$X$ 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?
Best Answer
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.