left( a\right) = \left( {{f}_{1}\left( a\right) ,{f}_{2}\left( a\right) }\right) .\n\]\n\nThen \( f \) is continuous if and only if the functions\n\n\[ {f}_{1} : A \rightarrow X\;\text{ and }\;{f}_{2} : A \rightar... | Proof. Let \( {\pi }_{1} : X \times Y \rightarrow X \) and \( {\pi }_{2} : X \times Y \rightarrow Y \) be projections onto the first and second factors, respectively. These maps are continuous. For \( {\pi }_{1}^{-1}\left( U\right) = U \times Y \) and \( {\pi }_{2}^{-1}\left( V\right) = X \times V \), and these sets ar... | Yes |
Theorem 19.5. Let \( \\left\\{ {X}_{\\alpha }\\right\\} \) be an indexed family of spaces; let \( {A}_{\\alpha } \\subset {X}_{\\alpha } \) for each \( \\alpha \) . If \( \\prod {X}_{\\alpha } \) is given either the product or the box topology, then\n\n\[ \n\\prod {\\bar{A}}_{\\alpha } = \\overline{\\prod {A}_{\\alpha ... | Proof. Let \( \\mathbf{x} = \\left( {x}_{\\alpha }\\right) \) be a point of \( \\prod {\\bar{A}}_{\\alpha } \) ; we show that \( \\mathbf{x} \\in \\overline{\\prod {A}_{\\alpha }} \) . Let \( U = \\prod {U}_{\\alpha } \) be a basis element for either the box or product topology that contains \( \\mathbf{x} \) . Since \... | Yes |
Theorem 19.6. Let \( f : A \rightarrow \mathop{\prod }\limits_{{\alpha \in J}}{X}_{\alpha } \) be given by the equation\n\n\[ f\left( a\right) = {\left( {f}_{\alpha }\left( a\right) \right) }_{\alpha \in J}, \]\n\nwhere \( {f}_{\alpha } : A \rightarrow {X}_{\alpha } \) for each \( \alpha \) . Let \( \prod {X}_{\alpha }... | Proof. Let \( {\pi }_{\beta } \) be the projection of the product onto its \( \beta \) th factor. The function \( {\pi }_{\beta } \) is continuous, for if \( {U}_{\beta } \) is open in \( {X}_{\beta } \), the set \( {\pi }_{\beta }^{-1}\left( {U}_{\beta }\right) \) is a subbasis element for the product topology on \( {... | Yes |
Theorem 20.1. Let \( X \) be a metric space with metric \( d \) . Define \( \bar{d} : X \times X \rightarrow \mathbb{R} \) by the equation\n\n\[ \bar{d}\left( {x, y}\right) = \min \{ d\left( {x, y}\right) ,1\} . \]\n\nThen \( \bar{d} \) is a metric that induces the same topology as \( d \) . | Proof. Checking the first two conditions for a metric is trivial. Let us check the triangle inequality:\n\n\[ \bar{d}\left( {x, z}\right) \leq \bar{d}\left( {x, y}\right) + \bar{d}\left( {y, z}\right) . \]\n\nNow if either \( d\left( {x, y}\right) \geq 1 \) or \( d\left( {y, z}\right) \geq 1 \), then the right side of ... | Yes |
Lemma 20.2. Let \( d \) and \( {d}^{\prime } \) be two metrics on the set \( X \) ; let \( \mathcal{T} \) and \( {\mathcal{T}}^{\prime } \) be the topologies they induce, respectively. Then \( {\mathcal{T}}^{\prime } \) is finer than \( \mathcal{T} \) if and only if for each \( x \) in \( X \) and each \( \epsilon > 0 ... | Proof. Suppose that \( {\mathcal{T}}^{\prime } \) is finer than \( \mathcal{T} \) . Given the basis element \( {B}_{d}\left( {x,\epsilon }\right) \) for \( \mathcal{T} \), there is by Lemma 13.3 a basis element \( {B}^{\prime } \) for the topology \( {\mathcal{T}}^{\prime } \) such that \( x \in {B}^{\prime } \subset {... | Yes |
Theorem 20.3. The topologies on \( {\mathbb{R}}^{n} \) induced by the euclidean metric \( d \) and the square metric \( \rho \) are the same as the product topology on \( {\mathbb{R}}^{n} \) . | Proof. Let \( \mathbf{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) and \( \mathbf{y} = \left( {{y}_{1},\ldots ,{y}_{n}}\right) \) be two points of \( {\mathbb{R}}^{n} \) . It is simple algebra to check that\n\n\[ \rho \left( {\mathbf{x},\mathbf{y}}\right) \leq d\left( {\mathbf{x},\mathbf{y}}\right) \leq \sqrt{n}\rho... | Yes |
Theorem 20.4. The uniform topology on \( {\mathbb{R}}^{J} \) is finer than the product topology and coarser than the box topology; these three topologies are all different if \( J \) is infinite. | Proof. Suppose that we are given a point \( \mathbf{x} = {\left( {x}_{\alpha }\right) }_{\alpha \in J} \) and a product topology basis element \( \prod {U}_{\alpha } \) about \( \mathbf{x} \) . Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be the indices for which \( {U}_{\alpha } \neq \mathbb{R} \) . Then for each \( ... | Yes |
Theorem 20.5. Let \( \bar{d}\left( {a, b}\right) = \min \{ \left| {a - b}\right| ,1\} \) be the standard bounded metric on \( \mathbb{R} \) . If \( \mathbf{x} \) and \( \mathbf{y} \) are two points of \( {\mathbb{R}}^{\omega } \), define\n\n\[ D\left( {\mathbf{x},\mathbf{y}}\right) = \sup \left\{ \frac{\bar{d}\left( {{... | Proof. The properties of a metric are satisfied trivially except for the triangle inequality, which is proved by noting that for all \( i \) ,\n\n\[ \frac{\bar{d}\left( {{x}_{i},{z}_{i}}\right) }{i} \leq \frac{\bar{d}\left( {{x}_{i},{y}_{i}}\right) }{i} + \frac{\bar{d}\left( {{y}_{i},{z}_{i}}\right) }{i} \leq D\left( {... | Yes |
Theorem 21.1. Let \( f : X \rightarrow Y \) ; let \( X \) and \( Y \) be metrizable with metrics \( {d}_{X} \) and \( {d}_{Y} \) , respectively. Then continuity of \( f \) is equivalent to the requirement that given \( x \in X \) and given \( \epsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \n{d}_{X}\left... | Proof. Suppose that \( f \) is continuous. Given \( x \) and \( \epsilon \), consider the set\n\n\[ \n{f}^{-1}\left( {B\left( {f\left( x\right) ,\epsilon }\right) }\right)\n\]\nwhich is open in \( X \) and contains the point \( x \) . It contains some \( \delta \) -ball \( B\left( {x,\delta }\right) \) centered at \( x... | Yes |
Lemma 21.2 (The sequence lemma). Let \( X \) be a topological space; let \( A \subset X \). If there is a sequence of points of \( A \) converging to \( x \), then \( x \in \bar{A} \); the converse holds if \( X \) is metrizable. | Proof. Suppose that \( {x}_{n} \rightarrow x \), where \( {x}_{n} \in A \). Then every neighborhood \( U \) of \( x \) contains a point of \( A \), so \( x \in \bar{A} \) by Theorem 17.5. Conversely, suppose that \( X \) is metrizable and \( x \in \bar{A} \). Let \( d \) be a metric for the topology of \( X \). For eac... | Yes |
Theorem 21.3. Let \( f : X \rightarrow Y \) . If the function \( f \) is continuous, then for every convergent sequence \( {x}_{n} \rightarrow x \) in \( X \), the sequence \( f\left( {x}_{n}\right) \) converges to \( f\left( x\right) \) . The converse holds if \( X \) is metrizable. | Proof. Assume that \( f \) is continuous. Given \( {x}_{n} \rightarrow x \), we wish to show that \( f\left( {x}_{n}\right) \rightarrow \) \( f\left( x\right) \) . Let \( V \) be a neighborhood of \( f\left( x\right) \) . Then \( {f}^{-1}\left( V\right) \) is a neighborhood of \( x \), and so there is an \( N \) such t... | Yes |
Lemma 21.4. The addition, subtraction, and multiplication operations are continuous functions from \( \mathbb{R} \times \mathbb{R} \) into \( \mathbb{R} \) ; and the quotient operation is a continuous function from \( \mathbb{R} \times \left( {\mathbb{R}-\{ 0\} }\right) \) into \( \mathbb{R} \) . | You have probably seen this lemma proved before; it is a standard “ \( \epsilon \) - \( \delta \) argument.” If not, a proof is outlined in Exercise 12 below; you should have no trouble filling in the details. | No |
Theorem 21.5. If \( X \) is a topological space, and if \( f, g : X \rightarrow \mathbb{R} \) are continuous functions, then \( f + g, f - g \), and \( f \cdot g \) are continuous. If \( g\left( x\right) \neq 0 \) for all \( x \), then \( f/g \) is continuous. | Proof. The map \( h : X \rightarrow \mathbb{R} \times \mathbb{R} \) defined by\n\n\[ h\left( x\right) = f\left( x\right) \times g\left( x\right) \]\n\nis continuous, by Theorem 18.4. The function \( f + g \) equals the composite of \( h \) and the addition operation\n\n\[ + : \mathbb{R} \times \mathbb{R} \rightarrow \m... | No |
Theorem 21.6 (Uniform limit theorem). Let \( {f}_{n} : X \rightarrow Y \) be a sequence of continuous functions from the topological space \( X \) to the metric space \( Y \) . If \( \left( {f}_{n}\right) \) converges uniformly to \( f \), then \( f \) is continuous. | Proof. Let \( V \) be open in \( Y \) ; let \( {x}_{0} \) be a point of \( {f}^{-1}\left( V\right) \) . We wish to find a neighborhood \( U \) of \( {x}_{0} \) such that \( f\left( U\right) \subset V \) .\n\nLet \( {y}_{0} = f\left( {x}_{0}\right) \) . First choose \( \epsilon \) so that the \( \epsilon \) -ball \( B\l... | Yes |
(1) If \( A \) is either open or closed in \( X \), then \( q \) is a quotient map. | Proof. Step 1. We verify first the following two equations:\n\n\[ \n{q}^{-1}\left( V\right) = {p}^{-1}\left( V\right) \;\text{ if }V \subset p\left( A\right) ; \n\]\n\n\[ \np\left( {U \cap A}\right) = p\left( U\right) \cap p\left( A\right) \;\text{ if }U \subset X. \n\]\n\nTo check the first equation, we note that sinc... | Yes |
Theorem 22.2. Let \( p : X \rightarrow Y \) be a quotient map. Let \( Z \) be a space and let \( g : X \rightarrow Z \) be a map that is constant on each set \( {p}^{-1}\left( {\{ y\} }\right) \), for \( y \in Y \). Then \( g \) induces a map \( f : Y \rightarrow Z \) such that \( f \circ p = g \). The induced map \( f... | Proof. For each \( y \in Y \), the set \( g\left( {{p}^{-1}\left( {\{ y\} }\right) }\right) \) is a one-point set in \( Z \) (since \( g \) is constant on \( {p}^{-1}\left( {\{ y\} }\right) ) \). If we let \( f\left( y\right) \) denote this point, then we have defined a map \( f : Y \rightarrow Z \) such that for each ... | Yes |
Give \( {X}^{ * } \) the quotient topology.\n\n(a) The map \( g \) induces a bijective continuous map \( f : {X}^{ * } \rightarrow Z \), which is a homeomorphism if and only if \( g \) is a quotient map. | Proof. By the preceding theorem, \( g \) induces a continuous map \( f : {X}^{ * } \rightarrow Z \) ; it is clear that \( f \) is bijective. Suppose that \( f \) is a homeomorphism. Then both \( f \) and the projection map \( p : X \rightarrow {X}^{ * } \) are quotient maps, so that their composite \( q \) is a quotien... | Yes |
Lemma 23.1. If \( Y \) is a subspace of \( X \), a separation of \( Y \) is a pair of disjoint nonempty sets \( A \) and \( B \) whose union is \( Y \), neither of which contains a limit point of the other. The space \( Y \) is connected if there exists no separation of \( Y \) . | Proof. Suppose first that \( A \) and \( B \) form a separation of \( Y \) . Then \( A \) is both open and closed in \( Y \) . The closure of \( A \) in \( Y \) is the set \( \bar{A} \cap Y \) (where \( \bar{A} \) as usual denotes the closure of \( A \) in \( X \) ). Since \( A \) is closed in \( Y, A = \bar{A} \cap Y ... | Yes |
Lemma 23.2. If the sets \( C \) and \( D \) form a separation of \( X \), and if \( Y \) is a connected subspace of \( X \), then \( Y \) lies entirely within either \( C \) or \( D \) . | Proof. Since \( C \) and \( D \) are both open in \( X \), the sets \( C \cap Y \) and \( D \cap Y \) are open in \( Y \) . These two sets are disjoint and their union is \( Y \) ; if they were both nonempty, they would constitute a separation of \( Y \) . Therefore, one of them is empty. Hence \( Y \) must lie entirel... | Yes |
Theorem 23.3. The union of a collection of connected subspaces of \( X \) that have a point in common is connected. | Proof. Let \( \left\{ {A}_{\alpha }\right\} \) be a collection of connected subspaces of a space \( X \) ; let \( p \) be a point of \( \bigcap {A}_{\alpha } \) . We prove that the space \( Y = \bigcup {A}_{\alpha } \) is connected. Suppose that \( Y = C \cup D \) is a separation of \( Y \) . The point \( p \) is in on... | Yes |
Theorem 23.4. Let \( A \) be a connected subspace of \( X \) . If \( A \subset B \subset \bar{A} \), then \( B \) is also connected. | Proof. Let \( A \) be connected and let \( A \subset B \subset \bar{A} \) . Suppose that \( B = C \cup D \) is a separation of \( B \) . By Lemma 23.2, the set \( A \) must lie entirely in \( C \) or in \( D \) ; suppose that \( A \subset C \) . Then \( \bar{A} \subset \bar{C} \) ; since \( \bar{C} \) and \( D \) are d... | Yes |
Theorem 23.5. The image of a connected space under a continuous map is connected. | Proof. Let \( f : X \rightarrow Y \) be a continuous map; let \( X \) be connected. We wish to prove the image space \( Z = f\left( X\right) \) is connected. Since the map obtained from \( f \) by restricting its range to the space \( Z \) is also continuous, it suffices to consider the case of a continuous surjective ... | Yes |
Theorem 23.6. A finite cartesian product of connected spaces is connected. | Proof. We prove the theorem first for the product of two connected spaces \( X \) and \( Y \) . This proof is easy to visualize. Choose a \ | No |
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