Does Cauchy Completeness imply the Heine-Borel theorem generally?

Question

I've been working through some reverse math with the completeness definitions of a metric space. More over, I've learned that in a metric space X that is ordered, The Least Upper Bound Property, Cauchy Criterion, Nested Interval Theorem, and Heine-Borel Theorem are all equivalent (provided that the Archimedean Property is true).

My Question: Let X be a metric space and suppose X is Cauchy complete. Does the Heine-Borel Theorem follow? In other words, is it true that if every Cauchy sequence converges to a limit in X, then every closed and bounded set in X is compact?

I've been able to show this is true for $\mathbb{R}^n$. Is this true for a general metric space?

Answer 1

No. Any infinite-dimensional Banach space is a counterexample to this. The closed unit ball in such a space is closed and bounded but not compact.

Answer 2

You already got a counterexample in another answer but let me add a potentially useful fact. If $X$ is a metric space then $Y\subseteq X$ is compact iff it is complete and totally bounded. In particular if $X$ is complete then $Y\subseteq X$ is compact iff it is closed and totally bounded. (Note that the closed unit ball in an infinite dimensional Banach space is bounded but not totally bounded)

Answer 3

A simple counter-example is the discrete metric, that is, $d(x,y)=1$ when $x\ne y,$ on an infinite set $X.$ Any $d$-Cauchy sequence is eventually constant and hence convergent. And every subset of $X$ is closed and bounded, but only the finite subsets of X are compact.

Another impediment is that if $d$ is a $complete$ metric on a set $X$ (that is, if every $d$-Cauchy sequence converges in $X$) then the metric $e(x,y)=\min (1,d(x,y))$ is also complete, and $e$ generates the same topology as $d$ does. Hence $d$ and $e$ generate the same collections of closed sets and of compact sets. So $X$ itself is closed and $e$-bounded, but not necessarily compact.

For example let $X=\Bbb R$ and $d(x,y)=|x-y|$ and $e(x,y)=\min (1,|x-y|).$