3.4 Compact Sets · Connectedness and Compactness

[3.4.1]Definition(Covering)#

A collection of subsets $\mathcal{A}$ of $X$ is said to be a cover for $X$ if the union of each set in $\mathcal{A}$ equals $X$ . $\mathcal{A}$ is said to be an open covering if each element of $\mathcal{A}$ is an open subset of $X$ .

Let $Y \subseteq X$ . Then a collection of subsets $\mathcal{A}$ of $X$ is said to be a cover for $Y$ if $Y$ is contained in the union of each set in $\mathcal{A}$ .

[3.4.2]Definition(Compact Set)#

Suppose $X$ is a topological space. Then $C \subseteq X$ is said to be compact if every open cover of $C$ has a finite subcover.

[3.4.3]Theorem(Compactness of Subspace)#

Suppose $X$ is a topological space, and $Y \subseteq X$ is a subspace. Then $Y$ is compact if and only if, for every open cover of $X$ that covers $Y$ , there is a finite subcover that also covers $Y$ .

Proof.

$(\Longrightarrow)$ : Suppose $\mathcal{A}$ is a collection of open subsets of $X$ that covers $Y$ . Then note $\mathcal{A} \cap Y$ , the intersection of each set in $\mathcal{A}$ by $Y$ , forms an open cover of $Y$ . Thus there is a finite subcover $\mathcal{B} \cap Y$ that covers $Y$ . Then $\mathcal{B}$ is a finite collection of open subsets of $X$ that covers $Y$ .

$(\Longleftarrow)$ : Let $\mathcal{B}$ be an open cover for $Y$ . Each $B \in \mathcal{B}$ can be written as $B = A \cap Y$ , where the set of all $A$ 's forms a collection of open subsets $\mathcal{A}$ of $X$ that covers $Y$ . There is a finite subcover of these that, when intersected with $Y$ , forms a finite subcover for $\mathcal{B}$ , completing the proof.

[3.4.4]Theorem(Closed Subspace of Compact Space is Compact)#

Any closed subspace of a compact space is compact.

Proof.

Let $X$ be a compact space, and let $Y \subseteq X$ be closed. Let $\mathcal{A}$ be an open cover of $Y$ . Then note the collection defined by $\mathcal{B} = \mathcal{A} \cup \{X \setminus Y\}$ is an open cover of $X$ ( $X \setminus Y$ is open, since $Y$ is closed). Since $X$ is compact, there is a finite subcover $B$ of $X$ from $\mathcal{B}$ . Then $B \setminus \{X \setminus Y\}$ is a finite subcover of $\mathcal{A}$ , completing the proof.

[3.4.5]Lemma(Separation of a Point and a Compact Set)#

Suppose $X$ is a Hausdorff space and $Y \subseteq X$ is compact. If $x_0 \notin Y$ , then there are disjoint open neighborhoods containing $x_0$ and $Y$ .

Proof.

Fix $x_0 \in X \setminus Y$ —we seek to find disjoint open neighborhoods containing $x_0$ and $Y$ . For each $y \in Y$ , it is obviously true that $x_0 \ne y$ . Thus, by Hausdorffness of $X$ , note that there are disjoint open neighborhoods $U_y \subseteq X$ of $x_0$ and $V_y \subseteq X$ of $y$ . Then note $\{V_y\}_{y \in Y}$ forms a cover of $Y$ by open sets from $X$ , and by compactness of $Y$ , we have that there exists a finite subcover $\{V_{y_k}\}_{k = 1}^n$ of these open sets for $Y$ . Thus we may write $Y \subseteq \bigcup_{k = 1}^n V_{y_k}$ .

Let $U = \bigcap_{k = 1}^n U_{y_k}$ and $V = \bigcup_{k = 1}^n V_{y_k}$ , which are open subsets of $X$ containing $x_0$ and $Y$ , respectively. Then any $v \in V$ belongs to $V_{y_k}$ for some $k \in \{1, 2, \dots, n\}$ . By how $V_{y_k}$ was chosen, it follows that $v \notin U_{y_k}$ , and so $v \notin U$ . Thus, $U$ and $V$ are disjoint open neighborhoods containing $x_0$ and $Y$ .

[3.4.6]Theorem(Compact Subspace of Hausdorff Space is Closed)#

Any compact subspace of a Hausdorff space is closed.

Proof.

Suppose $X$ is a Hausdorff space, and let $Y \subseteq X$ be compact. We will show $Y$ is closed by showing $X \setminus Y$ is open. Fix $x_0 \in X \setminus Y$ . By the preceding lemma, note that there are disjoint open neighborhoods $U$ and $V$ containing $x_0$ and $Y$ , respectively. Since $Y \subseteq V$ and $U$ and $V$ are disjoint, it follows that $U$ cannot intersect $Y$ . Thus $U \cap Y = \emptyset$ , and so $U \subseteq X \setminus Y$ . Thus $x_0 \in U \subseteq X \setminus Y$ , implying $X \setminus Y$ is open and $Y$ is closed.

[3.4.7]Theorem(Continuous Image of Compact Set is Compact)#

The continuous image of a compact set is compact.

Proof.

Suppose $X$ is a compact set, and $f: X \to Y$ is continuous. Let $\mathcal{A}$ be a cover for $f(X)$ with open subsets of $Y$ . Since the preimage of an open set is open, the preimage $f^{-1}(\mathcal{A})$ is an open cover of $X$ . By compactness, there is a finite subcover $\{f^{-1}(A_k)\}_{k = 1}^n$ with open sets of $X$ . Then $f\left( \{f^{-1}(A_k)\}_{k = 1}^n \right) = \{A_k\}_{k = 1}^n$ forms a finite subcover of $f(X)$ , completing the proof.

[3.4.8]Theorem#

Suppose $f: X \to Y$ is a bijective, continuous function. If $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.

Proof.

To show $f$ is a homeomorphism, we show $f$ sends closed sets to closed sets. Let $O \subseteq X$ be closed, and thus consequently compact since all closed subsets of a compact space is compact. Thus $f(O) \subseteq Y$ is closed, and thus consequently closed since all compact subsets of a Hausdorff space are closed. Thus $f$ sends closed sets to closed sets.

[3.4.9]Lemma(Tube Lemma)#

Suppose $X, Y$ are topological spaces, with $Y$ compact. Fix $x_0 \in X$ . If $N \subseteq X \times Y$ is an open set containing $\{x_0\} \times Y$ , then $N$ contains some tube $W \times Y$ about $\{x_0\} \times Y$ , where $W \subseteq X$ is an open neighborhood of $x_0$ .

Proof.

Fix $x_0 \in X$ and consider the set $\{x_0\} \times Y \subseteq X \times Y$ . Let $N \subseteq X \times Y$ be an open neighborhood of $\{x_0\} \times Y$ . For each $y \in Y$ , choose some open neighborhoods $U_y$ of $x_0$ and $V_y$ of $y$ such that $(x_0, y) \in U_y \times V_y \subseteq N$ . Then $\{U_y \times V_y\}_{y \in Y}$ forms an open cover for $\{x_0\} \times Y$ . Note that $\{x_0\} \times Y \cong Y$ , and so the left-hand side is compact, and so there is a finite subcover $\{U_k \times V_k\}_{k = 1}^n$ of $\{x_0\} \times Y$ . Define

W = \bigcap_{k = 1}^n U_k,

an open neighborhood in $X$ about $x_0$ . We will show $W \times Y \subseteq N$ , completing the proof.

Let $(x, y) \in W \times Y$ be arbitrary. Then note $(x_0, y) \in \{x_0\} \times Y$ , with $y$ being the same as chosen from $W \times Y$ . Thus there is some $j \in \{1, 2, \dots, n\}$ such that $y \in V_j$ , reason being that $\{U_k \times V_k\}_{k = 1}^n$ is a finite subcover for $\{x_0\} \times Y$ . Recall that $x \in U_j$ as well, since $x \in W$ , and so $(x, y) \in (U_j \times V_j) \subseteq N$ . Thus $W \times Y \subseteq N$ , completing the proof.

[3.4.10]Theorem(Finite Product of Compact Spaces is Compact)#

The product of finitely many compact spaces is compact.

Proof.

Suppose $X, Y$ are compact spaces, and fix $x \in X$ . Let $\mathcal{A}$ be an open cover for $X \times Y$ —we show it has a finite subcover.

Consider the slice $\{x_0\} \times Y$ : note that it is homeomorphic to the compact space $Y$ , so there is a finite subcover $\{A_k\}_{k = 1}^m$ for $\{x_0\} \times Y$ . Let $N = \bigcup_{k = 1}^m A_k$ , an open set in $X \times Y$ containing $\{x_0\} \times Y$ . By the Tube Lemma, there is some open neighborhood $W_x \subseteq X$ of $x$ such that $W_x \times Y \subseteq N$ . Thus $\{A_k\}_{k = 1}^m$ forms a finite cover for $W_x \times Y$ . Finally, note $\{W_x\}_{x \in X}$ is an open cover of $X$ . By compactness of $X$ , we have that $\{W_k\}_{k = 1}^n$ forms a finite subcover for $X$ . Thus we have that the union of the tubes

W_1 \times Y, W_2 \times Y, \dots, W_n \times Y

is all of $X \times Y$ ; since each tube has a finite cover of open sets from $\mathcal{A}$ , we have that $X \times Y$ is covered by finitely many sets from $\mathcal{A}$ , completing the proof.

The proof for more compact sets proceeds by induction on the number of sets.

[3.4.11]Definition(Finite Intersection Property)#

A collection of sets $\mathcal{C}$ is said to have the finite intersection property if, for every finite subcollection $\{C_k\}_{k = 1}^n$ of $\mathcal{C}$ , $\bigcap_{k = 1}^n C_k \ne \emptyset$ .

[3.4.12]Theorem(Compactness in Terms of Finite Intersection Property)#

Let $X$ be a topological space. Then $X$ is compact if and only if, for every collection $\mathcal{C}$ of closed subsets of $X$ that have the finite intersection property, the intersection $\bigcap_{C \in \mathcal{C}} C$ of all the elements of $\mathcal{C}$ is nonempty.

Proof.

Given a collection $\mathcal{A}$ of subsets of $X$ , we may define

\mathcal{C} = \{X \setminus A: A \in \mathcal{A}\}.

Then the following statements are true:

$\mathcal{A}$ is a collection of open subsets of $X$ if and only if $\mathcal{C}$ is a collection of closed subsets of $X$ .
$\mathcal{A}$ forms a cover for $X$ if and only if $\bigcap_{C \in \mathcal{C}} C = \emptyset$ .
$\{A_k\}_{k = 1}^n$ forms a finite subcover for $X$ if and only if $\bigcap_{k = 1}^n C_k = \emptyset$ , where $C_k = X \setminus A_k$ .

The first statement is by definition of a closed set, and the other two are a direct application of De Morgan's Laws. Namely that

X \setminus \left(\bigcup_{\lambda \in \Lambda} A_\lambda\right) = \bigcap_{\lambda \in \Lambda} (X \setminus A_\lambda).

Consider the original definition of compactness: $X$ is compact if and only if, for any open cover $\mathcal{A}$ , there is a finite subcover. Thus if $\mathcal{A}$ has no finite subcover for $X$ , then it follows that $\mathcal{A}$ does not cover the compact set $X$ . Now let $\mathcal{C} = \{X \setminus A: A \in \mathcal{A}\}$ as before. Immediately we see that compactness may be rephrased as such:

“We say that $X$ is compact if, for every collection of closed subsets $\mathcal{C}$ of $X$ with the finite intersection property, the intersection of all elements in $\mathcal{C}$ is nonempty.”

[3.4.13]Remark(Cantor's Intersection Theorem)#

Suppose $\{C_k\}_{k \in \bZ_+}$ is a sequence of closed subsets of a compact space $X$ , such that $C_k \supseteq C_{k + 1}$ for each $k \in \bZ_+$ . Immediately we see that any finite subcollection of $\{C_k\}$ has nonempty intersection, meaning this sequence has the finite intersection property. By compactness, we have that $\bigcap_{C \in \mathcal{C}} C \ne \emptyset$ . This result is familiar to us with $X = \bR$ .

[3.4.14]Recap#

For any topological space $X$ , a collection of subsets $\mathcal{A}$ of $X$ is said to cover $X$ if $\bigcup_{A \in \mathcal{A}} A = X$ . An open cover is a cover comprised of open sets. A subcover of $X$ is a subset of $\mathcal{A}$ that still manages to cover $X$ .
A space $X$ is said to be compact if, for every open cover $\mathcal{A}$ , there is a finite subcover of $X$ .
For a subspace $Y \subseteq X$ , a collection of subsets $\mathcal{A}$ of $X$ is said to cover $Y$ if $Y \subseteq \bigcup_{A \in \mathcal{A}} A$ . This is different from the traditional definition of a cover for one space.
A space $Y \subseteq X$ is compact if and only if, for every open cover of $Y$ from open subsets of $X$ , there is a finite subcover.
Intuition.
The idea behind this proof is that an open set in $Y$ is an open set in $X$ intersected with $Y$ . Thus open covers can be moved around seamlessly between $X$ and $Y$ : simply move an open cover to the space that is given as compact by hypothesis, take the finite subcover, then move it back to the other space.
Every closed subspace of a compact space is compact.
Intuition.
Let $X$ be compact and $Y \subseteq X$ closed. Of course, $X \setminus Y$ is open. For any open cover of $Y$ , we may adjoin $X \setminus Y$ to form an open cover for $X$ , use its compactness, then simply get rid of the $X \setminus Y$ term to get a finite subcover of $Y$ .
Every compact subspace of a Hausdorff space is closed.
Intuition.
Let $X$ be Hausdorff and $Y \subseteq X$ compact. We want to show $X \setminus Y$ is open. Fixing $x_0 \in X \setminus Y$ , we show the set is open by showing there exists an open neighborhood $U \subseteq X$ of $x_0$ lying in $X \setminus Y$ —that is, $U \cap Y = \emptyset$ . For each $y \in Y$ , we have that $y \ne x_0$ , so by Hausdorffness there exist disjoint open neighborhoods $U_y \subseteq X$ of $x_0$ and $V_y \subseteq X$ of $y$ . Then note $\{V_y\}_{y \in Y}$ forms an open cover for the compact space $Y$ , so there exists a finite subcover $\{V_{y_k}\}_{k = 1}^n$ of $Y$ . Then define $U = \bigcap_{k = 1}^n U_{y_k},$ an open neighborhood of $x_0$ . For any $y \in Y$ , there is some $k \in \{1, \dots, n\}$ such that $y \in V_{y_k}$ , implying $y \notin U_{y_k}$ , so $y \notin U$ .
We extract the key mechanism from the last proof: for any compact subspace $Y$ of a Hausdorff space $X$ , we can always construct disjoint open neighborhoods $U, V$ about $x_0, Y$ respectively—that is, $x_0 \in U \subseteq X \setminus Y$ , meaning $X \setminus Y$ is open.
The continuous image of a compact set is compact.
Intuition.
Let $f: X \to Y$ be a continuous map from the compact space $X$ to a target space $Y$ . Let $\mathcal{A}$ be an open covering for $f(X)$ . Then note $f^{-1}(\mathcal{A})$ is an open covering of $X$ by continuity of $f$ . Since $X$ is compact, we have a finite subcover $\{f^{-1}(A_k)\}_{k = 1}^n$ . Pushing forward yields that $f(X)$ is covered by the finite subcover $\{A_k\}_{k = 1}^n$ .
Let $f: X \to Y$ be a bijective, continuous function. If $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.
Intuition.
We show $f^{-1}$ is continuous by showing $f$ takes closed sets to closed sets. For some closed set $C \subseteq X$ , we note $C$ is also compact by a previous theorem. Since $f$ is continuous, it follows that $f(C) \subseteq Y$ is compact in a Hausdorff space, which makes $f(C)$ closed.
Let $X, Y$ be topological spaces, with $Y$ compact. For fixed $x_0 \in X$ and any open neighborhood $N \subseteq X \times Y$ of $\{x_0\} \times Y$ , there is some open neighborhood $W \subseteq X$ of $x_0$ such that $W \times Y \subseteq N$ .
Intuition.
Fix $x_0 \in X$ and let $N \subseteq X \times Y$ be an open neighborhood of $\{x_0\} \times Y$ . Form a cover for $\{x_0\} \times Y$ , of the form $\{U_y \times V_y\}_{y \in Y}$ , where $U_y \times V_y$ is a basic, open neighborhood about $(x_0, y)$ . Note that $\{x_0\} \times Y \cong Y$ , and since $Y$ is compact, we form an open cover $\{U_k \times V_k\}_{k = 1}^n$ . Then we define $W = \bigcap_{k = 1}^n U_k$ , an open neighborhood of $x_0$ . We show $\{U_k \times V_k\}_{k = 1}^n$ covers not only $\{x_0\} \times Y$ , but also $W \times Y$ . Let $(x, y) \in W \times Y$ be arbitrary, and let $(x_0, y) \in \{x_0\} \times Y$ be the corresponding element with the same $y$ -coordinate. Thus $y \in V_j$ for some $j \in \{1, \dots, n\}$ , and since $x \in U_k$ for every $k$ , $x \in U_j$ . Thus $(x, y) \in (U_j \times V_j) \subseteq N$ , completing the proof.
The product of finitely many compact sets is compact.
Intuition.
We prove this for two compact topological spaces $X, Y$ , and the full theorem is proven in a straightforward manner by induction. Suppose $\mathcal{A}$ is a cover of $X \times Y$ . Fix $x \in X$ , then note $\{x\} \times Y \cong Y$ is compact, so there is some finite subcover $\{A_k\}_{k = 1}^m$ of sets from $\mathcal{A}$ . Then $N = \bigcup_{k = 1}^m A_k$ is an open subset of $X \times Y$ containing $\{x\} \times Y$ . By the Tube Lemma, there is some open neighborhood $W_x \subseteq X$ of $x$ such that $W_x \times Y \subseteq N$ . Then $\{W_x\}_{x \in X}$ forms an open cover for $X$ , and since $X$ is compact, there is a finite subcover $\{W_k\}_{k = 1}^n$ . Thus
$\{W_1 \times Y, W_2 \times Y, \dots, W_n \times Y\}$
forms an open cover for $X \times Y$ . Each element of this set also has a finite subcover of elements from $\mathcal{A}$ , thus the entirety of $X \times Y$ has a finite subcover from elements of $\mathcal{A}$ .
A topological space $X$ is compact if, for every collection of closed subsets $\mathcal{C}$ with the finite intersection property, the intersection $\bigcap_{C \in \mathcal{C}} C \ne \emptyset$ .
Intuition.
Observe that, for any collection of subsets $\{\mathcal{A}_\lambda\}_{\lambda \in \Lambda}$ of $X$ , it is true that
$X \setminus \left(\bigcup_{\lambda \in \Lambda} A_\lambda\right) = \bigcap_{\lambda \in \Lambda} (X \setminus A_\lambda).$
Defining $\mathcal{C} = \{X \setminus A: A \in \mathcal{A}\}$ , where $\mathcal{A}$ is an open cover of $X$ , and reformulating compactness in terms of $\mathcal{C}$ and the relation above yields this theorem.

[3.4.15]Problem#

Let $\Tau, \Tau'$ be topologies on $X$ such that $\Tau \subseteq \Tau'$ . What does compactness of $X$ in one topology say about the other?
Show that if $X$ is compact and Hausdorff under topologies $\Tau, \Tau'$ , the topologies are either equal or incomparable.

Solution.

If $X$ is compact in $\Tau'$ , then all open covers of $X$ have finite subcovers—necessarily, this includes covers formed by sets in $\Tau$ , and so $X$ is compact in $\Tau$ . No statement can be made in the reverse direction.
Suppose the two topologies are comparable: without loss of generality, say $\Tau \subseteq \Tau'$ . Then define the map
$f: (X, \Tau') \to (X, \Tau) \quad f(x) = x.$
Note $f$ is a bijective, continuous map. Notably, it sends closed sets to closed sets: if $C \subseteq X$ is closed, then it is compact, and since $f$ is continuous, $f(C)$ is also compact (and thus closed since the codomain is Hausdorff). Thus $f$ is a homeomorphism and so $\Tau = \Tau'$ . Thus the topologies are either equal or incomparable.

[3.4.16]Problem#

Show that every compact subspace of a metric space is closed and bounded (with respect to the metric). Give an example of a closed and bounded subset of a metric space that is not necessarily compact.

Solution.

Suppose $(X, d)$ is a metric space in its metric topology, and suppose $C \subseteq X$ is compact. Note that the metric topology is Hausdorff, and since $C$ is a compact subset of a Hausdorff space, it is closed. To show it is bounded, let $\mathcal{A}$ be an open cover of $C$ with basic, open subsets from $X$ . Then we may take a finite subcover $\{A_1, \dots, A_n\}$ . Finitely many open balls can be contained in exactly one ball of finite radius, which means $C$ is bounded.

Consider the metric space $\bQ$ together with metric $d(x, y) = |x - y|$ . Then note that $[0, \sqrt2) \subseteq \bQ$ is closed and bounded. Then consider the open cover $\mathcal{A} = (-1, 1) \cup \{(0, \sqrt{2} - 1/n)\}_{n \in \bZ_+}$ , which has no finite subcover.

[3.4.17]Problem#

Suppose $A, B$ are disjoint, compact subspaces of the Hausdorff space $X$ . Show that there exist disjoint open sets $U, V$ such that $A \subseteq U$ and $B \subseteq V$ .

Solution.

Note that in a Hausdorff space, a compact space and a point outside of it can be separated by two disjoint open neighborhoods. Namely, for each $x \in A$ , there are disjoint open neighborhoods $U_x \subseteq X$ of $x$ and $V_x \subseteq X$ of $B$ . Then note $\{U_x\}_{x \in A}$ forms an open cover of $A$ , and by compactness of $A$ , we have a finite subcover $\{U_k\}_{k = 1}^n$ . Then simply define

U = \bigcup_{k = 1}^n U_k \quad V = \bigcap_{k = 1}^n V_k,

which are both open open sets containing $A$ and $B$ respectively. For each $u \in U$ , there is some $j \in \{1, \dots, n\}$ such that $u \in U_j$ , meaning $u \notin V_j$ and so $u \notin V$ . Thus $U, V$ are disjoint, completing the proof.

3.4Compact Sets

3.4.1Exercises#