Date modified: Mon 2023-07-03 . 01:31 PM
Date created: Sat 2023-10-07 . 09:02 PM
# Characterization of compactness in metric spaces. In a metric space, we have nice notions of sequences and convergence. We can use this to characterize compactness. > Characterization of compactness in metric spaces. > In a metric space $X$, a subset $K$ is (covering) compact if and only if it is sequentially compact. Namely, every sequence in $K$ has a convergent subsequence with limit in $K$. Let us recall its proof here. $\blacktriangleright$ $(\implies)$ Suppose $K$ is compact. Let $(a_n)$ be a sequence in $K$. Suppose to the contrary that every subsequence of $(a_n)$ do not converge in $K$. That is, each $x\in K$ is not the limit of any subsequence of $(a_n)$. Here we note that the sequence $(a_n)$ must have infinite range $R\subset K$, otherwise we can extract a convergent subsequence. For each $x\in K$, we construct open ball $B(x,r_x)$ such that the punctured ball $B(x,r_x)-\{x\}$ avoid $R$. This is possible. If this is not possible, by shrinking these balls we can produce a subsequence $(a_{n_K})$ of $(a_n)$ that converges to $x$. These open balls $B(x,r_x)$ then forms an open cover for $K$, which we can refine to a finite subcover. Say $K \subset B(x_1,r_{x_1})\cup \cdots\cup B(x_k,r_{x_k})$. But the range of the sequence $R$ lies in $K$, which lies in the union of these finitely many open balls. One of them must intersects $R$ infinitely. But each open ball was constructed to intersect $R$ in at most one point! A contradiction. $\blacksquare$ For the other direction, we establish two lemmas of sequential compactness. > Totally boundedness. > If $K$ is sequentially compact, then $K$ is totally bounded. Namely, for every positive $r > 0$, one can cover $K$ with finitely many open balls of radius $r$. $\blacktriangleright$ Let $K$ be sequentially compact and fix $r > 0$. Suppose not, that $K$ is not totally bounded. Then for some $x_1\in K$, the open ball $B(x_1,r)$ does not cover $K$. So take $x_2 \in K-B(x_1,r)$. The union $B(x_1,r)\cup B(x_2,r)$ also does not cover $K$. So we can produce $x_3 \in K-(B(x_1,r)\cup B(x_2,r))$. Repeat this procedure to get a sequence $(x_n)$ where each $x_n$ is not in the union of the open balls of radius $r$ centered at $x_1,\ldots,x_{n-1}$. In particular the pairwise distance between two points in this sequence is at least $r$. Now as $K$ is sequentially compact, we have some subsequence $(x_{n_k})$ that converges to some $x\in K$. Which means there are infinitely many such terms $x_{n_k}$ that are within $\frac{r}{4}$ of $x$, which in turn mean they are within $\frac{r}{2}$ of each other. Yet they are $r$ far away from each other, contradiction. $\blacksquare$ > Lebesgue number lemma. > If $K$ is sequentially compact, then for every open cover $\{U_{\lambda}\}_{\lambda \in\Lambda}$ of $K$, there exists a positive number $\delta > 0$ such that for every open ball $B(x,\delta)$ for $x\in K$, there exists some member $U_{\lambda}$ that contains $B(x,\delta)$. This number $\delta$ is sometimes called the Lebesgue number. $\blacktriangleright$ Take $K$ to be our whole space. Suppose no such $\delta > 0$ exist. Then for each $n$, there exists some $x_{n} \in K$ where the open ball $B(x_{n}, \frac{1}{n})$ is not covered by any single member $U_{\lambda}$ of the open cover. As $K$ is sequentially compact, we have a convergent subsequence $(x_{n_{k}})$ with limit $x\in K$. Hence $x$ lies in some one single $U_{\lambda}$ in the open cover of $K$. But $x$ is an interior point of $U_{\lambda}$, so we have some $r > 0$ such that $x\in B(x,r)\subset U_{\lambda}$. And as $x_{n_{k}}\to x$, there exists $k$ large enough such that $n_{k} > \frac{2}{r}$ with $d(x,x_{n_{k}}) < \frac{r}{2}$. This means the open ball $B(x_{n_{k}}, \frac{1}{n_{k}})$ lies in $B(x,r)$ which in turn lies in $U_{\lambda}$, a contradiction. $\blacksquare$ We now put everything together to show sequential compactness implies covering compactness. $\blacktriangleright$ $(\impliedby)$ Suppose $K$ is sequentially compact, and let $\{U_{\lambda}\}$ be an open cover for $K$. Then by lemma, we have some Lebesgue number $\delta > 0$ where every open ball $B(x,\delta)$ is contained in one $U_{\lambda}$ in the open cover, for $x \in K$. As $K$ is totally bounded, we can cover $K$ with finitely many such open $\delta$-balls, say $B(x_{1},\delta)\cup\cdots\cup B(x_{n},\delta)$. But each of these open balls is in one $U_{\lambda}$ of the open cover. This gives a finite subcovering for $K$. $\blacksquare$