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?

  • 4
    $\begingroup$ Maybe the sequence can converge to a element outside the given set. $\endgroup$
    – Tomás
    Jul 1, 2013 at 20:01
  • 5
    $\begingroup$ 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. $\endgroup$ Jul 1, 2013 at 20:02
  • $\begingroup$ It seems (and obviously so) that $\Bbb Q\cap[0,1]$ is the canonical counterexample! $\endgroup$
    – Asaf Karagila
    Jul 1, 2013 at 21:18

5 Answers 5


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.

  • $\begingroup$ Why is it obvious that this space is totally bounded? $\endgroup$
    – michek
    Mar 16, 2015 at 20:27
  • $\begingroup$ Because you can very easily cover $\mathbb Q\cap [0,1]$ by finitely many intervals of any prescribed length $r>0$. $\endgroup$ Mar 16, 2015 at 23:29

No, it "shouldn't", simply because it doesn't.

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.


The confusion is understandable. We restate the two definitions:

  • Total boundedness says that every uniform open cover has a finite subcover.
  • Covering compactness says that every open cover, including non-uniform covers, have a finite subcover.

(In metric spaces, open balls can be used in place of open sets/covers. This is because they form a base of the topology: that is, every open set can be decomposed into a union of open balls, implying that every open cover, being a union of open sets, can as well.)

A more illustrative counter-example would be $(0, 1)$, as user Daniel Fischer mentioned in a comment. It is not compact because I can find an open cover that has no finite subcover, namely

$$B\left(1, {1 \over 2}\right) \cup B\left({1 \over 2}, {1 \over 3}\right) \cup \cdots = \bigcup_{n = 1}^{+ \infty} B\left({1 \over n}, {1 \over n + 1}\right)$$

If we stop the construction of the union at any finite $n$, the interval $(0, {1 \over n} - {1 \over n + 1}]$ won't be covered.

It is totally bounded, though. There exists a finite, positive real number, namely $1$, such that, $\forall x, y \in (0, 1)$, $\ d(x, y) < 1$, making $(0, 1)$ bounded. On the real line, a bounded interval is totally bounded (given $\varepsilon > 0$ and an interval $(a, b)$, subdivide your interval into ${(b - a)\over\varepsilon} + 1$ open pieces, then union these pieces with open balls of radius $\varepsilon$ centered at the partition points).


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