MATH31052 Topology
5.1 Definition. A collection \(\F \) of open subsets of a topological space \(X\) is called an open cover for a subset \(A\subset X\) if \(A \subset \bigcup _{U\in \F }U\). If \(\F \) and \(\F '\) are open covers for \(A\) and \(\F '\subset \F \) then \(\F '\) is a subcover of \(\F \).
A subset \(K\) of a topological space \(X\) is compact if each open cover for \(K\) has a finite subcover. In particular, if \(X\) itself is a compact subset then we say that it is a compact topological space.
5.2 Example.
(a) A finite subset \(A = \{\,a_1,a_2,\ldots ,a_n\,\}\) of any topological space \(X\) is compact.
Proof. Given an open cover \(\F \) for \(A\) we can choose, for each \(a_i\), an open set \(U_i\in \F \) such that \(a_i\in U_i\). Then \(\{\,U_1,U_2,\dots ,U_n\,\}\) is the required finite subcover of \(\F \) for \(A\). Hence \(A\) is a compact subset. □
(b) A subset of a discrete topological space is compact if and only if it is finite
Proof. `\(\Leftarrow \)': This is a particular case of 5.2(a).
`\(\Rightarrow \)': We prove the contrapositive. Suppose that \(A\subset X\) is an infinite subset of a discrete topological space. Then \(\F =\bigl \{\,\{a\}\mid a\in A\,\bigr \}\) is an open cover for \(A\) with no finite subcover. Hence \(A\)
is not compact. □
(c) The subset \((0,1)\) of \(\R \) with the usual topology is not compact.
Proof. \(\F = \{\,(a,1)\mid a\in (0,1)\,\}\) is an open cover for \((0,1)\) since, given \(x\in (0,1)\), \(x\in (x/2,1)\). This has no finite subcover for \((0,1)\) since, given \(\{\,(a_1,1),(a_2,1),\ldots ,(a_n,1)\,\}\subset \F \), \(\bigcup _{i=1}^n(a_i,1) = (a,1)\) where \(a=\min \{a_i\}\) and \(a/2\notin (a,1)\). □
(d) \(\R \) with the usual topology is not compact.
Proof. \(\F = \{\,(-n,n)\mid n\in \N \,\}\) is an open cover for \(\R \) with no finite subcover since \(\bigcup _{i=1}^k(-n_i,n_i) = (-n,n)\) where \(n = \max \{n_i\}\) and \(n+1\notin (-n,n)\). □
(e) Given a non-compact topological space \((X,\tau )\) consider the set \(X^* = X \sqcup \{\infty \}\) and the topology
\[\tau ^* = \tau \cup \{X \setminus C \cup \{\infty \} \mid C \subset X \text { compact}\}.\]
Then \((X^*,\tau ^*)\) is a compact topological space (called the one-point compactification) [Excercise].
5.3 Proposition. Given a subset \(X_1\) of a topological space \(X\). The subspace \(X_1\) is compact if and only if the subset \(X_1\) is a compact subset of the topological space \(X\).
Proof. Exercise. □ class="theoremendmark" >
Proof. Suppose that \(\F \) is an open cover for \(f(K)\). Let \(f^{-1}(\F ) = \{f^{-1}(V)\mid V\in \F \}\). Then \(f^{-1}(\F )\) is an over cover for \(K\) since, given \(a\in K\), \(f(a)\in f(K)\) so that \(f(a)\in V\) for some \(V\in \F \). Hence \(a\in f^{-1}(V)\) for some \(V\in \F \).
Now, since \(K\) is compact, \(f^{-1}(\F )\) has a finite subcover for \(K\),
\[\{\,f^{-1}(V_1),f^{-1}(V_2),\ldots ,f^{-1}(V_n)\,\}.\]
. Thus, given \(b\in f(K)\), \(b=f(a)\) for some \(a\in K\). Then \(a\in f^{-1}(V_i)\) for some \(i\), \(1\leq i\leq n\), so that \(b=f(a)\in V_i\). Hence \(\{\,V_1,V_2,\ldots ,V_n\,\}\) is a finite subcover of \(\F \) for \(f(K)\).
Hence \(f(K)\) is compact. □
5.5 Proposition. Compactness is a topological property, i.e. if \(X\) and \(Y\) are homeomorphic spaces then \(X\) is compact if and only if \(Y\) is compact.
Proof. Suppose that \(X\cong Y\).
Suppose that \(X\) is compact. Then a homeomorphism is a continuous bijection \(f\colon X\to Y\) and so, in particular, a continuous surjection. Hence \(Y=f(X)\) is compact by Proposition 5.4.
Similarly, if \(Y\) is compact then \(X\) is compact. class="theoremendmark" >
5.6 Definition. Given a sequence \((a_n)\) in a topological space \(X\). Then \(a \in X\) is called an accumulation point if every open neighbourhood \(U \ni a\) contains infinitely many members of the sequence, i.e. \(a_n \in U\) for infinietly many \(n \in \N \).
5.7 Theorem. Every sequence in a compact topological space \(X\) has an accumulation point.
Proof. Assume there is no accumulation point. Then for every \(x \in X\) there is open neighbourhood \(U_x \ni x\), such that only finitely many sequence members are contained in \(U_x\). Clearly \(\{U_x \mid x \in X\}\) is an open cover of \(X\). Now, compactness implies the existence of a finite subcover \(\{U_{x_1}, \ldots , U_{x_\ell }\}\). Note, that only for finitely many \(n \in \N \) we have \(a_n \in U_{x_i}\). Hence, ther exists an \(N_i\) such that \(a_n \notin U_{x_i}\) for \(n > N_i\). Then \(a_n \notin \bigcup _{i=1}^\ell U_{x_i} = X\) for \(n > N=\max \{N_1, \ldots , N_\ell \}\), but this a contradiction, since \((a_n)\) was a sequence in \(X\), i.e. \(a_n \in X\) for every \(n \in \N \). □
Proof. Suppose that \(K\) is a compact subset of a Hausdorff space \(X\). To prove that \(K\) is closed we prove that \(X\setminus K\) is open, and we prove this by proving that it is a union of open subsets. Let \(x\in X\setminus K\). Then, by the Hausdorff condition, for each \(a\in K\) there are open subsets \(U_a\), \(V_a\) of \(X\) such that \(a\in U_a\), \(x\in V_a\) and \(U_a\cap V_a=\emptyset \).
Then \(\{\,U_a\mid a\in A\,\}\) is an open cover for \(K\). Hence, since \(K\) is compact, there is a finite subcover \(\{U_{a_i}\mid 1\leq i\leq n\}\) for \(K\). So \(K\subset \bigcup _{i=1}^n U_{a_i}\).
Put \(V_x = \bigcap _{i=1}^nV_{a_i}\). Then \(V_x\) is a finite intersection of open sets and so is open and, since \(x\in V_{a_i}\) for all \(i\), \(x\in V_x\). Furthermore, for \(1\leq i\leq n\), \(V_x\cap U_{a_i} \subset V_{a_i}\cap U_{a_i}=\emptyset \) and so \(V_x\cap U_{a_i}=\emptyset \). Hence \(V_x\cap \bigcup _{i=1}^nU_{a_i} = \emptyset \) and so \(X_x\cap K=\emptyset \) or, equivalently, \(V_x\subset X\setminus K\).
Thus \(X\setminus K=\bigcup _{x\in X\setminus K}V_x\) is a union of open sets and so is open. Hence \(K\) is closed. □
Proof. Using the notation in the proposition, let \(\F \) be an open cover for \(A\). To prove that \(\F \) has a subcover for \(A\), observe that \(\F \cup \{X\setminus A\}\) is an open cover for \(K\), since \(A\) is closed, and so has a finite subcover \(\{\,U_1,U_2,\ldots ,U_n,X\setminus A\,\}\) where \(U_i\in \F \) (we may as well assume \(X\setminus A\) is one of the open sets in the subcover since we could add it if it wasn't included). Then \(\{\,U_1,U_2,\ldots ,U_n\,\}\) is a finite subcover of \(\F \) for \(A\) as required to prove that \(A\) is compact. □
5.10 Theorem. Suppose that \(f\colon X\rightarrow Y\) is a continuous bijection from a compact space to a Hausdorff space. Then \(f\) is a homeomorphism.
Proof. Using the notation in the theorem, we prove that \(f^{-1}\colon Y\to X\) is continuous by using closed subsets (see Problems 2, Question 8), i.e. we make use of the observation that \(f^{-1}\colon Y\to X\) is continuous if and only if, when \(A\subset X\) is a closed subset of \(X\), \((f^{-1})^{-1}(A) = f(A)\) is a closed subset of \(Y\). This condition holds from results we have already proved as follows.
\(A\) is a closed subset of compact \(X\)
\(\Rightarrow \) \(A\) is a compact subset of \(X\) (by Proposition 5.9)
\(\Rightarrow \) \(f(A)\) is a compact subset of \(Y\) (by Proposition 5.4)
\(\Rightarrow \) \(f(A)\) is a closed subset of Hausdorff \(Y\) (by Proposition 5.8)
Hence \(f^{-1}\) is continuous and so \(f\) is a homeomorphism. □
Proof. Suppose that \(X_1\) and \(X_2\) are non-empty topological spaces.
`\(\Rightarrow \)': Suppose that the product space \(X_1\times X_2\) is compact. Then, \(p_1\colon X_1\times X_2 \to X_1\), the projection map given by \(p_1(x_1,x_2)=x_1\), is a surjection since \(X_2\) is non-empty. Hence \(X_1\)
is compact since the continuous image of a compact space is compact (Proposition 5.4). Similarly, \(X_2\) is compact. □
To prove the converse the following lemma is useful.
5.12 Lemma. Let \(\cal B\) be a basis for the topology of a topological space \(X\). Then \(K\subset X\) is compact if and only if every open cover for \(K\) by open sets in the basis \(\cal B\) has a finite subcover.
Proof. `\(\Rightarrow \)': This is a special case of the definition.
`\(\Leftarrow \)': Suppose that \(K\) is a subset satisfying the condition regarding covers by basic open sets. Let \(\cal F\) be an open cover for \(K\). The we can write each open set of \(\cal F\) as a union of basic open sets. Let \({\cal F}_1\)
be the set of all basic open sets which are used in this process. Then \(\bigcup _{V\in {\cal F}_1}V = \bigcup _{U\in {\cal F}}U\) and so \({\cal F}_1\) is an open cover for \(K\) by basic open sets. Hence, by the given condition, \({\cal
F}_1\) has a finite subcover \({\cal F}'_1\) for \(K\). For each basic open set \(V\) in \({\cal F}'_1\) we can choose an open set \(U\) in \(\cal F\) which contains it as a subset so that \(V\subset U\). This gives a finite subset \({\cal F}'\) of
\(\cal F\) such that \(\bigcup _{U\in {\cal F}'}U \supset \bigcup _{V\in {\cal F}'_1}V \supset K\) and so \({\cal F}'\) is a finite subcover for \(K\) as required to prove that \(K\) is compact. □
Proof of Theorem 5.11 (continued): `\(\Leftarrow \)': Suppose that \(X_1\) and \(X_2\) are compact. Let \(\cal F\) be an open cover for \(X_1\times X_2\) by basic open sets (i.e. sets of the form \(U\times V\), where \(U\) is open in \(X_1\) and \(V\) is open in \(X_2\)). Then, for \(x\in X_2\), \(\cal F\) is an open cover for \(X_1\times \{x\} \cong X_1\) (Remark 3.12(b)) which is compact. Hence \(\cal F\) has a finite subcover
\[ {\cal F}_x = \{\,U_1^x\times V_1^x, U_2^x\times V_2^x, \ldots , U_{n_x}^x\times V_{n_x}^x\,\} \]
for \(X_1\times \{x\}\), where each \(U_i^x\times V_i^x\) is in \(\cal F\) and \(x\in V_i^x\) for each \(i\).
Put \(V_x=V_1^x\cap V_2^x \cap \cdots \cap V_{n_x}^x\) which is open (finite intersection of open sets) and non-empty since it contains \(x\). Then \({\cal F}_x\) is an open cover for \(X_1\times V_x\).
Now \(\{\,V_x\mid x\in X_2\,\}\) is an open cover for \(X_2\). Hence, since \(X_2\) is compact, this has a finite subcover \(\{\,V_{x_1},V_{x_2},\ldots ,V_{x_m}\,\}\) for \(X_2\). Then \({\cal F}_{x_1}\cup {\cal F}_{x_2}\cup \cdots \cup {\cal F}_{x_m}\) is a finite subcover of \(\cal F\) for \(X_1\times X_2\).
Hence, by the lemma, \(X_1\times X_2\) is compact. □
5.13 Definition. A subset \(A\subset \R ^n\) is bounded if there is a real number \(M\) such that \(|a|\leq M\) for all \(a\in A\).
5.14 Theorem (Heine-Borel-Lebesgue Theorem). A subset of \(\R ^n\) with the usual topology is compact if and only if it is closed and bounded.
The proof depends on various results as follows.
5.15 Lemma. A compact subset of \(\R ^n\) with the usual topology is bounded.
Proof. Suppose that \(X\subset \R ^n\) is compact (usual topology). Then \(\{B_n(\0)\mid n\in \N \}\) is an open cover for \(\R ^n\) and so an open cover for \(X\). Hence, since \(X\) is compact, there is a finite subcover \(\{\,B_{n_1}(\0),B_{n_2}(\0),\ldots ,B_{n_k}(\0)\,\}\) for \(X\). Let \(n=\max \{n_i\}\). Then \(X \subset \bigcup _{i=1}^k B_{n_i}(\0) = B_n(\0)\) and so is bounded. □
5.16 Lemma. A compact subset of \(\R ^n\) is closed.
Proof. This follows from Proposition 5.8 since \(\R ^n\) is Hausdorff (Proposition 4.5). class="theoremendmark" □ >
5.17 Theorem (Heine-Borel Theorem). For \(a\leq b\), the subset \([a,b]\) of \(\R \) with the usual topology is compact.
Proof. Suppose for contradiction that \(\F \) is an open cover for \([a,b]\) with no finite subcover.
Write \(I_0=[a,b]\) and divide this interval into two subintervals \([a,(a+b)/2]\) and \([(a+b)/2,b]\). Then \(\F \) is an open cover for each of these subintervals. If \(\F \) has a finite subcover for each of the subintervals then their union would be a finite subcover for \(I_0\). So, there is no finite subcover for at least one of the subintervals; let \(I_1=[a_1,b_1]\) be such a subinterval. Notice that \(b_1-a_1=(b-a)/2\). Repeating this process we get a sequence of subintervals \(I_n=[a_n,b_n]\) with \(b_n-a_n=(b-a)/2^n\) such that
\[a \leq a_1 \leq \cdots \leq a_n \leq a_{n+1} \leq \cdots \leq b_{n+1} \leq b_n \leq \cdots \leq b_1 \leq b\]
and such that \(\F \) does not have a finite subcover for each \(I_n\).
Then \((a_n)_{n\geq 1}\) is an increasing sequence bounded above by \(b\) and so, from the theory of sequences in \(\R \), is convergent. Similarly \((b_n)_{n\geq 1}\) is a decreasing sequence bounded below by \(a\) and so is convergent. However, \(b_n-a_n=(b-a)/2^n\rightarrow 0\) as \(n\rightarrow \infty \) and so \(\lim _{n\rightarrow \infty }a_n=\lim _{n\rightarrow \infty }b_n\); let the common limit be \(\alpha \).
Since \(\alpha \in I_0\) it must lie in some open set \(U\) of the cover \(\F \). Since \(U\) is an open set in the usual topology on \(\R \) it is a neighbourhood of \(\alpha \) and so there is some \(\e >0\) such that \(B_{\e }(\alpha )=(\alpha -\e ,\alpha +\e )\subset U\) (by Definition 2.4). Now, choose \(n\in \N \) such that \(b_n-a_n=(b-a)/2^n<\e \). Then, since \(a_n\leq \alpha \leq b_n\), \(\alpha -a_n<\e \) and \(b_n-\alpha <\e \) so that \(I_n=[a_n,b_n]\subset (\alpha -\e ,\alpha +\e ) \subset U\). This shows that the singleton \(\{\,U\,\} \subset \F \) is a finite subcover of \(\F \) for \(I_n\) contradicting the choice of \(I_n\) as an interval for which \(\F \) does not have a subcover.
This contradiction shows that every open cover for the interval \(I_0=[a,b]\) does have a finite subcover and so \([a,b]\) is compact. □
Proof. This follows from the theorem using Theorem 5.11 and induction. □ class="theoremendmark" >
5.19 Corollary. A closed bounded subset of \(\R ^n\) with the usual topology is compact.
Proof. Suppose that \(X\subset \R ^n\) is closed and bounded. Since \(X\) is bounded there is a real number \(M\) such that \(|\x | \leq M\) for all \(\x \in X\). Then \(|x_i| \leq M\) for \(1\leq i\leq n\) and so \(X \subset [-M,M] \times \cdots \times [-M,M] = [-M,M]^n\). \([-M,M]^n\) is compact by Corollary 5.18. Hence, \(X\) is compact by Proposition 5.9. □
1 .
Prove that if \(A\) is a subset of a topological space \(X\) with the indiscrete topology then \(A\) is a compact subset.
2 .
Prove that if \(K_1\) and \(K_2\) are compact subsets of a topological space \(X\) then so is \(K_1\cup K_2\). Hence prove, by induction that a finite union of compact subsets of \(X\) is compact. Give an example to show that an infinite union of compact subsets need not be compact.
3 .
Prove that all the subsets of \(\R \) are compact in the cofinite topology (the topology of Problems 2, Question 2(b)).
4 .
In the topology on \(\R \) of Problems 2, Question 2(c), prove that
(a) a subset of \(\R \) is compact if it is compact with respect to the usual topology;
(b) the intervals \([a,b)\) (for \(a<b\)) and \([a,\infty )\) are compact;
(c) the intervals \((a,b]\) (for \(a<b)\) and \((-\infty ,b]\) are not compact.
5
. Prove that, given closed non-empty subsets \(A_n\), for \(n\geq 1\), of a topological space \(X\) such that \(A_1 \supset A_2 \supset \cdots
\supset A_n \supset A_{n+1} \supset \cdots \) and \(A_1\) is compact, then the intersection \(\bigcap _{n=1}^{\infty }A_n\) is non-empty.
[Hint: Give a proof by contradiction. Suppose that the intersection is empty and then use the sequence of closed subsets to construct an open cover of \(A_1\).]
6
. Suppose that \(X\) is a compact Hausdorff space with a closed subset \(A\subset X\) and a point \(b\in X\) such that \(b\not \in A\). Prove that
there are disjoint open subsets \(U\) and \(V\) such that \(A\subset U\) and \(b\in V\). A space with this property is called a regular space.
[Hint: Use the method used in proving Proposition 5.8.]
7 . Prove that the continuous bijections constructed in the solutions to Problems 4, Questions 2, 3, 4, 6 and 8 are homeomorphisms.
8 .
(a) Suppose that \(K\subset \R \) is a non-empty compact set in the usual topology (and so closed and bounded). Let \(b = \sup K\), the supremum of \(K\). Use the fact that \(K\) is closed to prove that \(b\in K\). [Recall the definition of the supremum: \(b\) is an upper bound for \(K\) (\(x\leq b\) of all \(x\in K\)) and is the least upper bound. It exists by the completeness property of the real numbers.]
(b) Prove that if a non-empty compact subset \(K\subset \R \) is path-connected then \(K\) is a closed interval \([a,b]\) for some real numbers \(a\) and \(b\) (\(a\leq b\)).
(c) Prove that if \(f\colon X\rightarrow \R \) is a continuous function on a non-empty path-connected compact space \(X\) then \(f(X)=[a,b]\) for some real numbers \(a\) and \(b\) (\(a\leq b\)).
9
. Where does the method of the proof of the Heine-Borel Theorem (Theorem 5.17) fail if the argument is applied to an open covering of an open
interval?
[Hint: See what happens if you apply the argument to the open cover \({\cal F} = \{\,(1/n,1)\mid n\geq 1\,\}\) of the open interval \((0,1)\) in \(\R \) with the usual topology.]
10 . Given a non-compact Hausdorff space \((X,\tau )\) consider the set \(X^* = X \sqcup \{\infty \}\) and the topology
\[\tau ^* = \tau \cup \{X \setminus C \cup \{\infty \} \mid C \subset X \text { compact}\}.\]
Show that \((X^*,\tau ^*)\) is a compact topological space (called the one-point compactification).