Compactness of Metric Space

Sequential Compactness

Theorem

Let $S$ be a subset in ${\mathbb{R}}^k$. Then $S$ is closed and bounded iff every sequence in $S$ has a subsequence that converges to a point in $S$.

Sequential Compactness

Let $(X,d)$ be a metric space. A set $S\subset X$ is called sequentially compact if every sequence in $S$ has a subsequence that converges to a point in $S$.

Theorem

In a metric space, every sequentially compact set is bounded and closed.

The Converse is Not True in General

However, the converse is not true in general. For example, consider the metric space $\text{Q}$ with the standard metric inherited from $\mathbb{R}$. Let a subset $A:=[0,1]\cap \text{Q}$. Clearly $A$ is closed as a intersection of closed sets, and bounded. However, $A$ is not sequentially compact since a subsequence of any sequence of rationals converging to $\sqrt{2}/2$ never converges to a point in $A$.

Theorem

Let $(X,d)$ be a metric space. If $A\subset X$ is sequentially compact, then for any $x\in X$ there is $y\in A$ such that $d(x,y)=d(x,A)$This $y$ is called a nearest point in $A$ to $x$.

Proof By the definition of $d(x,A)$, there is a sequence $(y_n)$ in $A$ such that $d(x,y_n)\to d(x,A)$. Since $A$ is sequentially compact, $(y_n)$ has a convergent subsequence $(y_{n_k})$ with $y_{n_k}\to y\in A$. Then $d(x,A)={\lim }_{k\to \infty }d(x,y_{n_k})$

Lemma Let $(X,d)$ be a metric space. If $A\subset X$ is sequentially compact, then for any $\epsilon >0$ there exists an integer $N$ and $N$ points $x_1,\dots ,x_N$ in $A$ such that $A\subset {\bigcup }_{i=1}^N{B}_{\epsilon }(x_i)$

Thrm Let $(X,d)$ be a metric space and let $A\subset X$ be sequentially compact. If $\mathcal{O}=\{U_{\alpha }{\}}_{\alpha \in I}$ is a family of open sets such that $A\subset {\bigcup }_{\alpha \in I}{U}_{\alpha }$then there exists finitely many open sets $U_1,\dots ,U_n$ from $\mathcal{O}$ such that $A\subset {\bigcup }_{i=1}^n{U}_{\alpha }$Proof

Covering and Compactness

Covering

Let $(X,d)$ be a metric space and $A\subset X$. A collection of sets $\{U_{\alpha }\}$ is called a covering of $A$ if $A\subset {\bigcup }_{\alpha \in I}{U}_{\alpha }$ . If in addition, each $U$ is open, then $\{U_{\alpha }\}$ is called an open covering of $A$.

Compactness

Let $(X,d)$ be a metric space. A set $K\subset X$ is called compact if for every open covering $\{U_{\alpha }\}$ of $K$ there exists a finite subcollection $\{U_i{\}}_{i=1}^n$ of $\{U_{\alpha }\}$ such that $K\subset {\bigcup }_{i=1}^n{U}_i$

Theorem

Let $(X,d)$ be a metric space. Then a set $K\subset X$ is compact iff it is sequentially compact.

Proof