Compactness and sequential compactness in metric spaces I've got a question: I'm trying to prove that every metric space is compact if and only if the space is sequentially compact. In all the proofs I have found, they used the Bolzano-Weierstrass theorem. Is there a way to prove this fact without using Bolzano-Weierstrass?
Thanks so much!!
 A: 
Theorem Let $(M,d)$ be a metric space. The following are equivalent:
  (a) $M$ is compact;
  (b) $M$ is sequentially compact;
  (c) $M$ is complete and totally bounded.

Proof: (a$\Rightarrow$b) Suppose $M$ is compact, and let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $M$. Suppose that the sequence $(x_n)$ did not have a convergent subsequence, that is, $(x_n)$ does not have a cluster point. Then for every $x\in M$, there exists some neighbourhood $U_x$ of $x$ such that $\left\{n:x_n\in U_x\right\}$ is finite. Then $\left\{U_x:x\in M\right\}$ is an open cover of $M$, so by compactness we can find a finite subcover $U_1,\ldots,U_{k}$. But notice that
$\mathbb{N}=\left\{n:x_n\in M\right\}=\cup_{i=1}^k\left\{n:x_n\in U_i\right\}$, and the latter set is finite, a contradiction.
Therefore, $(x_n)$ has a cluster point, hence a convergent subsequence.
(b$\Rightarrow$c) Suppose $M$ is sequentially compact. Since every Cauchy sequence has a convergent subsequence, it follows that $M$ is complete. Also, if $M$ was not totally bounded, there would exist some $\varepsilon>0$ such that no finite collection of open balls of radius $\varepsilon$ covers $M$. Let $B_1=B(x_1,\varepsilon)$ be one such ball. Then, let $x_2\in M\setminus B_1$, and let $B_2=B(x_2,\varepsilon)$. Then, let $x_3\in M\setminus (B_1\cup B_2)$. Continuing this way, we find a sequence $x_1,x_2,\ldots$ such that $x_{n+1}\in M\setminus\cup_{i=1}^nB(x_i,\varepsilon)$, hence $d(x_n,x_m)\geq\varepsilon$ whenever $n\neq m$, so $(x_n)$ cannot have a convergent subsequence, a contradiction.
Thus, $M$ is also totally bounded.
(c$\Rightarrow$a) Suppose that $M$ is complete and totally bounded. In order to find a contradiction, suppose that $M$ was not compact, so there exists some open cover $\left\{U_i:i\in I\right\}$ which does not admit a finite subcover.
We can cover $M$ by a finite number of sets $C^1_1,\ldots,C^1_{p_1}$ of diameter $\leq 1$ (since $M$ is totally bounded). One of these sets, say $C^1=C^1_{k_1}$, cannot be covered by a finite number of sets $U_i$ (if all could, we would find a finite subcover for $M$). Now, $C^1$ can be covered by a finite number of subsets $C^2_1,\ldots,C^2_{p_2}$ of diameter $\leq 1/2$. Again, one of the sets, say $C^2=C^2_{k_2}$, cannot be covered by a finite number of sets $U_i$.
Proceeding this way, we find (non-empty) sets $C^1\supseteq C^2\supseteq C^3\supseteq\cdots$ such that $C^k$ has diameter $\leq 1/k$ and $C^k$ cannot be covered by a finite number of sets $U_i$. Let $x_k\in C^k$ be an arbitrary element for every $k$. Then $(x_k)_k$ is Cauchy (by the condition on the diameters), so it converges to some $x\in M$. This $x$ belongs to some $U_i$, as $\left\{U_i:i\in I\right\}$ covers $M$, so there exists some $\delta$ such that $B(x,\delta)\subseteq U_i$. Letting $N$ be sufficiently large, so that $d(x,x_N)<\delta/2$ and $1/N<\delta/2$, we obtain $C^N\subseteq B(x_N,1/N)\subseteq B(x_N,\delta/2)\subseteq B(x,\delta)\subseteq U_i$, contradicting the construction of $C^N$.
Therefore, $M$ is compact.
