$\gdef\In\subset$

Let $(X,d)$ be a metric space, $K \In X$ a subspace. Recall that we had the following definitions:

(Compactness) : $K$ is compact if for any open cover of $K$, there exists a finite sub-cover.

(Sequential Compactness) : $K$ is sequentially compact, if for any sequences $(x_n)$ in $K$, there exists a subsequence $x_{n_k}$ that converges to $x$, for some $x$ in $K$.

In this note, we prove that they are equivalent.

Assume $K$ is compact. Let $(x_n)$ be a sequence in $K$, and assume $x_n$ has no subsequence that converges. That means, for any $x \in K$, there exists a radius $r_x>0$, such that the sequence $(x_n)$ visits $U_x = B_{r_x}(x)$ finitely many times. Since $K \In \cup_{x \in K} U_x$, and $K$ is compact, we may find a finite sub-cover, indexed by $\{x_1, \cdots, x_n\} \In K$ $$ K \In \bigcup_{i=1}^n U_{x_i}. $$ This is impossible, since the infinite sequence $(x_n)$ visits each $U_{x_i}$ finitely many times, and the sequences is contained in the $K$.

Assume $K$ is sequentially compact, we will prove the following two lemma.

Assume $K$ is sequentially compact, then for any $\epsilon>0$, there exists a finite subset $S \In K$, such that $K \In \bigcup_{x \in S} B_\epsilon(x). $

Proof: Suppose the Lemma is false, then there exists an $\epsilon>0$, such that there does not exist a finite subset $S \In K$ with $K \In B_\epsilon(S) := \bigcup_{x \in S} B_\epsilon(x) $. Then, we may produce a sequence iteratively: pick any $x_1 \in K$. Assume $x_k$ for $k \leq n$ is picked, then we may pick $x_{n+1} \in K \RM \cup_{i=1}^n B_\epsilon(x_i)$. This process can go on forever, and we have obtained a sequence $x_1, x_2, \cdots, $ with the property that $d(x_i, x_j) > \epsilon$. This sequence cannot have convergent subsequence, hence we have a contradition with sequential compactness.

Assume $K$ is sequentially compact, then for any open cover $\{U_\alpha\}$ of $K$, there exists $\delta>0$, such that for any $x \in K$, $B_\delta(x)$ is contained in some open set $U_\alpha$ in the given cover.

Proof: Assume the conclusion is false. Then for each positive integer $n$, let $V_n = \{ x \in K \mid B_{1/n}(x) \not \In U_\alpha \forall \alpha \in I \}$, and $V_n$ is non-empty. Pick $x_n \in V_n$ and form a sequence. By sequential compactness, we have a convergent subsequence $x_{n_k}$, converging to $x \in K$.

Say $U_{\alpha_0} \ni x$. Then, we may pick $\epsilon>0$, such that $B_\epsilon(x) \In U_{\alpha_0}$. For for $N$ large enough, such that $1/{n_k} < \epsilon /2$, and $d(x_{n_k}, x) < \epsilon/2$ for $k > N$, we have $$ B_{1/{n_k}} (x_{n_k}) \In B_\epsilon(x) \In U_{\alpha_0} $$ contradicting with the construction of $x_n$.

Now that we have the two lemmas, we can finish the proof. Take any open cover $\{U_\alpha\}$ of $K$. By Lemma 2, we find a $\delta$ such that any radius $\delta$ ball, with center in $K$, is contained in $U_\alpha$. Then, use Lemma 1, we can produce a finite covering of $K$ with radius $\delta$ balls. Then, for each such ball, we can find a 'parent' open set in $\{U_\alpha\}$, that forms the desired finite subcover of $\{U_\alpha\}$.