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?
Best Answer
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.