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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?

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    $\begingroup$ Maybe the sequence can converge to a element outside the given set. $\endgroup$
    – Tomás
    Jul 1, 2013 at 20:01
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    $\begingroup$ 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

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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.

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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.

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  • $\begingroup$ 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
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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.

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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.

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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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  • $\begingroup$ why the downvote? :'( $\endgroup$
    – James Chen
    Dec 14, 2020 at 8:54

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