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I'm afraid this question may turn out to be a stupid one. Though it is related to a previous question of mine, I'll write it down in full. Let $(X, d)$ be a metric space (MS). I have to prove the following statements are equivalent.

  1. $(X,d)$ is complete (i.e., every Cauchy sequence is convergent) and totally bounded (i.e., for every $\epsilon>0$, $(X,d)$ has a finite $\epsilon$-net).
  2. $(X,d)$ is sequentially compact (i.e., every sequence has a convergent subsequence) .
  3. $(X,d)$ is Bolzano - Weierstrass compact (i.e., every infinite subset of $X$ has a limit point).
  4. $(X,d)$ is compact (i.e., every open cover of $X$ has a finite subcover).
  5. Every infinite open cover of $(X,d)$ has a proper subcover.

To complete the equivalence I had to prove $2\implies4$. The proof is (roughly) as follows:

We take an open cover $\mathscr{U}=\{U_\alpha:\alpha \in A\}$ of a sequentially compact (SC) metric space $(X,d)$. Since every open cover of SC MS has a Lebesgue number (LN), $\mathscr{U}$ has a LN, say $\delta>0$. Also a SC MS is totally bounded and therefore $(X,d)$ has a finite $\delta/3$-net, say, $\{a_1,a_2,\ldots,a_n\}$ (i.e. $X=\bigcup_{i=1}^{n} B(a_i,\delta/3)$). Since for all $i \in \{1,2,\ldots, n\}$ $diam(B(a_i,\delta/3)) \leq 2\delta/3 <\delta$, and $\delta$ is a LN of $\mathscr{U}$ then for all $i \in \{1,\ldots,n\}$ there exist $\alpha_i \in A$ such that $B(a_i,\delta/3) \subseteq U_{\alpha_i}$. Thus we get a finite subcover $\{U_{\alpha_i}:i \in \{1,\ldots,n\}\}$ of $\mathscr{U}$, proving $(X,d)$ is compact.

My confusion starts when in this proof I use the argument that every SC MS is totally bounded. I had to prove this in $2 \implies 1$. So can I say in the proof of $2 \implies 4$ we are using condition 1? If not, then is it because there is an "and" in condition 1 and we are using only one part of it? Anyway, is there any proof of $2 \implies 4$ which manages to avoid the total boundedness argument? I'd appreciate any kind of help.

I'm not very sure about the appropriate tags. Regards.

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If you have already proved 2) implies 1), then you can certainly use that fact for your proof of 2) implies 4). –  David Mitra Nov 27 '11 at 18:02

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