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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\[ A = f\left( X\right) \cap \left( {-\infty, r}\right) \;\text{ and }\;B = f\left( X\right) \cap \left( {r, + \infty }\right) \]\n\nare disjoint, and they are nonempty because one contains \( f\left( a\right) \) and the other contains \( f\left( b\right) \). Eac... | 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 \(... | Yes |
Theorem 25.5. If \( X \) is a topological space, each path component of \( X \) lies in a component of \( X \) . If \( X \) is locally path connected, then the components and the path components of \( X \) are the same. | Proof. Let \( C \) be a component of \( X \) ; let \( x \) be a point of \( C \) ; let \( P \) be the path component of \( X \) containing \( x \) . Since \( P \) is connected, \( P \subset C \) . We wish to show that if \( X \) is locally path connected, \( P = C \) . Suppose that \( P \varsubsetneq C \) . Let \( Q \)... | Yes |
Lemma 26.1. Let \( Y \) be a subspace of \( X \) . Then \( Y \) is compact if and only if every covering of \( Y \) by sets open in \( X \) contains a finite subcollection covering \( Y \) . | Proof. Suppose that \( Y \) is compact and \( \mathcal{A} = {\left\{ {A}_{\alpha }\right\} }_{\alpha \in J} \) is a covering of \( Y \) by sets open in \( X \) . Then the collection\n\n\[ \left\{ {{A}_{\alpha } \cap Y \mid \alpha \in J}\right\} \]\n\nis a covering of \( Y \) by sets open in \( Y \) ; hence a finite sub... | Yes |
Theorem 26.2. Every closed subspace of a compact space is compact. | Proof. Let \( Y \) be a closed subspace of the compact space \( X \) . Given a covering \( \mathcal{A} \) of \( Y \) by sets open in \( X \), let us form an open covering \( \mathcal{B} \) of \( X \) by adjoining to \( \mathcal{A} \) the single open set \( X - Y \), that is,\n\n\[ \mathcal{B} = \mathcal{A} \cup \{ X - ... | Yes |
Theorem 26.3. Every compact subspace of a Hausdorff space is closed. | Proof. Let \( Y \) be a compact subspace of the Hausdorff space \( X \) . We shall prove that \( X - Y \) is open, so that \( Y \) is closed.\n\nLet \( {x}_{0} \) be a point of \( X - Y \) . We show there is a neighborhood of \( {x}_{0} \) that is disjoint from \( Y \) . For each point \( y \) of \( Y \), let us choose... | Yes |
Theorem 26.5. The image of a compact space under a continuous map is compact. | Proof. Let \( f : X \rightarrow Y \) be continuous; let \( X \) be compact. Let \( \mathcal{A} \) be a covering of the set \( f\left( X\right) \) by sets open in \( Y \) . The collection\n\n\[ \left\{ {{f}^{-1}\left( A\right) \mid A \in \mathcal{A}}\right\} \]\n\nis a collection of sets covering \( X \) ; these sets ar... | Yes |
Theorem 26.6. Let \( f : X \rightarrow Y \) be a bijective continuous function. If \( X \) is compact and \( Y \) is Hausdorff, then \( f \) is a homeomorphism. | Proof. We shall prove that images of closed sets of \( X \) under \( f \) are closed in \( Y \) ; this will prove continuity of the map \( {f}^{-1} \) . If \( A \) is closed in \( X \), then \( A \) is compact, by Theorem 26.2. Therefore, by the theorem just proved, \( f\left( A\right) \) is compact. Since \( Y \) is H... | Yes |
Theorem 26.7. The product of finitely many compact spaces is compact. | Proof. We shall prove that the product of two compact spaces is compact; the theorem follows by induction for any finite product.\n\nStep 1. Suppose that we are given spaces \( X \) and \( Y \), with \( Y \) compact. Suppose that \( {x}_{0} \) is a point of \( X \), and \( N \) is an open set of \( X \times Y \) contai... | No |
Theorem 26.9. Let \( X \) be a topological space. Then \( X \) is compact if and only if for every collection \( \mathcal{C} \) of closed sets in \( X \) having the finite intersection property, the intersection \( \mathop{\bigcap }\limits_{{C \in \mathcal{C}}}C \) of all the elements of \( \mathcal{C} \) is nonempty. | Proof. Given a collection \( \mathcal{A} \) of subsets of \( X \), let\n\n\[ \mathcal{C} = \{ X - A \mid A \in \mathcal{A}\} \]\n\nbe the collection of their complements. Then the following statements hold:\n\n(1) \( \mathcal{A} \) is a collection of open sets if and only if \( \mathcal{C} \) is a collection of closed ... | No |
Theorem 27.3. A subspace \( A \) of \( {\mathbb{R}}^{n} \) is compact if and only if it is closed and is bounded in the euclidean metric \( d \) or the square metric \( \rho \) . | Proof. It will suffice to consider only the metric \( \rho \) ; the inequalities\n\n\[ \rho \left( {x, y}\right) \leq d\left( {x, y}\right) \leq \sqrt{n}\rho \left( {x, y}\right) \]\n\nimply that \( A \) is bounded under \( d \) if and only if it is bounded under \( \rho \) .\n\nSuppose that \( A \) is compact. Then, b... | Yes |
Theorem 27.4 (Extreme value theorem). Let \( f : X \rightarrow Y \) be continuous, where \( Y \) is an ordered set in the order topology. If \( X \) is compact, then there exist points \( c \) and \( d \) in \( X \) such that \( f\left( c\right) \leq f\left( x\right) \leq f\left( d\right) \) for every \( x \in X \) . | Proof. Since \( f \) is continuous and \( X \) is compact, the set \( A = f\left( X\right) \) is compact. We show that \( A \) has a largest element \( M \) and a smallest element \( m \) . Then since \( m \) and \( M \) belong to \( A \), we must have \( m = f\left( c\right) \) and \( M = f\left( d\right) \) for some ... | Yes |
Lemma 27.5 (The Lebesgue number lemma). Let \( \mathcal{A} \) be an open covering of the metric space \( \left( {X, d}\right) \) . If \( X \) is compact, there is a \( \delta > 0 \) such that for each subset of \( X \) having diameter less than \( \delta \), there exists an element of \( \mathcal{A} \) containing it. | Proof. Let \( \mathcal{A} \) be an open covering of \( X \) . If \( X \) itself is an element of \( \mathcal{A} \), then any positive number is a Lebesgue number for \( \mathcal{A} \) . So assume \( X \) is not an element of \( \mathcal{A} \) .\n\nChoose a finite subcollection \( \left\{ {{A}_{1},\ldots ,{A}_{n}}\right... | Yes |
Theorem 27.6 (Uniform continuity theorem). Let \( f : X \rightarrow Y \) be a continuous map of the compact metric space \( \left( {X,{d}_{X}}\right) \) to the metric space \( \left( {Y,{d}_{Y}}\right) \) . Then \( f \) is uniformly continuous. | Proof. Given \( \epsilon > 0 \), take the open covering of \( Y \) by balls \( B\left( {y,\epsilon /2}\right) \) of radius \( \epsilon /2 \) . Let \( \mathcal{A} \) be the open covering of \( X \) by the inverse images of these balls under \( f \) . Choose \( \delta \) to be a Lebesgue number for the covering \( \mathc... | Yes |
Theorem 27.7. Let \( X \) be a nonempty compact Hausdorff space. If \( X \) has no isolated points, then \( X \) is uncountable. | Proof. Step 1. We show first that given any nonempty open set \( U \) of \( X \) and any point \( x \) of \( X \), there exists a nonempty open set \( V \) contained in \( U \) such that \( x \notin \bar{V} \). Choose a point \( y \) of \( U \) different from \( x \) ; this is possible if \( x \) is in \( U \) because ... | Yes |
Theorem 29.2. Let \( X \) be a Hausdorff space. Then \( X \) is locally compact if and only if given \( x \) in \( X \), and given a neighborhood \( U \) of \( x \), there is a neighborhood \( V \) of \( x \) such that \( \bar{V} \) is compact and \( \bar{V} \subset U \). | Proof. Clearly this new formulation implies local compactness; the set \( C = \bar{V} \) is the desired compact set containing a neighborhood of \( x \) . To prove the converse, suppose \( X \) is locally compact; let \( x \) be a point of \( X \) and let \( U \) be a neighborhood of \( x \) . Take the one-point compac... | Yes |
Corollary 29.3. Let \( X \) be locally compact Hausdorff; let \( A \) be a subspace of \( X \) . If \( A \) is closed in \( X \) or open in \( X \), then \( A \) is locally compact. | Proof. Suppose that \( A \) is closed in \( X \) . Given \( x \in A \), let \( C \) be a compact subspace of \( X \) containing the neighborhood \( U \) of \( x \) in \( X \) . Then \( C \cap A \) is closed in \( C \) and thus compact, and it contains the neighborhood \( U \cap A \) of \( x \) in \( A \) . (We have not... | Yes |
Corollary 29.4. A space \( X \) is homeomorphic to an open subspace of a compact Hausdorff space if and only if \( X \) is locally compact Hausdorff. | Proof. This follows from Theorem 29.1 and Corollary 29.3. | No |
Let \( X \) be a topological space.\n\n(a) Let \( A \) be a subset of \( X \) . If there is a sequence of points of \( A \) converging to \( x \) , then \( x \in \bar{A} \) ; the converse holds if \( X \) is first-countable.\n\n(b) Let \( f : X \rightarrow Y \) . If \( f \) is continuous, then for every convergent sequ... | The proof is a direct generalization of the proof given in \( §{21} \) under the hypothesis of metrizability, so it will not be repeated here. | No |
A subspace of a second-countable space is second-countable, and a countable product of second-countable spaces is second-countable. | Consider the second countability axiom. If \( \mathcal{B} \) is a countable basis for \( X \), then \( \{ B \cap A \mid B \in \mathcal{B}\} \) is a countable basis for the subspace \( A \) of \( X \) . If \( {\mathcal{B}}_{i} \) is a countable basis for the space \( {X}_{i} \), then the collection of all products \( \p... | Yes |
Theorem 30.3. Suppose that \( X \) has a countable basis. Then:\n\n(a) Every open covering of \( X \) contains a countable subcollection covering \( X \) . | Proof. Let \( \\left\\{ {B}_{n}\\right\\} \) be a countable basis for \( X \) .\n\n(a) Let \( \\mathcal{A} \) be an open covering of \( X \) . For each positive integer \( n \) for which it is possible, choose an element \( {A}_{n} \) of \( \\mathcal{A} \) containing the basis element \( {B}_{n} \) . The collection \( ... | Yes |
Lemma 31.1. Let \( X \) be a topological space. Let one-point sets in \( X \) be closed.\n\n(a) \( X \) is regular if and only if given a point \( x \) of \( X \) and a neighborhood \( U \) of \( x \) , there is a neighborhood \( V \) of \( x \) such that \( \bar{V} \subset U \) . | Proof. (a) Suppose that \( X \) is regular, and suppose that the point \( x \) and the neighborhood \( U \) of \( x \) are given. Let \( B = X - U \) ; then \( B \) is a closed set. By hypothesis, there exist disjoint open sets \( V \) and \( W \) containing \( x \) and \( B \), respectively. The set \( \bar{V} \) is d... | Yes |
Theorem 31.2. (a) A subspace of a Hausdorff space is Hausdorff; a product of Hausdorff spaces is Hausdorff. | (a) This result was an exercise in \( §{17} \) . We provide a proof here. Let \( X \) be Hausdorff. Let \( x \) and \( y \) be two points of the subspace \( Y \) of \( X \) . If \( U \) and \( V \) are disjoint neighborhoods in \( X \) of \( x \) and \( y \), respectively, then \( U \cap Y \) and \( V \cap Y \) are dis... | Yes |
Theorem 33.2. A subspace of a completely regular space is completely regular. A product of completely regular spaces is completely regular. | Proof. Let \( X \) be completely regular; let \( Y \) be a subspace of \( X \) . Let \( {x}_{0} \) be a point of \( Y \) , and let \( A \) be a closed set of \( Y \) disjoint from \( {x}_{0} \) . Now \( A = \bar{A} \cap Y \), where \( \bar{A} \) denotes the closure of \( A \) in \( X \) . Therefore, \( {x}_{0} \notin \... | Yes |
Theorem 34.2 (Imbedding theorem). Let \( X \) be a space in which one-point sets are closed. Suppose that \( {\left\{ {f}_{\alpha }\right\} }_{\alpha \in J} \) is an indexed family of continuous functions \( {f}_{\alpha } : X \rightarrow \) \( \mathbb{R} \) satisfying the requirement that for each point \( {x}_{0} \) o... | The proof is almost a copy of Step 2 of the preceding proof; one merely replaces \( n \) by \( \alpha \), and \( {\mathbb{R}}^{\omega } \) by \( {\mathbb{R}}^{J} \), throughout. One needs one-point sets in \( X \) to be closed in order to be sure that, given \( x \neq y \), there is an index \( \alpha \) such that \( {... | Yes |
Theorem 36.1 (Existence of finite partitions of unity). Let \( \\left\\{ {{U}_{1},\\ldots ,{U}_{n}}\\right\\} \) be a finite open covering of the normal space \( X \) . Then there exists a partition of unity dominated by \( \\left\\{ {U}_{i}\\right\\} \) . | Proof. Step 1. First, we prove that one can \ | No |
Theorem 36.2. If \( X \) is a compact \( m \) -manifold, then \( X \) can be imbedded in \( {\mathbb{R}}^{N} \) for some positive integer \( N \) . | Proof. Cover \( X \) by finitely many open sets \( \left\{ {{U}_{1},\ldots ,{U}_{n}}\right\} \), each of which may be imbedded in \( {\mathbb{R}}^{m} \) . Choose imbeddings \( {g}_{i} : {U}_{i} \rightarrow {\mathbb{R}}^{m} \) for each \( i \) . Being compact and Hausdorff, \( X \) is normal. Let \( {\phi }_{1},\ldots ,... | Yes |
Lemma 37.1. Let \( X \) be a set; let \( \mathcal{A} \) be a collection of subsets of \( X \) having the finite intersection property. Then there is a collection \( \mathcal{D} \) of subsets of \( X \) such that \( \mathcal{D} \) contains \( \mathcal{A} \), and \( \mathcal{D} \) has the finite intersection property, an... | Proof. As you might expect, we construct \( \mathcal{D} \) by using Zorn’s lemma. It states that, given a set \( A \) that is strictly partially ordered, in which every simply ordered subset has an upper bound, \( A \) itself has a maximal element.\n\nThe set \( A \) to which we shall apply Zorn’s lemma is not a subset... | No |
Lemma 37.2. Let \( X \) be a set; let \( \mathcal{D} \) be a collection of subsets of \( X \) that is maximal with respect to the finite intersection property. Then:\n\n(a) Any finite intersection of elements of \( \mathcal{D} \) is an element of \( \mathcal{D} \).\n\n(b) If \( A \) is a subset of \( X \) that intersec... | Proof. (a) Let \( B \) equal the intersection of finitely many elements of \( \mathcal{D} \) . Define a collection \( \mathcal{E} \) by adjoining \( B \) to \( \mathcal{D} \), so that \( \mathcal{E} = \mathcal{D} \cup \{ B\} \) . We show that \( \mathcal{E} \) has the finite intersection property; then maximality of \(... | Yes |
Theorem 37.3 (Tychonoff theorem). An arbitrary product of compact spaces is compact in the product topology. | Proof. Let\n\n\[ \nX = \mathop{\prod }\limits_{{\alpha \in J}}{X}_{\alpha } \n\]\n\nwhere each space \( {X}_{\alpha } \) is compact. Let \( \mathcal{A} \) be a collection of subsets of \( X \) having the finite intersection property. We prove that the intersection\n\n\[ \n\mathop{\bigcap }\limits_{{A \in \mathcal{A}}}\... | Yes |
Lemma 38.1. Let \( X \) be a space; suppose that \( h : X \rightarrow Z \) is an imbedding of \( X \) in the compact Hausdorff space \( Z \) . Then there exists a corresponding compactification \( Y \) of \( X \) ; it has the property that there is an imbedding \( H : Y \rightarrow Z \) that equals \( h \) on \( X \) .... | Proof. Given \( h \), let \( {X}_{0} \) denote the subspace \( h\left( X\right) \) of \( Z \), and let \( {Y}_{0} \) denote its closure in \( Z \) . Then \( {Y}_{0} \) is a compact Hausdorff space and \( {\bar{X}}_{0} = {Y}_{0} \) ; therefore, \( {Y}_{0} \) is a compactification of \( {X}_{0} \) . We now construct a sp... | Yes |
Theorem 38.2. Let \( X \) be a completely regular space. There exists a compactification \( Y \) of \( X \) having the property that every bounded continuous map \( f : X \rightarrow \mathbb{R} \) extends uniquely to a continuous map of \( Y \) into \( \mathbb{R} \) . | Proof. Let \( {\left\{ {f}_{\alpha }\right\} }_{\alpha \in J} \) be the collection of all bounded continuous real-valued functions on \( X \), indexed by some index set \( J \) . For each \( \alpha \in J \), choose a closed interval \( {I}_{\alpha } \) in \( \mathbb{R} \) containing \( {f}_{\alpha }\left( X\right) \) .... | Yes |
Lemma 38.3. Let \( A \subset X \) ; let \( f : A \rightarrow Z \) be a continuous map of \( A \) into the Hausdorff space \( Z \) . There is at most one extension of \( f \) to a continuous function \( g : \bar{A} \rightarrow Z \) . | Proof. This lemma was given as an exercise in \( §{18} \) ; we give a proof here. Suppose that \( g,{g}^{\prime } : \bar{A} \rightarrow X \) are two different extensions of \( f \) ; choose \( x \) so that \( g\left( x\right) \neq {g}^{\prime }\left( x\right) \) . Let \( U \) and \( {U}^{\prime } \) be disjoint neighbo... | Yes |
Theorem 38.4. Let \( X \) be a completely regular space; let \( Y \) be a compactification of \( X \) satisfying the extension property of Theorem 38.2. Given any continuous map \( f : X \rightarrow C \) of \( X \) into a compact Hausdorff space \( C \), the map \( f \) extends uniquely to a continuous map \( g : Y \ri... | Proof. Note that \( C \) is completely regular, so that it can be imbedded in \( {\left\lbrack 0,1\right\rbrack }^{J} \) for some \( J \) . So we may as well assume that \( C \subset {\left\lbrack 0,1\right\rbrack }^{J} \) . Then each component function \( {f}_{\alpha } \) of the map \( f \) is a bounded continuous rea... | Yes |
Theorem 38.5. Let \( X \) be a completely regular space. If \( {Y}_{1} \) and \( {Y}_{2} \) are two compact-ifications of \( X \) satisfying the extension property of Theorem 38.2, then \( {Y}_{1} \) and \( {Y}_{2} \) are equivalent. | Proof. Consider the inclusion mapping \( {j}_{2} : X \rightarrow {Y}_{2} \) . It is a continuous map of \( X \) into the compact Hausdorff space \( {Y}_{2} \) . Because \( {Y}_{1} \) has the extension property, we may, by the preceding theorem, extend \( {j}_{2} \) to a continuous map \( {f}_{2} : {Y}_{1} \rightarrow {... | Yes |
Lemma 39.1. Let \( \mathcal{A} \) be a locally finite collection of subsets of \( X \) . Then:\n\n(a) Any subcollection of \( \mathcal{A} \) is locally finite.\n\n(b) The collection \( \mathcal{B} = \{ \bar{A}{\} }_{A \in \mathcal{A}} \) of the closures of the elements of \( \mathcal{A} \) is locally finite.\n\n(c) \( ... | Proof. Statement (a) is trivial. To prove (b), note that any open set \( U \) that intersects the set \( \bar{A} \) necessarily intersects \( A \) . Therefore, if \( U \) is a neighborhood of \( x \) that intersects only finitely many elements \( A \) of \( \mathcal{A} \), then \( U \) can intersect at most the same nu... | Yes |
Lemma 39.2. Let \( X \) be a metrizable space. If \( \mathcal{A} \) is an open covering of \( X \), then there is an open covering \( \mathcal{E} \) of \( X \) refining \( \mathcal{A} \) that is countably locally finite. | Proof. We shall use the well-ordering theorem in proving this theorem. Choose a well-ordering \( < \) for the collection \( \mathcal{A} \) . Let us denote the elements of \( \mathcal{A} \) generically by the letters \( U, V, W,\ldots \) .\n\nChoose a metric for \( X \) . Let \( n \) be a positive integer, fixed for the... | No |
Lemma 40.1. Let \( X \) be a regular space with a basis \( \mathcal{B} \) that is countably locally finite. Then \( X \) is normal, and every closed set in \( X \) is a \( {G}_{\delta } \) set in \( X \) . | Proof. Step 1. Let \( W \) be open in \( X \) . We show there is a countable collection \( \left\{ {U}_{n}\right\} \) of open sets of \( X \) such that\n\n\[ W = \bigcup {U}_{n} = \bigcup {\bar{U}}_{n} \]\n\nSince the basis \( \mathcal{B} \) for \( X \) is countably locally finite, we can write \( \mathcal{B} = \bigcup... | Yes |
Lemma 40.2. Let \( X \) be normal; let \( A \) be a closed \( {G}_{\delta } \) set in \( X \) . Then there is a continuous function \( f : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( f\left( x\right) = 0 \) for \( x \in A \) and \( f\left( x\right) > 0 \) for \( x \notin A \) . | Proof. We gave this as an exercise in \( §{33} \) ; we provide a proof here. Write \( A \) as the intersection of the open sets \( {U}_{n} \), for \( n \in {\mathbb{Z}}_{ + } \) . For each \( n \), choose a continuous function \( {f}_{n} : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( f\left( x\right) =... | Yes |
Theorem 41.2. Every closed subspace of a paracompact space is paracompact. | Proof. Let \( Y \) be a closed subspace of the paracompact space \( X \) ; let \( \mathcal{A} \) be a covering of \( Y \) by sets open in \( Y \) . For each \( A \in \mathcal{A} \), choose an open set \( {A}^{\prime } \) of \( X \) such that \( {A}^{\prime } \cap Y = A \) . Cover \( X \) by the open sets \( {A}^{\prime... | Yes |
Theorem 41.5. Every regular Lindelöf space is paracompact. | Proof. Let \( X \) be regular and Lindelöf. Given an open covering \( \mathcal{A} \) of \( X \), it has a countable subcollection that covers \( X \) ; this subcollection is automatically countably locally finite. The preceding lemma applies to show \( \mathcal{A} \) has an open refinement that covers \( X \) and is lo... | Yes |
Lemma 43.1. A metric space \( X \) is complete if every Cauchy sequence in \( X \) has a convergent subsequence. | Proof. Let \( \left( {x}_{n}\right) \) be a Cauchy sequence in \( \left( {X, d}\right) \) . We show that if \( \left( {x}_{n}\right) \) has a subsequence \( \left( {x}_{{n}_{i}}\right) \) that converges to a point \( x \), then the sequence \( \left( {x}_{n}\right) \) itself converges to \( x \) .\n\nGiven \( \epsilon ... | Yes |
Theorem 43.2. Euclidean space \( {\mathbb{R}}^{k} \) is complete in either of its usual metrics, the euclidean metric \( d \) or the square metric \( \rho \) . | Proof. To show the metric space \( \left( {{\mathbb{R}}^{k},\rho }\right) \) is complete, let \( \left( {x}_{n}\right) \) be a Cauchy sequence in \( \left( {{\mathbb{R}}^{k},\rho }\right) \) . Then the set \( \left\{ {x}_{n}\right\} \) is a bounded subset of \( \left( {{\mathbb{R}}^{k},\rho }\right) \) . For if we choo... | Yes |
Lemma 43.3. Let \( X \) be the product space \( X = \prod {X}_{\alpha } \) ; let \( {\mathbf{x}}_{n} \) be a sequence of points of \( X \) . Then \( {\mathbf{x}}_{n} \rightarrow \mathbf{x} \) if and only if \( {\pi }_{\alpha }\left( {\mathbf{x}}_{n}\right) \rightarrow {\pi }_{\alpha }\left( \mathbf{x}\right) \) for eac... | Proof. This result was given as an exercise in \( §{19} \) ; we give a proof here. Because the projection mapping \( {\pi }_{\alpha } : X \rightarrow {X}_{\alpha } \) is continuous, it preserves convergent sequences; the \ | No |
Theorem 43.4. There is a metric for the product space \( {\mathbb{R}}^{\omega } \) relative to which \( {\mathbb{R}}^{\omega } \) is complete. | Proof. Let \( \bar{d}\left( {a, b}\right) = \min \{ \left| {a - b}\right| ,1\} \) be the standard bounded metric on \( \mathbb{R} \) . Let \( D \) be the metric on \( {\mathbb{R}}^{\omega } \) defined by\n\n\[ D\left( {\mathbf{x},\mathbf{y}}\right) = \sup \left\{ {\bar{d}\left( {{x}_{i},{y}_{i}}\right) /i}\right\} .\n\... | Yes |
Theorem 43.5. If the space \( Y \) is complete in the metric \( d \), then the space \( {Y}^{J} \) is complete in the uniform metric \( \bar{\rho } \) corresponding to \( d \) . | Proof. Recall that if \( \left( {Y, d}\right) \) is complete, so is \( \left( {Y,\bar{d}}\right) \), where \( \bar{d} \) is the bounded metric corresponding to \( d \) . Now suppose that \( {f}_{1},{f}_{2},\ldots \) is a sequence of points of \( {Y}^{J} \) that is a Cauchy sequence relative to \( \bar{\rho } \) . Given... | Yes |
Theorem 43.6. Let \( X \) be a topological space and let \( \left( {Y, d}\right) \) be a metric space. The set \( \mathcal{C}\left( {X, Y}\right) \) of continuous functions is closed in \( {Y}^{X} \) under the uniform metric. So is the set \( \mathcal{B}\left( {X, Y}\right) \) of bounded functions. Therefore, if \( Y \... | Proof. The first part of this theorem is just the uniform limit theorem (Theorem 21.6) in a new guise. First, we show that if a sequence of elements \( {f}_{n} \) of \( {Y}^{X} \) converges to the element \( f \) of \( {Y}^{X} \) relative to the metric \( \bar{\rho } \) on \( {Y}^{X} \), then it converges to \( f \) un... | Yes |
Theorem 45.1. A metric space \( \left( {X, d}\right) \) is compact if and only if it is complete and totally bounded. | Proof. If \( X \) is a compact metric space, then \( X \) is complete, as noted above. The fact that \( X \) is totally bounded is a consequence of the fact that the covering of \( X \) by all open \( \epsilon \) -balls must contain a finite subcovering.\n\nConversely, let \( X \) be complete and totally bounded. We sh... | Yes |
Lemma 45.2. Let \( X \) be a space; let \( \left( {Y, d}\right) \) be a metric space. If the subset \( \mathcal{F} \) of \( \mathcal{C}\left( {X, Y}\right) \) is totally bounded under the uniform metric corresponding to \( d \), then \( \mathcal{F} \) is equicontinuous under \( d \). | Proof. Assume \( \mathcal{F} \) is totally bounded. Given \( 0 < \epsilon < 1 \), and given \( {x}_{0} \), we find a neighborhood \( U \) of \( {x}_{0} \) such that \( d\left( {f\left( x\right), f\left( {x}_{0}\right) < \epsilon }\right. \) for \( x \in U \) and \( f \in \mathcal{F} \) .\n\nSet \( \delta = \epsilon /3 ... | Yes |
Theorem 46.1. A sequence \( {f}_{n} \) of functions converges to the function \( f \) in the topology of pointwise convergence if and only if for each \( x \) in \( X \), the sequence \( {f}_{n}\left( x\right) \) of points of \( Y \) converges to the point \( f\left( x\right) \) . | Proof. This result is just a reformulation, in function space notation, of a standard result about the product topology proved as Lemma 43.3. | No |
Lemma 46.3. If \( X \) is locally compact, or if \( X \) satisfies the first countability axiom, then \( X \) is compactly generated. | Proof. Suppose that \( X \) is locally compact. Let \( A \cap C \) be open in \( C \) for every compact subspace \( C \) of \( X \) . We show \( A \) is open in \( X \) . Given \( x \in A \), choose a neighborhood \( U \) of \( x \) that lies in a compact subspace \( C \) of \( X \) . Since \( A \cap C \) is open in \(... | Yes |
Lemma 46.4. If \( X \) is compactly generated, then a function \( f : X \rightarrow Y \) is continuous if for each compact subspace \( C \) of \( X \), the restricted function \( f \mid C \) is continuous. | Proof. Let \( V \) be an open subset of \( Y \) ; we show that \( {f}^{-1}\left( V\right) \) is open in \( X \) . Given any subspace \( C \) of \( X \), \[ {f}^{-1}\left( V\right) \cap C = {\left( f \mid C\right) }^{-1}\left( V\right) . \] If \( C \) is compact, this set is open in \( C \) because \( f \mid C \) is con... | Yes |
Theorem 46.5. Let \( X \) be a compactly generated space: let \( \left( {Y, d}\right) \) be a metric space. Then \( \mathcal{C}\left( {X, Y}\right) \) is closed in \( {Y}^{X} \) in the topology of compact convergence. | Proof. Let \( f \in {Y}^{X} \) be a limit point of \( \mathcal{C}\left( {X, Y}\right) \) ; we wish to show \( f \) is continuous. It suffices to show that \( f \mid C \) is continuous for each compact subspace \( C \) of \( X \) . For each \( n \), consider the neighborhood \( {B}_{C}\left( {f,1/n}\right) \) of \( f \)... | Yes |
Theorem 46.8. Let \( X \) be a space and let \( \left( {Y, d}\right) \) be a metric space. On the set \( \mathcal{C}\left( {X, Y}\right) \) , the compact-open topology and the topology of compact convergence coincide. | Proof. If \( A \) is a subset of \( Y \) and \( \epsilon > 0 \), let \( U\left( {A,\epsilon }\right) \) be the \( \epsilon \) -neighborhood of \( A \) . If \( A \) is compact and \( V \) is an open set containing \( A \), then there is an \( \epsilon > 0 \) such that \( U\left( {A,\epsilon }\right) \subset V \) . Indee... | Yes |
Theorem 46.10. Let \( X \) be locally compact Hausdorff; let \( \mathcal{C}\left( {X, Y}\right) \) have the compact-open topology. Then the map\n\n\[ e : X \times \mathcal{C}\left( {X, Y}\right) \rightarrow Y \]\n\ndefined by the equation\n\n\[ e\left( {x, f}\right) = f\left( x\right) \]\nis continuous.\n\nThe map \( e... | Proof. Given a point \( \left( {x, f}\right) \) of \( X \times \mathcal{C}\left( {X, Y}\right) \) and an open set \( V \) in \( Y \) about the image point \( e\left( {x, f}\right) = f\left( x\right) \), we wish to find an open set about \( \left( {x, f}\right) \) that \( e \) maps into \( V \) . First, using the contin... | Yes |
Lemma 48.1. \( X \) is a Baire space if and only if given any countable collection \( \left\{ {U}_{n}\right\} \) of open sets in \( X \), each of which is dense in \( X \), their intersection \( \bigcap {U}_{n} \) is also dense in \( X \) . | Proof. Recall that a set \( C \) is dense in \( X \) if \( \bar{C} = X \) . The theorem now follows at once from the two remarks:\n\n(1) \( A \) is closed in \( X \) if and only if \( X - A \) is open in \( X \) .\n\n(2) \( B \) has empty interior in \( X \) if and only if \( X - B \) is dense in \( X \) . | No |
Theorem 48.2 (Baire category theorem). If \( X \) is a compact Hausdorff space or a complete metric space, then \( X \) is a Baire space. | Proof. Given a countable collection \( \left\{ {A}_{n}\right\} \) of closed set of \( X \) having empty interiors, we want to show that their union \( \bigcup {A}_{n} \) also has empty interior in \( X \) . So, given the nonempty open set \( {U}_{0} \) of \( X \), we must find a point \( x \) of \( {U}_{0} \) that does... | Yes |
Lemma 48.3. Let \( {C}_{1} \supset {C}_{2} \supset \cdots \) be a nested sequence of nonempty closed sets in the complete metric space \( X \) . If \( \operatorname{diam}{C}_{n} \rightarrow 0 \), then \( \bigcap {C}_{n} \neq \varnothing \) . | Proof. We gave this as an exercise in \( §{43} \) . Here is a proof: Choose \( {x}_{n} \in {C}_{n} \) for each \( n \) . Because \( {x}_{n},{x}_{m} \in {C}_{N} \) for \( n, m \geq N \), and because \( \operatorname{diam}{C}_{N} \) can be made less than any given \( \epsilon \) by choosing \( N \) large enough, the sequ... | No |
Theorem 49.1. Let \( h : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Given \( \epsilon > 0 \), there is a function \( g : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \) with \( \left| {h\left( x\right) - g\left( x\right) }\right| < \epsilon \) for all \( x \), such tha... | Proof. Let \( I = \left\lbrack {0,1}\right\rbrack \) . Consider the space \( \mathcal{C} = \mathcal{C}\left( {I,\mathbb{R}}\right) \) of continuous maps from \( I \) to \( \mathbb{R} \), in the metric\n\n\[ \rho \left( {f, g}\right) = \max \{ \left| {f\left( x\right) - g\left( x\right) }\right| \} . \]\n\nThis space is... | Yes |
Theorem 50.1. Let \( X \) be a space having finite dimension. If \( Y \) is a closed subspace of \( X \), then \( Y \) has finite dimension and \( \dim Y \leq \dim X \) . | Proof. Let \( \dim X = m \) . Let \( \mathcal{A} \) be a covering of \( Y \) by sets open in \( Y \) . For each \( A \in \mathcal{A} \), choose an open set \( {A}^{\prime } \) of \( X \) such that \( {A}^{\prime } \cap Y = A \) . Cover \( X \) by the open sets \( {A}^{\prime } \), along with the open set \( X - Y \) . ... | Yes |
Lemma 50.4. Given a finite set \( \left\{ {{\mathbf{x}}_{1},\ldots ,{\mathbf{x}}_{n}}\right\} \) of points of \( {\mathbb{R}}^{N} \) and given \( \delta > 0 \) , there exists a set \( \left\{ {{\mathbf{y}}_{1},\ldots ,{\mathbf{y}}_{n}}\right\} \) of points of \( {\mathbb{R}}^{N} \) in general position in \( {\mathbb{R}... | Proof. We proceed by induction. Set \( {\mathbf{y}}_{1} = {\mathbf{x}}_{1} \) . Suppose that we are given \( {\mathbf{y}}_{1},\ldots ,{\mathbf{y}}_{p} \) in general position in \( {\mathbb{R}}^{N} \) . Consider the set of all planes in \( {\mathbb{R}}^{N} \) determined by subsets of \( \left\{ {{\mathbf{y}}_{1},\ldots ... | Yes |
Theorem 50.5 (The imbedding theorem). Every compact metrizable space \( X \) of topological dimension \( m \) can be imbedded in \( {\mathbb{R}}^{{2m} + 1} \) . | Proof. Let \( N = {2m} + 1 \) . Let us denote the square metric for \( {\mathbb{R}}^{N} \) by\n\n\[ \left| {\mathbf{x} - \mathbf{y}}\right| = \max \left\{ {\left| {{x}_{i} - {y}_{i}}\right| ;i = 1,\ldots, N}\right\} .\n\nThen we can use \( \rho \) to denote the corresponding sup metric on the space \( \mathcal{C}\left(... | Yes |
Theorem 50.6. Every compact subspace of \( {\mathbb{R}}^{N} \) has topological dimension at most \( N \) . | Proof. The proof is a generalization of the proof given in Example 3 for \( {\mathbb{R}}^{2} \) . Let \( \rho \) be the square metric on \( {\mathbb{R}}^{N} \) .\n\nStep 1. We begin by breaking \( {\mathbb{R}}^{N} \) up into \ | No |
Lemma 51.1. The relations \( \simeq \) and \( { \simeq }_{p} \) are equivalence relations. | Proof. Let us verify the properties of an equivalence relation.\n\nGiven \( f \), it is trivial that \( f \simeq f \) ; the map \( F\left( {x, t}\right) = f\left( x\right) \) is the required homotopy. If \( f \) is a path, \( F \) is a path homotopy.\n\nGiven \( f \simeq {f}^{\prime } \), we show that \( {f}^{\prime } ... | Yes |
Theorem 51.2. The operation \( * \) has the following properties:\n\n(1) (Associativity) If \( \left\lbrack f\right\rbrack * \left( {\left\lbrack g\right\rbrack * \left\lbrack h\right\rbrack }\right) \) is defined, so is \( \left( {\left\lbrack f\right\rbrack * \left\lbrack g\right\rbrack }\right) * \left\lbrack h\righ... | Proof. We shall make use of two elementary facts. The first is the fact that if \( k \) : \( X \rightarrow Y \) is a continuous map, and if \( F \) is a path homotopy in \( X \) between the paths \( f \) and \( {f}^{\prime } \), then \( k \circ F \) is a path homotopy in \( Y \) between the paths \( k \circ f \) and \(... | Yes |
Theorem 52.1. The map \( \widehat{\alpha } \) is a group isomorphism. | Proof. To show that \( \widehat{\alpha } \) is a homomorphism, we compute\n\n\[ \widehat{\alpha }\left( \left\lbrack f\right\rbrack \right) * \widehat{\alpha }\left( \left\lbrack g\right\rbrack \right) = \left( {\left\lbrack \bar{\alpha }\right\rbrack * \left\lbrack f\right\rbrack * \left\lbrack \alpha \right\rbrack }\... | Yes |
Corollary 52.2. If \( X \) is path connected and \( {x}_{0} \) and \( {x}_{1} \) are two points of \( X \), then \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is isomorphic to \( {\pi }_{1}\left( {X,{x}_{1}}\right) \) . | Suppose that \( X \) is a topological space. Let \( C \) be the path component of \( X \) containing \( {x}_{0} \) . It is easy to see that \( {\pi }_{1}\left( {C,{x}_{0}}\right) = {\pi }_{1}\left( {X,{x}_{0}}\right) \), since all loops and homotopies in \( X \) that are based at \( {x}_{0} \) must lie in the subspace ... | No |
Lemma 52.3. In a simply connected space \( X \), any two paths having the same initial and final points are path homotopic. | Proof. Let \( \alpha \) and \( \beta \) be two paths from \( {x}_{0} \) to \( {x}_{1} \) . Then \( \alpha * \bar{\beta } \) is defined and is a loop on \( X \) based at \( {x}_{0} \) . Since \( X \) is simply connected, this loop is path homotopic to the constant loop at \( {x}_{0} \) . Then\n\n\[ \left\lbrack {\alpha ... | Yes |
Theorem 52.4. If \( h : \left( {X,{x}_{0}}\right) \rightarrow \left( {Y,{y}_{0}}\right) \) and \( k : \left( {Y,{y}_{0}}\right) \rightarrow \left( {Z,{z}_{0}}\right) \) are continuous, then \( {\left( k \circ h\right) }_{ * } = {k}_{ * } \circ {h}_{ * } \). If \( i : \left( {X,{x}_{0}}\right) \rightarrow \left( {X,{x}_... | Proof. The proof is a triviality. By definition,\n\n\[ \n{\left( k \circ h\right) }_{ * }\left( \left\lbrack f\right\rbrack \right) = \left\lbrack {\left( {k \circ h}\right) \circ f}\right\rbrack , \n\]\n\n\[ \n\left( {{k}_{ * } \circ {h}_{ * }}\right) \left( \left\lbrack f\right\rbrack \right) = {k}_{ * }\left( {{h}_{... | Yes |
Corollary 52.5. If \( h : \left( {X,{x}_{0}}\right) \rightarrow \left( {Y,{y}_{0}}\right) \) is a homeomorphism of \( X \) with \( Y \), then \( {h}_{ * } \) is an isomorphism of \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) with \( {\pi }_{1}\left( {Y,{y}_{0}}\right) \) . | Proof. Let \( k : \left( {Y,{y}_{0}}\right) \rightarrow \left( {X,{x}_{0}}\right) \) be the inverse of \( h \) . Then \( {k}_{ * } \circ {h}_{ * } = {\left( k \circ h\right) }_{ * } = {i}_{ * } \) , where \( i \) is the identity map of \( \left( {X,{x}_{0}}\right) \) ; and \( {h}_{ * } \circ {k}_{ * } = {\left( h \circ... | Yes |
Theorem 53.1. The map \( p : \mathbb{R} \rightarrow {S}^{1} \) given by the equation\n\n\[ p\left( x\right) = \left( {\cos {2\pi x},\sin {2\pi x}}\right) \]\n\nis a covering map. | Proof. The fact that \( p \) is a covering map comes from elementary properties of the sine and cosine functions. Consider, for example, the subset \( U \) of \( {S}^{1} \) consisting of those points having positive first coordinate. The set \( {p}^{-1}\left( U\right) \) consists of those points \( x \) for which \( \c... | Yes |
Theorem 53.2. Let \( p : E \rightarrow B \) be a covering map. If \( {B}_{0} \) is a subspace of \( B \), and if \( {E}_{0} = {p}^{-1}\left( {B}_{0}\right) \), then the map \( {p}_{0} : {E}_{0} \rightarrow {B}_{0} \) obtained by restricting \( p \) is a covering map. | Proof. Given \( {b}_{0} \in {B}_{0} \), let \( U \) be an open set in \( B \) containing \( {b}_{0} \) that is evenly covered by \( p \) ; let \( \left\{ {V}_{\alpha }\right\} \) be a partition of \( {p}^{-1}\left( U\right) \) into slices. Then \( U \cap {B}_{0} \) is a neighborhood of \( {b}_{0} \) in \( {B}_{0} \), a... | Yes |
Theorem 53.3. It \( p : E \rightarrow B \) and \( {p}^{\prime } : {E}^{\prime } \rightarrow {B}^{\prime } \) are covering maps, then\n\n\[ p \times {p}^{\prime } : E \times {E}^{\prime } \rightarrow B \times {B}^{\prime } \]\n\n is a covering map. | Proof. Given \( b \in B \) and \( {b}^{\prime } \in {B}^{\prime } \), let \( U \) and \( {U}^{\prime } \) be neighborhoods of \( b \) and \( {b}^{\prime } \) , respectively, that are evenly covered by \( p \) and \( {p}^{\prime } \), respectively. Let \( \left\{ {V}_{\alpha }\right\} \) and \( \left\{ {V}_{\beta }^{\pr... | Yes |
Lemma 54.2. Let \( p : E \rightarrow B \) be a covering map; let \( p\left( {e}_{0}\right) = {b}_{0} \) . Let the map \( F : I \times I \rightarrow B \) be continuous, with \( F\left( {0,0}\right) = {b}_{0} \) . There is a unique lifting of \( F \) to a continuous map\n\n\[ \widetilde{F} : I \times I \rightarrow E \]\n... | Proof. Given \( F \), we first define \( \widetilde{F}\left( {0,0}\right) = {e}_{0} \) . Next, we use the preceding lemma to extend \( \widetilde{F} \) to the left-hand edge \( 0 \times I \) and the bottom edge \( I \times 0 \) of \( I \times I \) . Then we extend \( \widetilde{F} \) to all of \( I \times I \) as follo... | Yes |
Theorem 54.3. Let \( p : E \rightarrow B \) be a covering map; let \( p\left( {e}_{0}\right) = {b}_{0} \) . Let \( f \) and \( g \) be two paths in \( B \) from \( {b}_{0} \) to \( {b}_{1} \) ; let \( \widetilde{f} \) and \( \widetilde{g} \) be their respective liftings to paths in \( E \) beginning at \( {e}_{0} \) . ... | Proof. Let \( F : I \times I \rightarrow B \) be the path homotopy between \( f \) and \( g \) . Then \( F\left( {0,0}\right) = \) \( {b}_{0} \) . Let \( \widetilde{F} : I \times I \rightarrow E \) be the lifting of \( F \) to \( E \) such that \( \widetilde{F}\left( {0,0}\right) = {e}_{0} \) . By the preceding lemma, ... | Yes |
Theorem 54.4. Let \( p : E \rightarrow B \) be a covering map; let \( p\left( {e}_{0}\right) = {b}_{0} \) . If \( E \) is path connected, then the lifting correspondence\n\n\[ \phi : {\pi }_{1}\left( {B,{b}_{0}}\right) \rightarrow {p}^{-1}\left( {b}_{0}\right) \]\n\nis surjective. If \( E \) is simply connected, it is ... | Proof. If \( E \) is path connected, then, given \( {e}_{1} \in {p}^{-1}\left( {b}_{0}\right) \), there is a path \( \widetilde{f} \) in \( E \) from \( {e}_{0} \) to \( {e}_{1} \) . Then \( f = p \circ \widetilde{f} \) is a loop in \( B \) at \( {b}_{0} \), and \( \phi \left( \left\lbrack f\right\rbrack \right) = {e}_... | Yes |
Theorem 54.5. The fundamental group of \( {S}^{1} \) is isomorphic to the additive group of integers. | Proof. Let \( p : \mathbb{R} \rightarrow {S}^{1} \) be the covering map of Theorem 53.1, let \( {e}_{0} = 0 \), and let \( {b}_{0} = p\left( {e}_{0}\right) \) . Then \( {p}^{-1}\left( {b}_{0}\right) \) is the set \( \mathbb{Z} \) of integers. Since \( \mathbb{R} \) is simply connected, the lifting correspondence\n\n\[ ... | Yes |
Lemma 55.1. If \( A \) is a retract of \( X \), then the homomorphism of fundamental groups induced by inclusion \( j : A \rightarrow X \) is injective. | Proof. If \( r : X \rightarrow A \) is a retraction, then the composite map \( r \circ j \) equals the identity map of \( A \) . It follows that \( {r}_{ * } \circ {j}_{ * } \) is the identity map of \( {\pi }_{1}\left( {A, a}\right) \), so that \( {j}_{ * } \) must be injective. | Yes |
Theorem 55.2 (No-retraction theorem). There is no retraction of \( {B}^{2} \) onto \( {S}^{1} \) . | Proof. If \( {S}^{1} \) were a retract of \( {B}^{2} \), then the homomorphism induced by inclusion \( j : {S}^{1} \rightarrow {B}^{2} \) would be injective. But the fundamental group of \( {S}^{1} \) is nontrivial and the fundamental group of \( {B}^{2} \) is trivial. | Yes |
Lemma 55.3. Let \( h : {S}^{1} \rightarrow X \) be a continuous map. Then the following conditions are equivalent:\n\n(1) \( h \) is nulhomotopic.\n\n(2) \( h \) extends to a continuous map \( k : {B}^{2} \rightarrow X \) .\n\n(3) \( {h}_{ * } \) is the trivial homomorphism of fundamental groups. | Proof. (1) \( \Rightarrow \) (2). Let \( H : {S}^{1} \times I \rightarrow X \) be a homotopy between \( h \) and a constant map. Let \( \pi : {S}^{1} \times I \rightarrow {B}^{2} \) be the map\n\n\[ \pi \left( {x, t}\right) = \left( {1 - t}\right) x. \]\n\nThen \( \pi \) is continuous, closed and surjective, so it is a... | Yes |
Corollary 55.4. The inclusion map \( j : {S}^{1} \rightarrow {R}^{2} - \mathbf{0} \) is not nulhomotopic. The identity map \( i : {S}^{1} \rightarrow {S}^{1} \) is not nulhomotopic. | Proof. There is a retraction of \( \mathbb{R} - \mathbf{0} \) onto \( {S}^{1} \) given by the equation \( r\left( x\right) = x/\parallel x\parallel \) . Therefore, \( {j}_{ * } \) is injective, and hence nontrivial. Similarly, \( {i}_{ * } \) is the identity homomorphism, and hence nontrivial. | Yes |
Theorem 55.5. Given a nonvanishing vector field on \( {B}^{2} \), there exists a point of \( {S}^{1} \) where the vector field points directly inward and a point of \( {S}^{1} \) where it points directly outward. | Proof. A vector field on \( {B}^{2} \) is an ordered pair \( \left( {x, v\left( x\right) }\right) \), where \( x \) is in \( {B}^{2} \) and \( v \) is a continuous map of \( {B}^{2} \) into \( {\mathbb{R}}^{2} \) . In calculus, one often uses the notation\n\n\[ \mathbf{v}\left( x\right) = {v}_{1}\left( x\right) \mathbf... | Yes |
Theorem 55.6 (Brouwer fixed-point theorem for the disc). If \( f : {B}^{2} \rightarrow {B}^{2} \) is continuous, then there exists a point \( x \in {B}^{2} \) such that \( f\left( x\right) = x \) . | Proof. We proceed by contradiction. Suppose that \( f\left( x\right) \neq x \) for every \( x \) in \( {B}^{2} \) . Then defining \( v\left( x\right) = f\left( x\right) - x \) gives us a nonvanishing vector field \( \left( {x, v\left( x\right) }\right) \) on \( {B}^{2} \) . But the vector field \( v \) cannot point dir... | Yes |
Theorem 56.1 (The fundamental theorem of algebra). A polynomial equation\n\n\\[ \n{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \\cdots + {a}_{1}x + {a}_{0} = 0 \n\\]\n\nof degree \\( n > 0 \\) with real or complex coefficients has at least one (real or complex) root. | Proof. Step 1. Consider the map \\( f : {S}^{1} \\rightarrow {S}^{1} \\) given by \\( f\\left( z\\right) = {z}^{n} \\), where \\( z \\) is a complex number. We show that the induced homomorphism \\( {f}_{ * } \\) of fundamental groups is injective.\n\nLet \\( {p}_{0} : I \\rightarrow {S}^{1} \\) be the standard loop in... | No |
Theorem 57.2. There is no continuous antipode-preserving map \( g : {S}^{2} \rightarrow {S}^{1} \) . | Proof. Suppose \( g : {S}^{2} \rightarrow {S}^{1} \) is continuous and antipode preserving. Let us take \( {S}^{1} \) to be the equator of \( {S}^{2} \) . Then the restriction of \( g \) to \( {S}^{1} \) is a continuous antipode-preserving map \( h \) of \( {S}^{1} \) to itself. By the preceding theorem, \( h \) is not... | Yes |
Theorem 57.3 (Borsuk-Ulam theorem for \( {S}^{2} \) ). Given a continuous map \( f : {S}^{2} \rightarrow \) \( {\mathbb{R}}^{2} \), there is a point \( x \) of \( {S}^{2} \) such that \( f\left( x\right) = f\left( {-x}\right) \) . | Proof. Suppose that \( f\left( x\right) \neq f\left( {-x}\right) \) for all \( x \in {S}^{2} \) . Then the map\n\n\[ g\left( x\right) = \left\lbrack {f\left( x\right) - f\left( {-x}\right) }\right\rbrack /\parallel f\left( x\right) - f\left( {-x}\right) \parallel \]\n\nis a continuous map \( g : {S}^{2} \rightarrow {S}... | Yes |
Theorem 57.4 (The bisection theorem). Given two bounded polygonal regions in \( {\mathbb{R}}^{2} \), there exists a line in \( {\mathbb{R}}^{2} \) that bisects each of them. | Proof. We take two bounded polygonal regions \( {A}_{1} \) and \( {A}_{2} \) in the plane \( {\mathbb{R}}^{2} \times 1 \) in \( {\mathbb{R}}^{3} \) , and show there is a line \( L \) in this plane that bisects each of them.\n\nGiven a point \( u \) of \( {S}^{2} \), let us consider the plane \( P \) in \( {\mathbb{R}}^... | Yes |
Lemma 58.1. Let \( h, k : \left( {X,{x}_{0}}\right) \rightarrow \left( {Y,{y}_{0}}\right) \) be continuous maps. If \( h \) and \( k \) are homotopic, and if the image of the base point \( {x}_{0} \) of \( X \) remains fixed at \( {y}_{0} \) during the homotopy, then the homomorphisms \( {h}_{ * } \) and \( {k}_{ * } \... | Proof. The proof is immediate. By assumption, there is a homotopy \( H : X \times I \rightarrow Y \) between \( h \) and \( k \) such that \( H\left( {{x}_{0}, t}\right) = {y}_{0} \) for all \( t \) . It follows that if \( f \) is a loop in \( X \) based at \( {x}_{0} \), then the composite\n\n\[ I \times I\xrightarrow... | Yes |
Theorem 58.2. The inclusion map \( j : {S}^{n} \rightarrow {\mathbb{R}}^{n + 1} - \mathbf{0} \) induces an isomorphism of fundamental groups. | Proof. Let \( X = {\mathbb{R}}^{n + 1} - \mathbf{0} \) ; let \( {b}_{0} = \left( {1,0,\ldots ,0}\right) \) . Let \( r : X \rightarrow {S}^{n} \) be the map \( r\left( x\right) = x/\parallel x\parallel \) . Then \( r \circ j \) is the identity map of \( {S}^{n} \), so that \( {r}_{ * } \circ {j}_{ * } \) is the identity... | Yes |
Lemma 58.4. Let \( h, k : X \rightarrow Y \) be continuous maps; let \( h\left( {x}_{0}\right) = {y}_{0} \) and \( k\left( {x}_{0}\right) = {y}_{1} \). If \( h \) and \( k \) are homotopic, there is a path \( \alpha \) in \( Y \) from \( {y}_{0} \) to \( {y}_{1} \) such that \( {k}_{ * } = \widehat{\alpha } \circ {h}_{... | Proof. Let \( f : I \rightarrow X \) be a loop in \( X \) based at \( {x}_{0} \). We must show that\n\n\[ \n{k}_{ * }\left( \left\lbrack f\right\rbrack \right) = \widehat{\alpha }\left( {{h}_{ * }\left( \left\lbrack f\right\rbrack \right) .}\right.\n\]\n\nThis equation states that \( \left\lbrack {k \circ f}\right\rbra... | Yes |
Corollary 58.6. Let \( h : X \rightarrow Y \) . If \( h \) is nulhomotopic, then \( {h}_{ * } \) is the trivial homomorphism. | Proof. The constant map induces the trivial homomorphism. | No |
Theorem 58.7. Let \( f : X \rightarrow Y \) be continuous; let \( f\left( {x}_{0}\right) = {y}_{0} \). If \( f \) is a homotopy equivalence, then\n\n\[ \n{f}_{ * } : {\pi }_{1}\left( {X,{x}_{0}}\right) \rightarrow {\pi }_{1}\left( {Y,{y}_{0}}\right)\n\]\n\nis an isomorphism. | Proof. Let \( g : Y \rightarrow X \) be a homotopy inverse for \( f \). Consider the maps\n\n\[ \n\left( {X,{x}_{0}}\right) \overset{f}{ \rightarrow }\left( {Y,{y}_{0}}\right) \overset{g}{ \rightarrow }\left( {X,{x}_{1}}\right) \overset{f}{ \rightarrow }\left( {Y,{y}_{1}}\right) ,\n\]\n\nwhere \( {x}_{1} = g\left( {y}_... | Yes |
Theorem 59.3. If \( n \geq 2 \), the \( n \) -sphere \( {S}^{n} \) is simply connected. | Proof. Let \( p = \left( {0,\ldots ,0,1}\right) \in {\mathbb{R}}^{n + 1} \) and \( q = \left( {0,\ldots ,0, - 1}\right) \) be the \ | No |
Theorem 60.1. \( {\pi }_{1}\left( {X \times Y,{x}_{0} \times {y}_{0}}\right) \) is isomorphic with \( {\pi }_{1}\left( {X,{x}_{0}}\right) \times {\pi }_{1}\left( {Y,{y}_{0}}\right) \) . | Proof. Let \( p : X \times Y \rightarrow X \) and \( q : X \times Y \rightarrow Y \) be the projection mappings. If we use the base points indicated in the statement of the theorem, we have induced homomorphisms\n\n\[ \n{p}_{ * } : {\pi }_{1}\left( {X \times Y,{x}_{0} \times {y}_{0}}\right) \rightarrow {\pi }_{1}\left(... | Yes |
Theorem 60.3. The projective plane \( {P}^{2} \) is a compact surface, and the quotient map \( p : {S}^{2} \rightarrow {P}^{2} \) is a covering map. | Proof. First we show that \( p \) is an open map. Let \( U \) be open in \( {S}^{2} \) . Now the antipodal map \( a : {S}^{2} \rightarrow {S}^{2} \) given by \( a\left( x\right) = - x \) is a homeomorphism of \( {S}^{2} \) ; hence \( a\left( U\right) \) is open in \( {S}^{2} \) . Since\n\n\[ \n{p}^{-1}\left( {p\left( U... | Yes |
Corollary 60.4. \( {\pi }_{1}\left( {{P}^{2}, y}\right) \) is a group of order 2 . | Proof. The projection \( p : {S}^{2} \rightarrow {P}^{2} \) is a covering map. Since \( {S}^{2} \) is simply connected, we can apply Theorem 54.4, which tells us there is a bijective correspondence between \( {\pi }_{1}\left( {{P}^{2}, y}\right) \) and the set \( {p}^{-1}\left( y\right) \) . Since this set is a two-ele... | Yes |
Theorem 60.6. The fundamental group of the double torus is not abelian. | Proof. The double torus \( T\# T \) is the surface obtained by taking two copies of the torus, deleting a small open disc from each of them, and pasting the remaining pieces together along their edges. We assert that the figure eight \( X \) is a retract of \( T\# T \) . This fact implies that inclusion \( j : X \right... | Yes |
Lemma 61.1. Let \( C \) be a compact subspace of \( {S}^{2} \) ; let \( b \) be a point of \( {S}^{2} - C \) ; and let \( h \) be a homeomorphism of \( {S}^{2} - b \) with \( {\mathbb{R}}^{2} \) . Suppose \( U \) is a component of \( {S}^{2} - C \) . If \( U \) does not contain \( b \), then \( h\left( U\right) \) is a... | Proof. We show first that if \( U \) is a component of \( {S}^{2} - C \), then \( U - b \) is connected. This result is trivial if \( b \notin U \), so suppose that \( b \in U \) and suppose the sets \( A \) and \( B \) form a separation of \( U - b \) . Choose a neighborhood \( W \) of \( b \) disjoint from \( C \) su... | Yes |
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