0
$\begingroup$

I have a proof that I would like some hints in solving:

Let $X$ be a metric space. Show, if there is an $r > 0$ and a sequence $(x_n)$ from $X$ such that $d(x_n,x_m) \geqslant r$ for $n≠m$, then $X$ is not compact.

$\endgroup$
1
$\begingroup$

Hint. If $X$ is compact then $(x_n)$ has convergent subsequence. In particular it is Cauchy.

$\endgroup$
  • $\begingroup$ Am I going to have to use the triangle inequality combined with proving the contrapositive? $\endgroup$ – Jim Darson Oct 26 '13 at 15:33
  • $\begingroup$ @JimDarson Yes. $\endgroup$ – Hanul Jeon Oct 26 '13 at 15:42
  • $\begingroup$ I'm having trouble using the triangle inequality in this. I understand that I need to use it and why, and I understand what it means for a sequence to have a convergent subsequence, but I'm having trouble actually writing out an appropriate use of the triangle inequality. $\endgroup$ – Jim Darson Oct 26 '13 at 15:53
  • $\begingroup$ @JimDarson That is, there is increasing sequence $(n_k)$ of natural numbers satisfy that $\lim_{n\to\infty} a_{n_k}\to L$ for some $L$. The proof of `every convergnet sequence is Cauchy' needs the triangle inequality. $\endgroup$ – Hanul Jeon Oct 26 '13 at 16:01
  • $\begingroup$ I'm afraid that I'm going to need more hints. I am assuming that X is compact. Since X is compact, then X is sequentially compact. Therefore, every sequence from X has a subsequence which converges in X. Thus, if (xn) is a sequence from X, then there exists an element p of X and a subsequence (xnj) of (xn) which converges to p. Since this subsequence converges, this subsequence is a cauchy sequence. What then? $\endgroup$ – Jim Darson Oct 26 '13 at 16:09
0
$\begingroup$

HINT: Let $A$ be the closure of the $\{x_n\mid n\in\Bbb N\}$. Show that $\{X\setminus A\}\cup\{B(x_n,r/2)\mid n\in\Bbb N\}$ is an open cover of $X$ which does not have a finite subcover.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.