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Let $M$ be a metric space

I'm asked to prove the statement

"Every closed ball of $M$ is complete $\implies$ $M$ is complete".

My attempt at this is as follows:

Let $\{y_i\}$ be a cauchy sequence in $M$.

Since cauchy sequences are infinite(?) there exists a subsequence of the cauchy sequence which can be enclosed in a closed ball with diameter $d=\text{diam} (x_i,...x_j)$ such that $d<\epsilon$, where $\epsilon$ is arbitrarily close to $0$.

This subsequence converges to a $y\in M$ by our statement and if a subsequence of a cauchy sequence converges to $y$ then the whole cauchy sequence converges to $y$.

Is this correct? If it is not, can I modify it to be correct?. If I can't, how can I prove it?

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Cauchy sequences are bounded, so eventually the Cauchy sequence is in some ball.

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  • $\begingroup$ So is my proof correct? $\endgroup$ – Heuristics Feb 23 at 23:01
  • $\begingroup$ Yes. It's just that you can skip taking the subsequence part. A Cauchy sequence can always be enclosed in a ball of finite radius. $\endgroup$ – TheManWhoNeverSleeps Feb 24 at 0:16
  • $\begingroup$ but isn't it true that some points of the cauchy sequence can be at the border of the metric space and thus can't be enclosed? $\endgroup$ – Heuristics Feb 24 at 17:01
  • $\begingroup$ Usually the way you define a ball of some radius $R$ around a point $x$ in a metric space $M$ is by $B_R(x)=\lbrace y\in M, d(x, y)<R \rbrace$, so even if the points are at the border there is no problem. It might just be that, say you have a metric space where all points satisfy $d(x, y)<R$ for some $R$. Then the entire metric space is in $B_R(x)$ for any $x$. If you take $R_1>R$ then what will happen is $B_{R_1}(x)=B_R(x)$ $\endgroup$ – TheManWhoNeverSleeps Feb 25 at 2:16
  • $\begingroup$ If you look at the answer the other person gave, he gives a proof of why a Cauchy sequence is bounded. $\endgroup$ – TheManWhoNeverSleeps Feb 25 at 2:23
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It's quite simple really: let $(x_n)$ be a Cauchy sequence.

Appply the definition of Cauchy for $\varepsilon=1$ and find $N$ such that $n,m \ge N$ implies $d(x_n,x_m) < 1$. Now defining

$$ R= 2 \max \{ d(x_1,x_N), d(x_2,x_N), \ldots , d(x_{N-1},x_{N}), 1 \}$$

we see that the sequence lies as a whole in the closed ball with radius $R$ and centre $x_N$. So the sequence converges by assumption (it's Cauchy in any subspace it lies in if we keep the metric) in this ball, and thus also in $X$.

If fond of subsequences: the tail (better than subsequence) $(x_n)_{n \ge N}$ lies in $D(x_N, 1)$ (the closed ball); hence converges to $x$ in that ball. And if a subsequence converges so does the whole sequence; so that idea of your original proof idea can be kept.

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  • $\begingroup$ But is my proof wrong? $\endgroup$ – Heuristics Feb 23 at 23:12
  • $\begingroup$ @Heuristics it's sloppy and incomplete. $\endgroup$ – Henno Brandsma Feb 24 at 7:05

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