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

Question

$X$ 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.

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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?

Answer 1

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.

Answer 2

Here is a modified version of the proof that I believe fills all the holes:

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.