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.
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Hint. If $X$ is compact then $(x_n)$ has convergent subsequence. In particular it is Cauchy.
answered Oct 26 '13 at 15:25
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$\begingroup$ Am I going to have to use the triangle inequality combined with proving the contrapositive? $\endgroup$ – Jim Darson Oct 26 '13 at 15:33
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$\begingroup$ 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
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$\begingroup$ @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
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$\begingroup$ 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
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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.
answered Oct 26 '13 at 15:57
