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Corollary 6.2.20. Let \( M \) be a manifold. If \( {H}_{1}\left( {M;\mathbb{Z}/2\mathbb{Z}}\right) = 0 \), then \( M \) is orientable. | Recall we have the universal coefficient theorem, Theorem 5.5.12. Since \( {H}_{0}\left( {M;\mathbb{Z}}\right) = \mathbb{Z} \), that theorem gives an isomorphism\n\n\[ e : {H}^{1}\left( {M;\mathbb{Z}/2\mathbb{Z}}\right) \rightarrow \operatorname{Hom}\left( {{H}_{1}\left( M\right) ,\mathbb{Z}/2\mathbb{Z}}\right) . \] | No |
Theorem 6.2.26. Let \( M \) be an oriented manifold with boundary. Then \( \partial M \) has a well-defined induced orientation given by the construction in Definition 6.2.25. | Proof. This is simply a matter of checking that the local orientations \( \left\{ {\bar{\varphi }}_{x}\right\} \) are indeed compatible, and that they are independent of the choice of coordinate patches \( \left( {{U}_{\alpha },{\varphi }_{\alpha }}\right) \) used in the construction. | No |
Lemma 6.2.28. Let \( G = \mathbb{F} \) be a field of characteristic 0 or odd characteristic. Then a manifold \( M \) is \( G \) -orientable if and only if it is orientable. If \( G = \mathbb{F} \) is a field of characteristic 2, then every manifold \( M \) is \( G \) -orientable. | Proof. We do the more interesting case of a field \( \mathbb{F} \) of characteristic \( \neq 2 \) . Consider the diagram in Definition 6.2.3.\n\nIf we let \( V = {\varphi }_{x}\left( D\right) \) and replace \( G \) in that diagram by \( {H}_{n}\left( {M, M - V;G}\right) \) and the two vertical maps by the isomorphisms ... | Yes |
Corollary 6.2.31. Let \( M \) be a compact connected \( n \) -manifold with boundary.\n\n(1) For any such \( M,{H}_{n}\left( {M,\partial M;\mathbb{Z}/2\mathbb{Z}}\right) \cong \mathbb{Z}/2\mathbb{Z} \) and \( {H}^{n}\left( {M,\partial M;\mathbb{Z}/2\mathbb{Z}}\right) \cong \mathbb{Z}/2\mathbb{Z} \) . | Proof. The statements on homology are a direct consequence of Theorems 6.2.6 and 6.2.30.\n\nThe statements for cohomology then follow from the universal coefficient theorem and Theorem 6.1.12. | No |
Corollary 6.2.36. Let \( M \) be a \( G \) -oriented \( n \) -manifold with boundary with fundamental class \( \left\lbrack {M,\partial M}\right\rbrack \), and let \( \partial M \) have the induced \( G \) -orientation with fundamental class \( \left\lbrack {\partial M}\right\rbrack \) . If \( i : \partial M \rightarro... | Proof. We have the exact sequence of the pair \( \left( {M,\partial M}\right) \) :
\[
{H}_{n}\left( {M,\partial M;G}\right) \overset{\partial }{ \rightarrow }{H}_{n - 1}\left( {\partial M;G}\right) \overset{{i}_{ * }}{ \rightarrow }{H}_{n - 1}\left( {M;G}\right) .
\]
But \( \left\lbrack {\partial M}\right\rbrack = \p... | Yes |
We shall show that \( {S}^{1} \) is orientable. | We take a rather strange looking description and parameterization of \( {S}^{1} \), but we do so to use this as a \ | No |
Let \( G = \mathbb{Z} \). Choose an orientation of \( {S}^{p} \), and let \( {S}^{p} \) have fundamental homology class \( \left\lbrack {S}^{p}\right\rbrack \) and fundamental cohomology class \( \left\{ {S}^{p}\right\} \). Also choose an orientation of \( {S}^{q} \) and let \( {S}^{q} \) have fundamental homology clas... | By Poincaré duality, \( \cap \left\lbrack M\right\rbrack : {H}^{q}\left( M\right) \rightarrow {H}_{p}\left( M\right) \) is an isomorphism. Hence \( \widetilde{\beta } \cap \left\lbrack M\right\rbrack = \pm \widetilde{a} \), and we choose the orientation on \( M \) so that the sign is positive. Then \[ 1 = e\left( {\wid... | Yes |
To define an orientation of \( \mathbb{C}{P}^{1} \) it suffices to give a local orientation \( {\bar{\varphi }}_{{z}_{0}} \) at a single point \( {z}_{0} \), and we choose \( {z}_{0} \) to be the point with homogeneous coordinates \( \left\lbrack {0,1}\right\rbrack \) . We specify \( {\bar{\varphi }}_{{z}_{0}} \) by le... | \[ {H}_{1}\left( {S}^{1}\right) \rightarrow {H}_{1}\left( {\mathbb{C}-\{ 0\} }\right) \rightarrow {H}_{2}\left( {\mathbb{C},\mathbb{C}-\{ 0\} }\right) \rightarrow {H}_{2}\left( {\mathbb{C}{P}^{1},\mathbb{C}{P}^{1}-\{ \left\lbrack {0,1}\right\rbrack \} }\right) . \] Here the first isomorphism is induced by inclusion, th... | Yes |
Theorem 6.4.5. Let \( M \) be a compact \( n \) -dimensional manifold with \( n \) odd. Then the Euler characteristic \( \chi \left( M\right) = 0 \) . | Proof. We may use any coefficients to compute the Euler characteristic, so we choose \( \mathbb{Z}/2\mathbb{Z} \) . This means that \( M \) is orientable with these coefficients. Also, they form a field, so for any \( j,{H}_{j}\left( {M;\mathbb{Z}/2\mathbb{Z}}\right) \) and \( {H}^{j}\left( {M;\mathbb{Z}/2\mathbb{Z}}\r... | Yes |
Theorem 6.4.6. Let \( M \) be a compact \( n \) -manifold with odd Euler characteristic. Then \( M \) is not the boundary of a compact \( \left( {n + 1}\right) \) -manifold. | Proof. Suppose that \( M \) is the boundary of the compact \( \left( {n + 1}\right) \) -manifold \( X \) . Consider the exact sequence of the pair \( \left( {X, M}\right) \) :\n\n\[ 0 \rightarrow {H}_{n + 1}\left( {X;\mathbb{Z}/2\mathbb{Z}}\right) \rightarrow {H}_{n + 1}\left( {X, M;\mathbb{Z}/2\mathbb{Z}}\right) \righ... | Yes |
Corollary 6.4.11. Let \( M \) be a compact connected oriented manifold of dimension \( {2n}, n \) odd. Then for \( G = \mathbb{Z} \) or any field \( \mathbb{F} \) of characteristic not equal to 2, \( \operatorname{rank}\left( {{K}^{n}\left( {M;G}\right) }\right) \) is even. Also, the Euler characteristic \( \chi \left(... | Proof. By Theorem 6.4.8, \( \langle \) , \( \rangle {isanonsingularskew} - {symmetricbilinearformon} \) \( {K}^{n}\left( {M;G}\right) \), so by Theorem B.2.1, \( {K}^{n}\left( {M;G}\right) \) must have even rank.\n\nWe may use any field to compute Euler characteristic. Choosing \( \mathbb{F} = \mathbb{Q} \), say, and u... | Yes |
Theorem 6.4.15. Let \( M \) be a compact connected oriented manifold of dimension \( {2n} \) with \( n \) even. If the signature \( \sigma \left( M\right) \neq 0 \), then \( M \) is not the boundary of an oriented \( \left( {{2n} + 1}\right) \) -manifold. | Proof. Suppose that \( M \) is the boundary of \( {X}^{{2n} + 1} \) . Let \( V = {H}^{n}\left( {{M}^{2n};\mathbb{R}}\right) \), and let \( V \) have dimension \( t \) . We will use Lefschetz duality to find a subspace \( {V}_{0} \) of \( V \) of dimension \( t/2 \) with the restriction of the intersection form on \( M ... | Yes |
Theorem 6.4.17. Let \( M \) and \( N \) be compact connected oriented \( n \) -manifolds, \( n > 0 \) , with fundamental classes \( \left\lbrack M\right\rbrack \) and \( \left\lbrack N\right\rbrack \) respectively. Then\n\n\[ \n{H}_{n}\left( {M\# N}\right) \cong {H}^{n}\left( {M\# N}\right) \cong \mathbb{Z} \]\n\n\[ \n... | Proof. We work in cohomology as we wish to obtain the cup product structure. The argument in homology is very similar.\n\nConsider the disjoint union \( M \cup N \) of \( M \) and \( N \) . Then it is certainly true that \( {H}^{j}\left( {M \cup N}\right) \cong {H}^{j}\left( M\right) \oplus {H}^{j}\left( N\right) \) fo... | Yes |
For \( n \geq 2,{\pi }_{n}\left( {X,{x}_{0}}\right) \) is an abelian group. For \( n \geq 3,{\pi }_{n}\left( {X, A,{x}_{0}}\right) \) is an abelian group. | Proof. Here is a picture of a homotopy between \( {\alpha \beta } \) and \( {\beta \alpha } \) in case \( n = 2 \), for \( {\pi }_{n}\left( {X,{x}_{0}}\right) \) .\n\n\n\nSimilarly for \( n > 2 \) for \( {\pi }_{n}\l... | Yes |
Theorem 7.1.9. (1) If \( f : \left( {X, A,{x}_{0}}\right) \rightarrow \left( {X, A,{x}_{0}}\right) \) is the identity map, then \( {f}_{ * } \) : \( {\pi }_{n}\left( {X, A,{x}_{0}}\right) \rightarrow {\pi }_{n}\left( {X, A,{x}_{0}}\right) \) is the identity map. | Proof. Parts (1), (2), (3), and (5) are immediate. We leave the proof of (4) as an exercise. | No |
Theorem 7.1.11. Let \( X \) be a path-connected space and let \( {x}_{0},{x}_{1} \in X \) . Let \( \alpha : I \rightarrow X \) be a path from \( {x}_{0} \) to \( {x}_{1} \), i.e., \( \alpha \left( 0\right) = {x}_{0} \) and \( \alpha \left( 1\right) = {x}_{1} \) . Then \( \alpha \) induces an isomorphism \( {\alpha }_{ ... | Proof. Let \( f : \left( {{I}^{n},\partial {I}^{n}}\right) \rightarrow \left( {X,{x}_{0}}\right) \) represent an element of \( {\pi }_{n}\left( {X,{x}_{0}}\right) \) . The following picture shows how to obtain \( {\alpha }_{ * }\left( f\right) \in {\pi }_{n}\left( {X,{x}_{1}}\right) \) for \( n = 2 \), with the general... | No |
Corollary 7.1.12. Let \( X \) be a path-connected space and let \( {x}_{0} \in X \) . Then the construction of Theorem 7.1.11 gives an action of \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) on \( {\pi }_{n}\left( {X,{x}_{0}}\right) \) for every \( n \) . The set of equivalence classes of elements of \( {\pi }_{n}\left( {X... | Proof. The only point to note is that we are considering homotopies \( F : {I}^{n} \times I \rightarrow X \) with the property that for every \( t \in I, F \mid \partial {I}^{n} \times \{ t\} \) is a map to a single point, so this gives us homotopies of maps \( f : {S}^{n} \rightarrow X \) where the point 1 is allowed ... | Yes |
Theorem 7.2.1. Let \( X \) and \( Y \) be path-connected spaces. Let \( {x}_{0} \in X \) and \( {y}_{0} \in Y \) . Let \( p : X \times Y \rightarrow X \) and \( q : X \times Y \rightarrow Y \) be projection on the first and second factors respectively. Then \( {p}_{ * } \times {q}_{ * } : {\pi }_{n}\left( {X \times Y,\... | Proof. First we show \( {p}_{ * } \times {q}_{ * } \) is onto. Let \( f : \left( {{S}^{n},1}\right) \rightarrow \left( {X,{x}_{0}}\right) \) represent an element \( \alpha \) of \( {\pi }_{n}\left( {X,{x}_{0}}\right) \) and let \( g : \left( {{S}^{n},1}\right) \rightarrow \left( {Y,{y}_{0}}\right) \) represent an eleme... | Yes |
Theorem 7.2.2. Let \( X \) be a path-connected space and let \( \widetilde{X} \) be a connected covering space of \( X \) . Let \( p : \widetilde{X} \rightarrow X \) be the covering projection. Let \( {\widetilde{x}}_{0} \in \widetilde{X} \) and let \( {x}_{0} \in X \) with \( p\left( {\widetilde{x}}_{0}\right) = {x}_{... | Proof. First we show \( {p}_{ * } \) is onto. Let \( f : \left( {{S}^{n},1}\right) \rightarrow \left( {X,{x}_{0}}\right) \) represent an element of \( {\pi }_{n}\left( {X,{x}_{0}}\right) \) . We wish to show there is an \( \widetilde{f} : \left( {{S}^{n},1}\right) \rightarrow \left( {\widetilde{X},{\widetilde{x}}_{0}}\... | Yes |
Corollary 7.2.3. For every \( n \geq 2,{\pi }_{n}\left( {{S}^{1},1}\right) = 0 \) . | Proof. We know from Example 2.2.3 that \( p : \mathbb{R} \rightarrow {S}^{1} \) by \( p\left( t\right) = \exp \left( {2\pi it}\right) \) is a covering map, so for \( n \geq 2,{\pi }_{n}\left( {{S}^{1},1}\right) \cong {\pi }_{n}\left( {\mathbb{R},0}\right) = 0 \) as \( \mathbb{R} \) is contractible. | Yes |
The projection \( p : X \times Y \rightarrow X \) is a locally trivial fiber bundle with fiber \( Y \). | Indeed, we call this a globally trivial fiber bundle. | No |
Theorem 7.2.9. Let \( p : E \rightarrow B \) be a locally trivial fiber bundle. Let \( {b}_{0} \in B \) , \( F = {p}^{-1}\left( {b}_{0}\right) \), and \( {e}_{0} \in F \) . Then for every \( n \) , \[ {p}_{ * } : {\pi }_{n}\left( {E, F,{e}_{0}}\right) \rightarrow {\pi }_{n}\left( {B,{b}_{0}}\right) \] is an isomorphism... | Proof. First we show \( {p}_{ * } \) is onto. Let \( g : \left( {{I}^{n},\partial {I}^{n}}\right) \rightarrow \left( {B,{b}_{0}}\right) \) represent an element of \( {\pi }_{n}\left( {B,{b}_{0}}\right) \) . Regard \( {I}^{n} \) as \( I \times {I}^{n - 1} \) . Then \( \{ 0\} \times {I}^{n - 1} \subset \partial {I}^{n - ... | Yes |
Corollary 7.2.10. Let \( p : E \rightarrow B \) be a locally trivial fiber bundle with fiber \( F = \) \( {p}^{-1}\left( {b}_{0}\right) \) and let \( {f}_{0} \in F \) . Then there is an exact sequence\n\n\[ \cdots \rightarrow {\pi }_{n}\left( {F,{f}_{0}}\right) \rightarrow {\pi }_{n}\left( {E,{f}_{0}}\right) \overset{{... | Proof. The first claim follows immediately from Theorems 7.1.9 and 7.2.9.\n\nAs for the second claim, if \( s \) is a section, then \( {s}_{ * } : {\pi }_{n}\left( {B,{b}_{0}}\right) \rightarrow {\pi }_{n}\left( {E,{f}_{0}}\right) \) splits \( {p}_{ * } \) , so this long exact sequence breaks up into a series of split ... | Yes |
Theorem 7.2.14. \( {\pi }_{i}\left( {S}^{n}\right) = 0 \) for \( i < n \) . | Proof. Give \( {S}^{i} \) a CW-structure with one cell in dimension \( i \) and one cell in dimension 0, and give \( {S}^{n} \) a CW-structure with one cell in dimension \( n \) and one cell in dimension 0 . Let \( f : {S}^{i} \rightarrow {S}^{n} \) represent an element of \( {\pi }_{i}\left( {S}^{n}\right) \) . Then b... | Yes |
Corollary 7.2.16. For any \( n \geq 1,{\pi }_{n}\left( {S}^{n}\right) \cong \mathbb{Z} \) . | Proof. By Hopf's theorem, we have an isomorphism\n\n\[ \n{\pi }_{n}\left( {S}^{n}\right) \rightarrow \left\{ {\text{ degrees of maps from }{S}^{n}\text{ to }{S}^{n}}\right\} \n\] \n\nBut this latter set is \( \mathbb{Z} \) by Theorem 4.2.31. | Yes |
Lemma 7.2.18. The following diagram commutes: | \[ \cdots \rightarrow {\pi }_{n}\left( {A,{x}_{0}}\right) \rightarrow {\pi }_{n}\left( {X,{x}_{0}}\right) \rightarrow {\pi }_{n}\left( {X, A,{x}_{0}}\right) \overset{\partial }{ \rightarrow }{\pi }_{n - 1}\left( {A,{x}_{0}}\right) \rightarrow \cdots \] | No |
Theorem 7.2.21. (a) For any space \( X,{\sum X} \) is path-connected. | Proof. (a) is trivial | No |
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 \( n \) for which this statement holds. Then \( A \) contains 1, since if \( n = 1 \), the only n... | 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 doe... | 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 6.5. 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\n is 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(2) \( \Rightarrow \) (3). If \( f : \{ 1,\ldots, n\} \rightarrow B \) is surjective, define \( g : B \rightarrow \{ 1,\ldots, n\} \) by the equation\n\n\[ g\left( ... | 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 \( g \) arbitraril... | 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\[ f\left( 1\right) = \text{smallest element of}C = g\left( 1\right) \text{,} \]\n\nby \( \left( *\right) \). Now for all \( j < i \), we have \( f\left( j\right) = g\... | Yes |
Theorem 8.3. 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\nsuch that \( c\left( B\right) \) is an element of \( B \), for each \( B \in \math... | 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 \( B \), a... | 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 ele... | 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 \( {S}_{\Omega } \) ; let... | 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. (2) \( \Rightarrow \) (1). 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 \( x \in B \subset U \) . Condition (2) tells us ther... | 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 \( \left( {a, b}\right) \), th... | 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\n is 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\[ \nS = \left\{ {{\pi }_{1}^{-1}\left( U\right) \mid U\text{ open in }X}\right\} \cup \left\{ {{\pi }_{2}^{-1}\left( V\right) \mid V\text{ open in }Y}\right\} \n\]\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 \( S \) . Because every element of \( S \) belongs to \( \mathcal{T} \), so do arbitrary unions of finite intersections of elements of \( \mathcal{S} \) . Thus \( {\mathcal{T}}^{\... | 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 |
Consider the subset \( Y = \left\lbrack {0,1}\right\rbrack \) of the real line \( \mathbb{R} \), in the subspace topology. The subspace topology has as basis all sets of the form \( \left( {a, b}\right) \cap Y \), where \( \left( {a, b}\right) \) is an open interval in \( \mathbb{R} \) Such a set is of one of the follo... | \[ \left( {a, b}\right) \cap Y = \left\{ \begin{array}{ll} \left( {a, b}\right) & \text{ if }a\text{ and }b\text{ are in }Y, \\ \lbrack 0, b) & \text{ if only }b\text{ is in }Y, \\ (a,1\rbrack & \text{ if only }a\text{ is in }Y, \\ Y\text{ or }\varnothing & \text{ if neither }a\text{ nor }b\text{ is in }Y. \end{array}\... | 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 \( Q \) ). Wntten out, it is the following.\n\n\( x \notin \bar{A} \Leftr... | 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}^{\prime ... | 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, \( v \) t 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... | 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(1) \( \Rightarrow \) (2). Assume that \( f \) is continuous. Let \( A \) be a subset of \( X \) . We ... | 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 \( X \), th... | 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}^{-t}\left( C... | 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 are ... | Yes |
Theorem 19.1 (Comparison of the box and product topologies). The box topology on \( \prod {X}_{\alpha } \) has as basis all sets of the form \( \prod {U}_{\alpha } \), where \( {U}_{\alpha } \) is open in \( {X}_{\alpha } \) for each \( \alpha \) . The product topology on \( \prod {X}_{\alpha } \) has as basis all sets... | Two things are immediately clear First, for finte products \( \mathop{\prod }\limits_{{\alpha = 1}}^{n}{X}_{\alpha } \) the two topologies are precisely the same. Second, the box topology is in general finer than the product topology. | No |
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 functon \( {\pi }_{\beta } \) 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 \( {X... | 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 \( d \) .\n\nThe metnc \( \bar{d} \) is... | 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 th... | Yes |
Lemma 20.2. Let \( d \) and \( {d}^{\prime } \) be two metncs 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 1s by Lemma 13.3 a basis element \( {B}^{\prime } \) for the topology \( {\mathcal{T}}^{\prime } \) such that \( x \in {B}^{\prime } \subset {B}... | 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 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\chi }\lef... | 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 ... | 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 \ | 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 |
Theorem 22.1. Let \( p : X \rightarrow Y \) be a quotient map; let \( A \) be a subspace of \( X \) that is saturated with respect to \( p \) ; let \( q : A \rightarrow p\left( A\right) \) be the map obtained by restricting \( p \). (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 |
(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 \( \widetilde{A} \subset \widetilde{C} \) ; since \( \widetilde{C} \)... | 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 |
Theorem 24.1. If \( L \) is a linear continuum in the order topology, then \( L \) is connected, and so are intervals and rays in \( L \) . | Proof. Recall that a subspace \( Y \) of \( L \) is said to be convex if for every pair of points \( a, b \) of \( Y \) with \( a < b \), the entire interval \( \left\lbrack {a, b}\right\rbrack \) of points of \( L \) lies in \( Y \) . We prove that if \( Y \) is a convex subspace of \( L \), then \( Y \) is connected.... | Yes |
Theorem 24.3 (Intermediate value theorem). Let \( f : X \rightarrow Y \) be a continuous map, where \( X \) is a connected space and \( Y \) is an ordered set in the order topology. If \( a \) and \( b \) are two points of \( X \) and if \( r \) is a point of \( Y \) lying between \( f\left( a\right) \) and \( f\left( ... | Proof. Assume the hypotheses of the theorem. The sets\n\n\[ \nA = f\left( X\right) \cap \left( {-\infty, r}\right) \;\text{ and }\;B = f\left( X\right) \cap \left( {r, + \infty }\right) \n\] \n\nare disjoint, and they are nonempty because one contains \( f\left( a\right) \) and the other contains \( f\left( b\right) \)... | Yes |
Theorem 25.1. The components of \( X \) are connected disjoint subspaces of \( X \) whose union is \( X \), such that each nonempty connected subspace of \( X \) intersects only one of them. | Proof. Being equivalence classes, the components of \( X \) are disjoint and their union is \( X \) . Each connected subspace \( A \) of \( X \) intersects only one of them. For if \( A \) intersects the components \( {C}_{1} \) and \( {C}_{2} \) of \( X \), say in points \( {x}_{1} \) and \( {x}_{2} \), respectively, ... | Yes |
Theorem 25.3. A space \( X \) is locally connected if and only if for every open set \( U \) of \( X \), each component of \( U \) is open in \( X \) . | Proof. Suppose that \( X \) is locally connected; let \( U \) be an open set in \( X \) ; let \( C \) be a component of \( U \) If \( x \) is a point of \( C \), we can choose a connected neighborhood \( V \) of \( x \) such that \( V \subset U \) . Since \( V \) is connected, it must lie entirely in the component \( C... | Yes |
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