Exercise 12.5.11 in Tao's Analysis 2. Let (X,d) be a metric space.

Claim: For every open cover of (X,d) there is a finite subcover $\implies$ X compact.

Proof: If X is not compact, then there is a sequence $(x^n)$ with no limitpoint. In other words, the sequence doesn't have convergent subsequences. Suppose the given hint below and that X has a finite subcover, then there can be at most finitely many elements in the sequence. By contradiction is shown that X is compact.

Tao´s hint is that $\forall x \in X$ there consist an open ball around x containing only finitely many elements of the sequence. An open ball is per definition bounded, as an set is bounded when the set can be contained in an open ball.

The hint seems plausible to me, but i don't know how to prove it. It remembers me of the Bolzano-Weierstraß theorem, but that is real-analysis. And this question gives an uncanny counterexample.

  • 2
    $\begingroup$ Bounded sequences don't have convergent subsequences in general. Take e.g. $d(x,y)=\min\{|x-y|,1\}$ on $\mathbb{R}$. It's a metric that gives the standard topology. Also, every sequence is bounded with respect to this metric but there are plenty of sequences on $\mathbb{R}$ that do not have convergent subsequences. $\endgroup$
    – T. Eskin
    Mar 13 '16 at 16:26

We have a metric space $X$ with an infinite number of points and I assume you are taking as a definition of compactness that every sequence has a convergent subsequence.

Now, suppose there is a sequence $a_n$ of points in $X$ with no convergent subsequence. This means that for each $x\in X$, we can find open sets $U_x$ such that only finitely many $a_n$ are contained in $U_x$. Then, $\left \{ U_x \right \}_{x\in X}$ is an open cover of $X$ with no finite subcover, since any $\left \{ U_{x_{i}} \right \}_{1\leq i\leq n}$ can contain only finitely many $a_n$.

  • $\begingroup$ Ok, why then can we find an open set $U_x$ such that within only finitely many $a_n$ are contained? $\endgroup$ Mar 13 '16 at 17:22
  • $\begingroup$ hint: suppose not. then, there are $x\in X, a_{n_{k}}\in \left \{ a_n \right \}$ and open sets $B_{1/k}(x)$ such that $a_{n_{k}}\in B_{1/k}(x)$ for each $k$ $\endgroup$ Mar 13 '16 at 19:09

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