Show that a sequentially compact metric space is totally bounded.

Well since sequentially compact $\implies$ compact, any open cover of sequentially compact metric space $X$ has a finite subcover. So $X$ sequentially compact $\implies$ $\exists$ finite subcover. Can this subcover consist of balls of radius $\epsilon$? How would I go about showing this?

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    $\begingroup$ Well, what happens if you put a ball of radius $\epsilon$ around every point? Does that cover the space? $\endgroup$ – Arturo Magidin Feb 13 '12 at 4:15
  • $\begingroup$ So let $\{B_{\epsilon}(x)\}$ be the collection of balls of radius $\epsilon>0$ around points $x\in X$. $\bigcup B_{\epsilon}(x)$ cover $X$ since $X$ is compact $\implies$ $X$ totally bounded? $\endgroup$ – Emir Feb 13 '12 at 4:22
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    $\begingroup$ Since $X$ is compact, there is a finite subcover. Since there is a finite subcover, for every $\epsilon$, $X$ can be covered with finitely many open balls of radius $\epsilon$, hence $X$ is totally bounded. $\endgroup$ – Arturo Magidin Feb 13 '12 at 4:23

As total boundedness is usually used as a stepping stone towards showing a sequentially compact metric space is compact, I do not think assuming compactness of $X$ is what is intended for this problem.

The contrapositive is easily proved:

Suppose $X$ is not totally bounded. Then there is an $\epsilon>0$ so that no finite collection of open balls covers $X$. So, for any finite collection of points $\{x_1,\ldots ,x_n\}$, there is a point $x$ in $X$ not in any of open balls $B_\epsilon(x_k)$, $k=1,\ldots n$; whence, $d(x,x_k)>\epsilon$ for each admissible $k$.

Using the above observation, one may construct, by induction, a sequence $(y_n)$ in $X$ such that $d(y_i,y_j)>\epsilon$ whenever $i\ne j$.

I'll leave the rest of the argument for you...

  • $\begingroup$ then this sequence does not have a subsequence which converges to an element in $X$, which contradicts the assumption that $X$ is sequentially compact? $\endgroup$ – Emir Feb 13 '12 at 4:54
  • $\begingroup$ @Emir No contradiction. We're proving the contrapositive statement. $\endgroup$ – David Mitra Feb 13 '12 at 4:55

For a direct proof, see e.g. Theorem 34 here. A key intermediate point is the fact that a metric space is totally bounded iff every sequence admits a Cauchy subsequence.


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