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# Heine-Borel theorem.
In a metric space $X$, if a subset $K\subset X$ is compact, then $K$ is necessarily closed and bounded. Without loss, $X$ is not empty.
$\blacktriangleright$ Take such compact $K$, cover it with the collection of open balls $\{B(x_0,n)\}_{n\in \mathbb{N}}$ of radius $n$, all centered at some arbitrary $x_0\in X$. Since $K$ compact, we can refine it to a subcover, whose union is one open ball $B(x_0,N)$ for some $N$ that contains $K$. Hence $K$ is bounded.
We show $K$ is closed by showing it contains all its adherent points. Take $p$ to be an adherent point of $K$, namely for every open neighborhood $U$ about $p$, we will have $U$ intersecting $K$. Suppose $p$ is not in $K$. For each $x\in K$, we have $d(x,p) > 0$. So consider the distance $r_x = \frac{1}{2}d(x,p)$ for each $x \in K$. The collection of open balls $B(x,r_x)$ is an open cover of $K$, none of which contains $p$. Since $K$ is compact, we can refine this to a finite subcover, so that $K \subset B(x_1,r_{x_1})\cup\cdots\cup B(x_m, r_{x_m})$, for some finitely many $x_1,\ldots,x_m$. But if we take $r = \min r_{x_i}$, the open ball $B(p,r)$ is disjoint for each open ball $B(x_k,r_{x_i})$, and hence disjoint from $K$. This contradicts $p$ being an adherent point. Hence $p \in K$ and that $K$ is closed. $\blacksquare$
However, if a subset $S$ in a metric space $X$ is closed and bounded, it need not be compact!. Here is an example. Let $X=\mathbb{Z}$ with discrete metric $d(a,b)=1$ if $a\neq b$, otherwise $0$. Then $X$ itself is bounded: $X\subset B(0,2)$. Also, $X$ is closed as it is the whole metric space. But $X$ is not compact, as the open cover $B(n, \frac{1}{2})$ covers $X$ but cannot be refined to a finite subcovering!
Nevertheless, we have
> **Heine-Borel theorem.**
> In Euclidean spaces, a subset is compact if and only if it is closed and bounded.
First we take a lemma.
> A closed set in a compact set is compact.
$\blacktriangleright$ Suppose $F$ is a closed set inside a compact set $K$. Consider any open cover $\{U_\lambda\}_{\lambda\in\Lambda}$ for $F$. Now add to this collection $U=F^c$, which is open. Then this new collection of open sets $\{U_\lambda\}_{\lambda\in\Lambda} \cup \{U\}$ is an open cover for $K$. But $K$ is compact, so we can refine it to a finite subcover. This subcover also covers $F$, as $F\subset K$. Now if this subcover contains the open set $U$, we remove it. Then this yields a finite subcover of the original open cover for $F$. Whence $F$ is compact. $\blacksquare$
So to prove Heine-Borel, note that any bounded set is contained in closed cube. So it suffices to show closed cubes are compact. Then any closed and bounded subset of it will also be compact.
Let us prove the 1-dimensional case, but the argument can be generalized to $n$-dimension.
> The closed interval $[a,b]$ is compact.
$\blacktriangleright$ Suppose to the contrary that $I_0=[a,b]$ is not compact. So there exists some open cover $\{U_\lambda\}_{\lambda\in\Lambda}$ for $I_0$ that has no finite subcovering. Divide the interval $I_0$ into two closed intervals with half the length of $I_0$. Then one of them cannot be finite subcovered, call that one $I_1$. Repeat this procedure to get a sequence of nested closed intervals $I_0 \supset I_1 \supset\cdots$ where $|I_k|=\frac{1}{2^k}|I_0|$, and each $I_k$ is not finite coverable by the open cover. By nested interval theorem, there exists some $x \in I_k$ for all $k$. Since $x \in I_0$,, which is further covered by the open cover, $x\in U_\lambda$ for some one $\lambda$. This means there is an open ball $B(x,r)$ of some radius $r > 0$ such that $x \in B(x,r)\subset U_\lambda$. But for large enough $k$, we have $\frac{1}{2^k} \ll r$ and $x\in I_k \subset B(x,r)\subset U_\lambda$. This means $I_k$ can be covered by one single open set in the open cover, a contradiction. Hence $I_0$ is compact. $\blacksquare$
Note, here we used
> Nested closed interval theorem.
> If $I_0 \supset I_1\supset I_2\supset\cdots$ is a sequence of nested closed intervals in the reals, then their common intersection $\bigcap I_k$ is not empty.
$\blacktriangleright$ If we denote $I_k=[a_k,b_k]$, then we have the sequence of real numbers $$
a_0 \le a_1\le a_2 \le\cdots\le b_2 \le b_1 \le b_0.
$$Then by the axioms of the reals, the set $A=\{a_k\}$ , being bounded above by $b_0$, has a least upper bound $a=\sup a_k$ in the reals. Since each $b_k$ is an upperbound of $A$, we have $a_k \le a \le b_k$ for each $k$. Whence $a\in I_k$ for each $k$. $\blacksquare$