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I have this question:

If every closed and bounded subset of a metric space $M$ is compact, does it follow that $M$ is complete?

I don't know how Cauchy sequences interact with compact sets. But by definition,

Since $U$ is a subset of $M$ and $U$ is closed and bounded and compact, therefore every subsequence $(a_{n_k})$ on $M$ converges to some limit $a$ in $U$.

Now I need to prove that every Cauchy sequence is convergent. But it seems impossible. What to do?

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  • $\begingroup$ No, you need to prove that every Cauchy sequence is bounded. $\endgroup$ – Giuseppe Negro Jan 25 '16 at 1:22
  • $\begingroup$ Do you know the Hopf Rinow theorem? $\endgroup$ – Tsemo Aristide Jan 25 '16 at 1:23
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    $\begingroup$ @Giuseppe maybe I'm missing something but it seems to me that Cauchy sequences are bounded in any metric space by the triangle inequality. $\endgroup$ – Matt Samuel Jan 25 '16 at 1:24
  • $\begingroup$ en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem $\endgroup$ – Tsemo Aristide Jan 25 '16 at 1:25
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    $\begingroup$ If a Cauchy sequence has a subsequence that converges, then, also the whole sequence converges. $\endgroup$ – L.F. Cavenaghi Jan 25 '16 at 1:32
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HINT:

  • Prove (if you’ve not already done so) that a Cauchy sequence in any metric space is bounded.
  • Prove that if a Cauchy sequence in any metric space has a convergent subsequence, then the Cauchy sequence itself converges.
  • Prove that if $S\subseteq M$ is bounded, so is $\operatorname{cl}S$.

Now if $M$ is not complete, it has a Cauchy sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ that does not converge. Let $S=\{x_n:n\in\Bbb N\}$; $S$ is a bounded subset of $M$, so $\operatorname{cl}S$ is a closed, bounded subset of $M$ and is therefore compact and hence sequentially compact. Therefore ... ?

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  • $\begingroup$ In this case isn't S = cl S since the only convergent sequences in S are those sequences that are eventually constant? $\endgroup$ – user85798 Dec 18 '17 at 16:14
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Let $d$ be the metric for $M$. Let $(x_i)_{i\in N}$ be a Cauchy sequence. Then $\{x_i:i \in N\}$ is bounded because $$\lim_{n\to \infty}\sup_{n\leq m<m'}d(x_m,x_{m'})=0\implies \exists n\;\forall m>n\;(d(x_n,x_m)<1)\implies$$ $$\implies \exists n\; (\{x_i:i\in N\}\subset B_d(x_n,r))$$ $$\text {where }\; r=2+\max \{d(x_i,x_n):i\leq n\}.$$Now every metric for a compact metrizable space is a complete metric. Take any $r>0$ and any $x$ such that $B_d(x,r)\supset \{x_i:i\in N\}.$ Let $S$ be the closure of $B_d(x,r)$ in $M.$ We have $$S\subset \{y:d(y,x)\leq r\}$$ so $S$ is bounded. The metric $d|S,$ which is $d$ restricted to the subspace $S$, is a metric for the subspace $S$. Since $S$ is closed and bounded in $M$,it is compact, so $d|S$ is a complete metric for $S.$ So $(x_i)_{i\in N}$ converges to a limit point in $S$ .So $d$ is a complete metric for $M$ because every Cauchy sequence converges to a point in $M$.... The reason every metric $e$ for a compact metrizable space $S$ is a complete metric is that if $(x_i)_{i\in N}$ is an $e$-Cauchy sequence with no limit point in $S$, then $\{x_n:n\geq i\}$ is closed for each $i\in N$, and hence $\{ S\backslash \{x_n:n\geq i\}\;\}_{i\in N}$ is an open cover of $S$ with no finite sub-cover.

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Yes, it follows.

Take a Cauchy sequence $(x_n)$. If $(x_n)$ has a convergent subsequence, we are done, since a Cauchy sequence with a convergent subsequence must itself converge. Therefore, our strategy is to suppose that $(x_n)$ has no convergent subsequence and arrive at a contradiction.

Consider the set of its image $\{x_n\}$. It is easy to verify that $\{x_n\}$ is bounded. Now, suppose $(x_n)$ has no convergent subsequence. Therefore, $\{x_n\}$ must have no limit points (some care must be taken here, due to index-issues. I leave that to you). We then conclude that $\{x_n\}$ is closed. Using the hypothesis on the space, $\{x_n\}$ is compact. Hence, every sequence has a convergent subsequence. Taking the sequence $(x_n)$ yields a contradiction with our assumption.

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  • $\begingroup$ How did you get that $\{x_n\}$ is closed if it had no limit points? $\endgroup$ – Charlie Tian Sep 15 '17 at 13:57
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This problem is really a corollary of the Heine Borel theorem and can be proved in a similar way. Let {x_i} be a Cauchy sequence in M, then we can pick some open ball G(r) , r>0 which contains all the points in the sequence and is bounded. We need to construct a closed set around G(r), then the property of the space M can be applied. Consider a closed ball F(2r), that is a closed ball 'centred' on G(r), but having radius 2r, then clearly G(r) is completely contained in F(2r) and F(2r) is closed, completely contained in M, hence there is a subsequence of {x_i} which converges to a point x of F(2r) and hence to a point in M, hence M is complete. x is unique - if we could find a different subsequence {y_ik} converging to y, we could surround x and y by disjoint open balls, we could find a infinite number of points for which d(x_ij, y_ik) > eps, so would not be Cauchy.

( please look at this Wiki entry for Heine Borel in the paragraph to show "If a set is closed and bounded, then it is compact" for the original proof https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem

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