Every argument that I can see right now to show that (5) implies (4) either essentially goes through one of the other equivalent forms or uses a much more sophisticated result about metric spaces, namely, that every metric space is paracompact. This means that every open cover $\mathscr{U}$ of $X$ has a locally finite open refinement $\mathscr{V}$ covering $X$. That is,
- $\mathscr{V}$ is an open cover of $X$;
- for each $V\in\mathscr{V}$ there is a $U\in\mathscr{U}$ such that $V\subseteq U$; and
- each $x\in X$ has an open nbhd $N_x$ such that $\{V\in\mathscr{V}:N_x\cap V\ne\varnothing\}$ is finite.
Note that the third condition implies that each point of $X$ is in only finitely many members of $\mathscr{V}$, i.e., that $\mathscr{V}$ is point-finite. This is actually all that I need. (A space in which every open cover has a point-finite open refinement is said to be metacompact, so I’m actually using only the weaker result that every metric space is metacompact.)
Theorem: Every point-finite open cover of $X$ has an irreducible subcover, meaning one with no proper subcover.
Proof: Let $\mathfrak{R}=\{\mathscr{R}\subseteq\mathscr{V}:\mathscr{R}\text{ covers }X\}$; $\mathfrak{R}$ is partially ordered by $\supseteq$. Let $\mathfrak{C}$ be a chain in $\mathfrak{R}$, and let $\mathscr{C}=\bigcap\mathfrak{C}$; I claim that $\mathscr{C}\in\mathfrak{R}$, i.e., that $\mathscr{C}$ still covers $X$.
Proof of Claim: Suppose that some $x\in X$ is not covered by $\mathscr{C}$. Let $V_1,\dots,V_n$ be the finitely many members of $\mathscr{V}$ containing $x$. Then none of these $V_k$ can belong to $\mathscr{C}$ (or else $x$ would be covered by $\mathscr{C}$). But $\mathscr{C}$ is the intersection of the collections in the chain $\mathfrak{C}$, so for each $k=1,\dots,n$ there is some $\mathscr{C}_k\in\mathfrak{C}$ such that $V_k\notin\mathscr{C}_k$. Because $\mathfrak{C}$ is a chain, the collections $\mathscr{C}_1,\dots,\mathscr{C}_n$ are nested, and without loss of generality we may assume that the indexing has been chosen so that $\mathscr{C}_1\supseteq\dots\supseteq\mathscr{C}_n$. But then $\mathscr{C}_n$ contains none of the sets $V_1,\dots,V_n$, so $\mathscr{C}_n$ does not cover $x$, and hence $\mathscr{C}_n\notin\mathfrak{R}$, a contradiction.
We can now apply Zorn’s lemma to the partial order $\langle\mathfrak{R},\supseteq\rangle$ to conclude that $\mathfrak{R}$ has a maximal element $\mathscr{M}$ with respect to $\supseteq$: that is, $\mathscr{M}$ is in $\mathfrak{R}$, but no proper subcollection of $\mathscr{M}$ belongs to $\mathfrak{R}$. But then $\mathscr{M}$ is an open cover of $X$ with no proper subcover, i.e., an irreducible cover of $X$.$\dashv$
Now it’s easy to show that (5) implies (4). Suppose that every infinite open cover of $X$ has a proper subcover; this amounts to saying that every irreducible open cover of $X$ is finite. Let $\mathscr{U}$ be an open cover of $X$. By what we just showed, $\mathscr{U}$ has an irreducible subcover $\mathscr{V}$, and being irreducible, $\mathscr{V}$ must be finite. Thus, $X$ is compact.
Def'n 1. is the general def'n of compactness in topology, whether or not the topology can be generated by a metric.
In a metric space, a set $X$ is compact iff every sequence in $X$ has a limit point that BELONGS to $X$ iff every infinite subset of $X$ has a limit point that belongs to $X.$
The n-dimensional generalization of the (1-dimensional) Bolzano-Weierstrass theorem is that a subset of $R^n$ is compact iff it is closed and bounded. (But this does not hold for all metric spaces.)
It is useful to know various equivalents to compactness, and to know that there are equivalents specific to metric spaces, just as it is useful to know various equivalents of "continuous function", some of which are specific to metric spaces.
Another term is "pre-compact" which I have seen only in the context of Hausdorff spaces : $X$ is pre-compact iff $\overline X$ is compact.
Best Answer
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.