[Math] Totally bounded and closed implies compact?

analysisgeneral-topology

Is there a fault in exercice 9.3.1 b) from Analysis by Zorich? The exercice asks to prove that a subset of a metric space is compact if and only if it is totally bounded and closed. But I have a counterexample for it: Consider the open unit ball $B(0,1)$ in $\mathbb{R}^n$ as a metric space itself. Then it is closed in itself and totally bounded, but not compact.

Am I right?

Best Answer

You are right.

For a subset $A$ of metric space $(X,d)$ to be compact, it is not enough that $A$ is totally bounded and closed (since $X$ is always closed). However, the correct assumptions to conclude that $A$ is compact are that it is totally bounded and complete. You can follow the link I included in my previous answer to a similar question to see that proof of this fact.

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