[Math] Some equivalent formulations of compactness of a metric space

metric-spaces

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.

Best Answer

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.

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