# Heine-Borel Theorem and arbitrary topological space

We know that Heine-Borel Theorem does not hold in arbitrary metric space. (If $A$ be an infinite subset of a discrete metric space then $A$ is not compact though it is closed and bounded.) How do we characterise the class of metric spaces that satisfies the Heine-Borel Theorem?

I have read the proof of Heine-Borel Theorem in the Real line. It considers any closed and bounded subset $[a,b]$. Then it attains an open cover $U$ (say). Let $P=\{x\in[a,b]:U$ has a finite subcover for $[a,x]\}$. We take it $V$. Clearly $P$ is bounded above and hence it attains supremum...say $c$. Let $c$ contain in $U_1$ (element of $U$). We take an open ball centered at $c$ that is contained in $U_1$. Then the open ball centered in $c$ with the union of $V$ is also finite open sets. Then $c$ coincides with $b$ and hence $[a,b]$ becomes compact.

Which step does not hold in arbitrary metric space? (I have written the procedure of the proof briefly.)

• $P$ already makes no sense in a general metric space, even $\mathbb{R}^n$ (where Heine-Borel still holds). You might try imitating the proof of Heine-Borel in $\mathbb{R}^n$ instead to see where that fails. – Ian May 23 '17 at 20:10
• In most metric spaces there is not a given total order the way there is in $\Bbb{R}$. The step of taking a supremum thus fails. – ForgotALot May 23 '17 at 20:11
• #ForgotALot Do you mean that any metric spaces except ordered sets don't hold this property? – Subhajit Saha May 23 '17 at 20:42
• @SubhajitSaha It doesn't mean that the property fails, but this proof makes no sense outside of an ordered set, because there's no analogue of $P$. – Ian May 23 '17 at 20:54
• en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem – Mirko May 24 '17 at 1:39