Compact Sets

We now consider how a collection of open sets can cover a set $$A$$ as a whole. Such a collection may contain infinitely many sets, and this raises a natural question. Can finitely many members of the collection already cover $$A$?$ Compactness formalizes the sets for which the answer is always yes.

Definition 2.2.1. Let $$M$$ be a metric space and let $A\subseteq M.$ A collection $\lbrace U_{i}\rbrace_{i \in I}$ of subsets of $$M$$ is a cover of $$A$$ if

\[A\subseteq\bigcup_{i \in I}U_{i}.\]

A cover of $$A$$ whose elements are open sets is called an open cover of $$A$.$ A subcollection of a cover of $$A$$ that still covers $$A$$ is called a subcover of $$A$,$ and a subcover of $$A$$ whose index set is finite is called a finite subcover of $$A$.$

Since openness depends on the ambient space, it is natural to ask whether compactness does as well. We therefore keep track of the ambient space in the initial definition and then show that this apparent dependence disappears.

Definition 2.2.2. Let $$M$$ be a metric space and let $A\subseteq N\subseteq M.$ The set $$A$$ is compact in $$N$$ if every cover of $$A$$ whose elements are open in $$N$$ has a finite subcover of $$A$.$

Theorem 2.2.3. Let $$M$$ be a metric space and let $A\subseteq N\subseteq M.$ Then, $$A$$ is compact in $$M$$ if and only if $$A$$ is compact in $$N.$$

Proof. Suppose that $$A$$ is compact in $$M.$$ Let $\lbrace U_{i}\rbrace_{i \in I}$ be a cover of $$A$$ whose elements are open in $$N$.$ By Theorem 2.1.18, there is an open set $$V_i$$ such that $V_i \cap N=U_i$ for each $$i\in I$;$ and since $$A$$ is compact in $$M,$$ we have

\[A\subseteq \bigcup_{k=1}^{n}V_{i_k}\tag{$\ast$}\]

for some finitely many indices $i_1,\dots,i_n\in I.$ Since $A\subseteq N,$ $(\ast)$ implies

\[A\subseteq \bigcup_{k=1}^{n}U_{i_k}.\tag{$\ast\ast$}\]

Therefore $$A$$ is compact in $$N.$$ Conversely, suppose that $$A$$ is compact in $$N$.$ Let $\lbrace V_i\rbrace_{i\in I}$ be a collection of open sets which covers $$A,$$ and let $U_i=V_i\cap N$ for each $i\in I.$ Then $$U_i$$ is open in $$N$$ for each $i\in I,$ and $\lbrace U_i\rbrace_{i\in I}$ covers $$A.$$ Since $$A$$ is compact in $$N$,$ $(\ast\ast)$ holds for some finitely many indices $i_1,\dots,i_n\in I.$ Since $U_i\subseteq V_i$ for each $i\in I,$ $(\ast\ast)$ implies $(\ast).$ Therefore $$A$$ is compact in $$M.$$ $\square$

Thus compactness is intrinsic to the space $$A$:$ it does not depend on whether openness is measured in $$M$$ or in an intermediate subspace $$N$.$ We may therefore drop the reference to the ambient space.

Definition 2.2.4. Let $$M$$ be a metric space and let $$A\subseteq N\subseteq M$.$ Then, $$A$$ is compact if it is compact in $$N$.$ Equivalently, $$A$$ is compact if every cover of $$A$$ whose elements are open in $$N$$ has a finite subcover. In particular, a metric space $$M$$ is said to be compact if it is compact as a subset of itself.

We now look into some relationship between compactness and closedness.

Theorem 2.2.5. Let $$M$$ be a metric space and let $K\subseteq M.$ If $$K$$ is compact then $$K$$ is closed.

Proof. Suppose $$x\in K^c$.$ For $$y\in K$,$ since $$x\ne y$,$ we have $$d(x,y)>0$.$ Let

\[0< r_y < \frac 1 2 d(x,y).\]

Then the collection $\mathcal U=\lbrace B_{r_y}(y) : y\in K\rbrace$ is an open cover of $$K$.$ Since $$K$$ is compact, there are finitely many points $y_1,\dots,y_n\in K$ such that

\[K \subseteq \bigcup_{i=1}^n B_{r_{y_i}}(y_i) = U.\]

Since $$r_{y}<\frac 1 2 d(x,y)$,$ we have $B_{r_{y_i}}(y_i) \cap B_{r_{y_i}}(x) = \varnothing$ for each $$i=1,\dots,n$.$ Therefore, if

\[r = \min \lbrace r_{y_1},\dots,r_{y_n}\rbrace,\]

then $$B_r(x)$$ does not intersect $$U$.$ Since $$K\subseteq U$,$ we have $K \cap B_r(x)=\varnothing.$ It follows that $$B_r(x)\subseteq K^c$,$ and thus $$K^c$$ is open. By Theorem 2.1.14, $$K$$ is closed. $\square$

Remark 2.2.6. The proof illustrates one of the main ways compactness is used in analysis—turning pointwise information into a uniform one. For each point $$x$$ in a set $$A$,$ we choose an open neighborhood $$N(x)$$ on which a desired property holds. The resulting neighborhoods cover $$A$,$ but there may be infinitely many of them, and the corresponding local choices need not combine into a single global conclusion of $$A$.$ It is compactness that removes this difficulty. Once only finitely many local choices remain, we can take their minimum, maximum, or a common bound and thereby pass from local control to global control of the set.

Theorem 2.2.7. Let $$M$$ be a metric space and let $K\subseteq M.$ If $$K$$ is compact, then every closed subset of $$K$$ is compact.

Proof. Let $$F$$ be a closed subset of $$K$,$ and let $\mathcal U=\lbrace U_i\rbrace_{i\in I}$ be an open cover of $$F$.$ Let $$\mathcal V = \mathcal U \cup \lbrace F^c\rbrace$.$ Since each $$U_i$$ and $$F^c$$ are open, $\mathcal V$ is an open cover of $$K$.$ Since $$K$$ is compact, there is a finite subcollection $\mathcal W$ of $\mathcal V$ which covers $$K$.$ Then $\mathcal W$ covers $$F$.$ If $$F^c$$ is a member of $$\mathcal W$,$ we may just need to remove it from $$\mathcal W$,$ and $\mathcal W \setminus \lbrace F^c\rbrace$ is still an open cover of $$F$.$ Hence $$F$$ is compact. $\square$

Corollary 2.2.8. Let $$M$$ be a metric space and let $K,F\subseteq M.$ If $$K$$ is compact and $$F$$ is closed, then $K\cap F$ is compact.

Proof. Since $$K$$ is closed by Theorem 2.2.5 and $$F$$ is closed, $K\cap F$ is closed by (2) of Theorem 2.1.15. Since $K\cap F$ is a closed subset of $$K$,$ Theorem 2.2.7 shows that $K\cap F$ is compact. $\square$

The finite-subcover condition has a dual formulation in terms of intersections of closed sets. To state it explicitly, we introduce the finite intersection property.

Definition 2.2.9. A collection $\{A_i\}_{i\in I}$ is said to have the finite intersection property if every finite subcollection of $\{A_i\}_{i\in I}$ has nonempty intersection.

Theorem 2.2.10. A metric space $$M$$ is compact if and only if every collection of closed subsets of $$M$$ satisfying the FIP has nonempty intersection.

Proof. Suppose that $$M$$ is compact, and let $\lbrace F_i\rbrace_{i\in I}$ be a collection of closed subsets of $$M$$ satisfying the FIP. Let $$U_i=M\setminus F_i$.$ If $\bigcap_{i\in I} F_i = \varnothing,$ then $\lbrace U_i\rbrace_{i\in I}$ is an open cover of $$M$.$ Since $$M$$ is compact, $M = \bigcup_{k=1}^n U_{i_k}$ for some finitely many indices $$i_1,\dots,i_n$.$ It follows that $$\bigcap_{k=1}^n F_{i_k} = \varnothing$,$ which contradicts the FIP. Hence $$\bigcap_{i\in I} F_i \ne \varnothing$.$ Conversely, suppose that every collection of closed subsets of $$M$$ satisfying the FIP has nonempty intersection. Let $\lbrace U_i\rbrace_{i\in I}$ be an open cover of $$M$,$ and put $$F_i = M\setminus U_i$.$ Then every $$F_i$$ is closed in $$M$,$ and $$\bigcap_{i\in I} F_i =\varnothing$.$ Consequently, $\lbrace F_i\rbrace_{i\in I}$ cannot have the finite intersection property. Hence there exists a finite subcollection $\lbrace F_{i_k}\rbrace_{k=1}^{n}$ whose intersection is empty. Taking complements yields $$M \subseteq \bigcup_{k=1}^n U_{i_k}$.$ Therefore $$M$$ is compact. $\square$

The preceding theorem states that a collection of closed subsets of a compact space has a common point whenever it satisfies the FIP. The compactness of the entire ambient space, however, is stronger than necessary: it suffices for one member of the collection to be compact.

Theorem 2.2.11. Let $$M$$ be a metric space. If $\{F_i\}_{i\in I}$ is a collection of closed subsets of $$M$$ satisfying the FIP, and $F_{i_0}$ is compact for some $$i_0\in I$,$ then $\bigcap_{i\in I}F_i$ is nonempty.

Proof. Define $K_i=F_i\cap F_{i_0}$ for each $$i\in I$.$ Then, we have

\[\bigcap_{k=1}^nK_{i_k} = F_{i_1}\cap\cdots\cap F_{i_n} \cap F_{i_0}\]

for any finitely many indices $$i_1,\dots,i_n\in I$.$ Since $\{F_i\}_{i\in I}$ has the FIP, the right-hand side is nonempty. Therefore $\{K_i\}_{i\in I}$ has the FIP. Since $F_{i_0}$ is compact and $$K_i$$ is closed in $$F_{i_0}$,$ Theorem 2.2.10 now gives $$\bigcap_{i\in I}K_i\neq\varnothing$.$ Since

\[\bigcap_{i\in I}K_i = \lowparen{\bigcap_{i\in I}F_i}\cap F_{i_0} =\bigcap_{i\in I}F_i,\]

we finally have $$\bigcap_{i\in I}F_i \ne \varnothing$.$ $\square$

Corollary 2.2.12. Let $$M$$ be a metric space. If $$(K_n)$$ is a sequence of nonempty compact subsets of $$M$$ such that $K_{n+1}\subseteq K_n$ for $n=1,2,3,\dots$ , then $\bigcap_{n=1}^{\infty} K_n$ is nonempty.

Proof. Let $\{K_{n_1},\dots,K_{n_k}\}$ be a finite subcollection, and put

\[N=\max\{n_1,\dots,n_k\}.\]

Since $$K_{n+1}\subseteq K_n$,$ we have

\[\bigcap_{j=1}^{k} K_{n_j} = K_{N} \ne \varnothing.\]

Hence $\{K_n\}_{n\in\mathbb N^{+}}$ is a collection of closed sets satisfying the FIP. Since $$K_1$$ is compact, Theorem 2.2.11 now gives

\[\bigcap_{n=1}^{\infty} K_n = \bigcap_{n\in \mathbb N^{+}\!\!} K_n\ne \varnothing.\tag*{$\square$}\]

This result assumes compactness from the outset. For closed intervals in $$\mathbb R$,$ however, compactness has not yet been established. We therefore prove the next intersection theorem directly from the least-upper-bound property of $$\mathbb R$.$

Theorem 2.2.13 (Nested intervals theorem). If $$(I_n)$$ is a sequence of bounded closed intervals in $\mathbb R$ such that $I_{n+1}\subseteq I_n,$ then $\bigcap_{n=1}^{\infty} I_n$ is nonempty.

Proof. Let $$I_n = [a_n,b_n]$$ and let $$A$$ be the set of all $$a_n$.$ Then, $$A$$ is nonempty and bounded above by $$b_1$.$ Let $\alpha$ be the supremum for $$A$.$ If $m,n\in \mathbb N^{+}$ then we have

\[a_n \le a_{m+n} < b_{m+n} \le b_m,\]

so that $$b_m$$ is an upper bound for $$A$.$ Therefore $a_m\le \alpha \le b_m$ for every $$m$,$ so that $$\alpha \in \bigcap_{n=1}^{\infty} I_n$.$ $\square$

We next extend this theorem to $$\mathbb R^k$.$ The appropriate higher-dimensional analogue of a closed interval is a $$k$-$ cell.

Definition 2.2.14. Let $k\in \mathbb N^{+}$ and let $$a_i$$ and $$b_i$$ be reals such that $$a_i<b_i$$ for $$i=1,\dots,k$.$ The set of all points $\mathbf x = (x_1,\dots,x_k)\in \mathbb R^k$ whose coordinates satisfy $a_i\le x_i\le b_i$ is called a $$k$-$ cell.

Theorem 2.2.15. Let $$k\in \mathbb N^{+}$.$ If $$(I_n)$$ is a sequence of $$k$-$ cells such that $$I_{n+1}\subseteq I_n$,$ then $\bigcap_{n=1}^{\infty} I_n$ is nonempty.

Proof. Let $$I_n$$ consist of all points $\mathbf x=(x_1,\dots,x_k)$ such that $x_i \in I_{n,i}$ where $1\le i\le k$ and $$I_{n,i}=[a_{n,i}, b_{n,i}]$.$ By means of Theorem 2.2.13, we may let $\mathbf c_i$ be a point of $\bigcap_{n=1}^{\infty} I_{n,i}$ for each $$i$.$ If $$\mathbf c = (c_1,\dots,c_k)$,$ then $$\mathbf c \in \bigcap_{n=1}^{\infty} I_n$.$ $\square$

We now prove a fundamental theorem of compactness in Euclidean space: Every cell is compact. The preceding nested-cell theorem provides the key step in a bisection argument for compactness.

Theorem 2.2.16. Every $$k$-$ cell is compact.

Proof. Let $$I$$ be a $$k$-$ cell, consisting of all points $\mathbf x=(x_1,\dots,x_k)\in \mathbb R^k$ such that $x_i\in [a_i, b_i]$ for $$1\le i\le k$.$ Let $\mathbf a=(a_1,\dots,a_k)$ and $$\mathbf b=(b_1,\dots,b_k)$,$ and define $$\delta = \|\mathbf a-\mathbf b\|$.$ Then, we have $\|\mathbf x-\mathbf y\|\le \delta$ for every $$\mathbf x,\mathbf y\in I$.$ Suppose, for contradiction, that there exists an open cover $\mathcal U$ of $$I$$ that has no finite subcover of $$I$.$ Put $$c_i = (a_i+b_i)/2$,$ then the intervals $$[a_i,c_i]$$ and $$[c_i,b_i]$$ determine $$2^k$$ $$k$-$ cells, whose union is $$I$.$ At least one of these sub-cells, call it $$I_1$,$ cannot be covered by any finite subcollection of $\mathcal U$ (If every sub-cell had a finite subcover, the union of these $$2^k$$ finite subcovers would be a finite subcover of $$I$).$ We subdivide $$I_1$$ and continue the process. Then we obtain a sequence $$(I_n)$$ with the following properties:

$I\supseteq I_1 \supseteq I_2 \supseteq \cdots$ ;
$$I_n$$ is not covered by any finite subcollection of $$\mathcal U$;$
if $\mathbf x,\mathbf y\in I_n$ then $$\| \mathbf x-\mathbf y\|\le 2^{-n}\delta$.$

By Theorem 2.2.15, there is a point $\mathbf c$ which lies in every $$I_n$.$ Then, there exists $U_0\in \mathcal U$ that contains $$\mathbf c$.$ Since $$U_0$$ is open, there exists $$r>0$$ such that $$B_r(\mathbf c)\subseteq U_0$.$ If $$n$$ is so large that $$2^{-n}\delta<r$,$ then $$I_n\subseteq B_r(\mathbf c)\subseteq U_0$,$ which contradicts property (2). $\square$

We next examine the relation between having a limit point and compactness. Note that the local-to-global principle appears again in the proof of the following theorem: compactness convert the local finiteness of $$A$$ into the global finiteness of $$A$.$

Theorem 2.2.17. If $$A$$ is an infinite subset of a compact set $$K$,$ then $$A$$ has a limit point in $$K$.$

Proof. If no point of $$K$$ is a limit point of $$A$,$ then every $x\in K$ would have an open ball $B_{r_x}(x)$ that contains at most one point of $$A$.$ These balls form an open cover of $$K$.$ Since $$K$$ is compact, there exists $x_1,\dots,x_n$ such that

\[K\subseteq\bigcup_{i=1}^n B_{r_{x_i}}(x_i).\]

Since $A\subseteq K$ and each ball contains at most one point of $$A$,$ the set $$A$$ has at most $$n$$ points. This contradicts the assumption that $$A$$ is infinite. $\square$

The theorem states that every infinite subset of a compact space must accumulate somewhere. From this topological consequence, we derive a corollary about Euclidean space: In Euclidean space, boundedness places a set inside a compact cell, so every bounded infinite set inherits the accumulation property.

Corollary 2.2.18 (Bolzano–Weierstrass theorem). Every bounded infinite subset of $\mathbb R^k$ has a limit point in $$\mathbb R^k$.$

Proof. Let $$A$$ be a bounded infinite subset of $$\mathbb R^k$.$ Since $$A$$ is bounded, some $$k$-$ cell $$I$$ contains $$A$.$ Since $$I$$ is compact by Theorem 2.2.16, Theorem 2.2.17 shows that $$A$$ has a limit point in $$I$.$ $\square$

In Euclidean space, the somewhat abstract notion of compactness can be characterized geometrically by closedness and boundedness.

Definition 2.2.19. Let $$M$$ be a metric space, and let $$A\subseteq M$.$ The set $$A$$ is said to be bounded if there exists $$R>0$$ such that

\[d(x,y) < R\]

for all $$x,y\in A$.$

Theorem 2.2.20 (Heine–Borel theorem). Let $$A\subseteq \mathbb R^k$.$ The set $$A$$ is compact if and only if $$A$$ is closed and bounded.

Proof. Suppose that $$A$$ is compact. Fix $$\mathbf p\in \mathbb R^k$.$ Then, $\lbrace B_n(\mathbf p): n\in \mathbb N^{+}\rbrace$ is an open cover of $$A$,$ and a finite subcover places $$A$$ inside $B_{n_0}(\mathbf p)$ for sufficiently large $$n_0\in \mathbb N$.$ It follows that $\|\mathbf x - \mathbf y\|<2n_0$ for all $$\mathbf x,\mathbf y\in A$.$ Thus $$A$$ is bounded. Also, it is clear that $$A$$ is closed, by Theorem 2.2.5. Conversely, suppose that $$A$$ is closed and bounded, and let $$\mathbf x=(x_1,\dots,x_k)\in A$.$ Since $$A$$ is bounded, for each $$i$,$ there exists a closed interval $$[a_i,b_i]$$ that contains every $$x_i$.$ Thus, we have $A\subseteq I$ for some $$k$-$ cell $$I$.$ Since $$A$$ is closed, Theorem 2.2.16 and Theorem 2.2.7 show that $$A$$ is compact. $\square$

References

Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.

External

2.2. Compact Sets

References