I really need help with this question: Prove that a metric space which contains a sequence with no convergent subsequence also contains an cover by open sets with no finite subcover.

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    $\begingroup$ Do you know what a contrapositive is? Are you familiar with the various characterizations of compactness? $\endgroup$ – Potato Jun 10 '12 at 4:54
  • $\begingroup$ Just so you know, a topological space is sequentially compact if every sequence has a convergent subsequence, and compact if every open cover admits a finite subcover. In the case of a metric space, these are equivalent. $\endgroup$ – talmid Jun 10 '12 at 5:43

Let $(a_n)$ be a sequence in the metric space $M$ that doesn't have any convergent subsequence. The set $\{a_n\}$ consists of isolated points (that is, it doesn't have any accumulation points; otherwise you could take a convergent subsequence), and it's infinite (because if it wasn't, one of the points would repeat infinitely in the sequence and we could get a constant subsequence). Now, for each $n$ take an open ball $B_{r_n}(a_n)$ around $a_n$ with radius $r_n$ small enough not to contain any other terms in the sequence, constructing an (infinite) family of open sets $\{B_{r_n}(a_n)\}$.

Now if you know some facts about compactness you could argue: The set $\{a_n\}$ is closed, and then since it's a subset of the metric space $M$, if $M$ is compact then so is $\{a_n\}$. However, note that $\{B_{r_n}(a_n)\}$ is an open cover of $\{a_n\}$, but contains no finite subcover. Then $\{a_n\}$ isn't compact, and neither is $M$ (that is, not every subcover of $M$ admits a finite subcover).

Or otherwise, like suggested by Martin Sleziak (and showing a direct proof): consider the open cover $\{B_{r_n}(a_n)\} \cup (M \setminus \{a_n\})$, which doesn't admit a finite subcover.

It's a nice exercise trying to fill in the details.

An example is given in the space of bounded sequences of real numbers, with the supremum norm: for $i \in \mathbb{N}$, let $e_i$ to be the sequence with a one at the $i$-th position and zeroes elsewhere; then $(e_i)$ is a sequence with no convergent subsequence and $\{e_i\}$ doesn't have accumulation points (even more: it's uniformly discrete).

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    $\begingroup$ If you want open cover of the whole space, you can slightly modify the solution in this answer by adding the open set $M\setminus\{a_n; n\in\mathbb N\}$. $\endgroup$ – Martin Sleziak Jun 10 '12 at 5:09
  • $\begingroup$ @MartinSleziak I was thinking of that. I'll add it. Thanks. $\endgroup$ – talmid Jun 10 '12 at 5:12

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