Prove that closed and bounded subsets of metric spaces are compact

My Attempted Proof

Let $(X, d)$ be a metric space. Suppose $A \subset X$ is closed and bounded, since $A$ is bounded $\exists \ r > 0$ such that $d(x_1, x_2) \leq r$ for all $x_1, x_2 \in A$.

Now let $B_d(x, r)$ be a neighbourhood of $x \in A$. Since $B_d(x, r)$ is open in the topology, $\mathcal{T}_d$ induced by the metric $d$, we have $$\bigcup_{x\ \in\ A} B_d(x, r) \in \mathcal{T}_d$$.

Put $V = \bigcup_{x\ \in\ A} B_d(x, r)$, then $V$ is an open cover of $A$, and $A \subset V$ so that $A$ is comapct. $\ \square$

Is my proof correct? If so how rigorous is it?

  • $\begingroup$ $V$ is an open cover of $A$, yes, but this is not what you're intended to prove.You must take any open cover of $A$ and show it only requires a finite number of these open sets to cover $A$. $\endgroup$ – Daniel Jan 24 '17 at 13:39
  • $\begingroup$ Hmm, this is not true. There is nothing special about boundedness in a metric space. Given a metric space $(X,d)$, there is always an equivalent bounded metric $\hat{d}$ which generates the same topology on $X$. So $(X,d)$ and $(X,\hat{d})$ have exactly the same topology (so the same compact subsets), but all sets are bounded in the latter. $\endgroup$ – MPW Jan 24 '17 at 13:44

You are trying to prove a false proposition. It is not true that a closed and bounded subset of an arbitrary metric space is compact. For instance, let $M$ be an infinite set and define $d(x,y)=1$ whenever $x\ne y$; the whole space $M$ is closed and bounded, but not compact, because the cover by singletons is an open cover with no finite subcover.

  1. Your proof is not correct. Your conclusion "... so that $A$ is compact " is not true.

  2. Let $X$ be a normed space and $A$ the closed unit- ball in $X$. Then $A$ is closed and bounded, but we have

$A$ is compact iff $ \dim X < \infty$.

Hence, if $ \dim X = \infty$, then $A$ is not compact.


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