# A metric space is complete when every closed and bounded subset of it is compact

$X$ is a metric space such that every closed and bounded subset of $X$ is compact . Then $X$ is complete.

Let $(x_{n})_{n}$ be a Cauchy sequence of $X$. Now every Cauchy sequence in a metric space is bounded. If $(x_{n})$ is convergent with limit, say, $x_{0}$, then $\{x_{n} : n \}\cup \{x_{0}\}$ is closed, otherwise the set $\{x_{n} : n \}$ is a closed set. If the sequence was convergent there was nothing to do. So when the sequence's convergent is not known, being closed and bounded, $\{x_{n} : n \}$ is a compact set. In metric spaces, compactness is equivalent to sequential compactness. Hence, the sequence $(x_{n})_n$ has a convergence subsequence $(x_{n_{k}})_k$ converging to, say, $x_{0}$. For being Cauchy, $(x_{n}$ also converges to $x_{0}$.

Thus we see that every Cauchy sequence in $X$ is convergent . Hence $X$ is complete.

Is my proof correct? For I have doubts regarding the step where I assumed that the range of the sequence, i.e. $\{x_{n} : n \}$ is closed. This would be true if I were in $\mathbb R$ but can this be assured for an arbitrary metric space? Or does that step require modification in some other respect?

• Why the set $\{x_n\}$ is closed? Sep 22, 2015 at 16:07
• I'm pretty sure the sequence $x_n = 1/n$ does not form a closed subset of $\Bbb R$.
– user98602
Sep 22, 2015 at 16:08
• @MikeMiller : I will edit my question . I had in mind if the sequence were not assumed to be convergent . But $\{{{1}\over {n}}\}\cup \{0\}$ is closed in $\mathbb R$ right $?$ Sep 22, 2015 at 16:13
• Ah, that makes more sense. Important detail; with it, your proof is valid. To show this you need to exploit the fact that it's a Cauchy sequence to show that the only possible accumulation point would be the limit of the sequence, and then the fact that it has no limit can be used to show the sequence forms a closed subset. E: Dominik's answer is much clearer and less painful than mine.
– user98602
Sep 22, 2015 at 16:15
• I'm struggling here. Why must the closure $\overline{\{x_n\}}$ be in X? As far as I can see $X = (0,1)$ and $x_n = 1/n$ provides a counterexample. Every closed subset of this is bounded and compact, yet X is not complete? Nov 25, 2015 at 9:28

The set $\{x_n\}$ is in general not closed, but its closure $\overline{\{x_n\}}$ is. You just need to prove that the closure of a bounded set is again bounded to correct your proof.
Consider a Cauchy sequence $$(x_{k} : k)$$ in $$X$$. Let the limit points of this sequence be denoted $$L$$. I claim that $$L$$ is nonempty and in fact contains exactly one point. Then clearly the only point in $$L$$ is the limit, so it converges.
First note that $$(x_{k}:k)$$ is bounded because $$X$$ is a metric space. As such, there must exist an open ball $$B$$ containing the sequence entirely. The closure $$\mathrm{cl}(B)$$ is compact by assumption, and therefore by sequential compactness $$(x_{k} : k)$$ must have at least a single limit point. In other words, $$L$$ is nonempty.
Next, why can $$L$$ not contain more than one point? Assume that it contains distinct points $$y$$ and $$z$$, and let $$\epsilon = d(y,z)$$. Since they are limit points, for each $$\delta > 0$$ there must be infinitely many points of $$\{ x_{k} : k \}$$ that are within distance $$\delta$$ of $$y$$ and of $$z$$. However, this is impossible if we choose $$\delta < \epsilon/4$$. That is, if $$x_{y}$$ is a point close to $$y$$ and $$x_{z}$$ is a point close to $$z$$, then $$d(x_{y}, x_{z}) > \epsilon - 2\delta = \epsilon / 2$$ by the triangle inequality. That is, no matter how far along in the sequence we look, we can find points of distance $$\epsilon/2$$ away from each other. This violates the Cauchy assumption. Therefore, $$L$$ has exactly one point.