# Why does Totally bounded need Complete in order to imply Compact?

Why does "Totally bounded" need "Complete" in order to imply "Compact"? Shouldn't the definition of totally bounded imply the existence of a convergent subsequence of every sequence?

• Maybe the sequence can converge to a element outside the given set. – Tomás Jul 1 '13 at 20:01
• Consider $(0,\,1] \subset \mathbb{R}$. As a subset of a compact set, it is totally bounded. But $\frac1n$ has no limit in that set. – Daniel Fischer Jul 1 '13 at 20:02
• It seems (and obviously so) that $\Bbb Q\cap[0,1]$ is the canonical counterexample! – Asaf Karagila Jul 1 '13 at 21:18
• Wow. We had a little catastrophe here: four essentially equal answers :-) – Mariano Suárez-Álvarez Jul 1 '13 at 21:24

No, total boundedness of $\langle X,d\rangle$ implies that every sequence in $X$ has a Cauchy subsequence. Completeness of $\langle X,d\rangle$ implies that every Cauchy sequence in $X$ actually has a limit point in $X$ and therefore converges. The two together therefore imply that every sequence in $X$ has a convergent subsequence, i.e., that $X$ is sequentially compact. Finally, there is a theorem that a metric space is sequentially compact if and only if it’s compact, so total boundedness plus completeness imply compactness.
No, totally bounded does not imply the existence of convergent subsequences. For instance, the space $\mathbb Q \cap [0,1]$, with the usual metric, is totally bounded but is not compact (nor complete of course) as any sequence of rationals converging to an irrational number will show.
• Because you can very easily cover $\mathbb Q\cap [0,1]$ by finitely many intervals of any prescribed length $r>0$. – Ittay Weiss Mar 16 '15 at 23:29
For example the set $\mathbb Q\cap[0,1]$ (with its usual metric inherited from $\mathbb R$) is totally bounded and you will surely find sequences in it which do not have convergent subsequences.
Not at all. Consider $X=\Bbb Q\cap[0,1]$ as a subspace of the real line. This is totally bounded, but not complete/compact.