Let $(X, d)$ be a metric space. 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.

I have proved the following: 1$\Longleftrightarrow$2, 2$\Longleftrightarrow$3, 4$\implies$3, 2$\implies$4, 4$\implies$5 and 5$\implies$3. Although they are enough to prove the equivalence of all five statements, I tried to prove it from the other possible directions. But I'm stuck with the proof of 3$\implies$4 and 5$\implies$4. Does anybody know how to prove them without using any of the above statements? Please help.

  • $\begingroup$ For those who might be interested, I just archived (in a Math Forum sci.math post) two topology handouts of mine from June 2003. One is a chart of compactness notions in topological spaces and the other is a chart of countability notions (in the sense of separability, second countability, etc.) in topological spaces. Each of these was assigned to a group of 3 to 4 students in the class, who were responsible for later presenting proofs to our class of all the implied results in the chart they were assigned. The handouts are at mathforum.org/kb/message.jspa?messageID=7611461 $\endgroup$ – Dave L. Renfro Nov 18 '11 at 17:02

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.

  • $\begingroup$ That's very nice, as usual. I think Mary Ellen Rudin's proof of paracompactness of metric spaces is not that much more involved than the argument for metacompactness you give here. Funny enough, Ornstein's proof was received by the editors just a day earlier. $\endgroup$ – t.b. Nov 18 '11 at 18:40
  • $\begingroup$ @Brian M. Scott: It took me some time to fully understand your proof, but I can see it now. It's excellent. Thanks for taking the trouble. $\endgroup$ – Sayantan Nov 18 '11 at 23:58
  • 3
    $\begingroup$ @t.b.: One of the rare occasions when Mary Ellen did something the easy way! (That’s an exaggeration, but we used to tease her that her unique way of looking at things at times produced excessively complicated arguments.) $\endgroup$ – Brian M. Scott Nov 19 '11 at 18:22

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