# A complete subset of a metric space is closed?

Supposing $A$ is a subset of a metric space $S$, it is simple enough to show that if $S$ is complete and $A$ is closed, that $A$ is complete.

However, without being given that $S$ is complete, what would the proof of the converse be?

(i.e. proving that if $A$ is complete, $A$ is closed)

Suppose $A \subset S$ is complete. To prove that $A$ is closed, it suffices to prove that if $(x_n)$ is a sequence of points of $A$ which converges to $x \in S$, then $x \in A$. So let $x_n$ be such a sequence. Since $x_n$ converges in $S$, it is a Cauchy sequence in $S$. Therefore $(x_n)$ is also a Cauchy sequence in $A$, so by completeness of $A$, there exists some $y \in A$ such that $x_n \to y$. Then by uniqueness of limits, we must have $y=x$, so $x \in A$. Thus $A$ is closed.
• proof is not correct mathematically ;assume $x_n$ converges to some $y\in A$ then use uniqueness of limit Apr 14, 2015 at 3:49
We just need to prove every limit point of $$Y$$ is in $$Y$$. Consider any limit point $$p$$ of $$Y$$. For every positive integer $$n$$, there is a point $$p_n \in Y$$ s.t. $$d(p_n, p) < \frac{1}{n}$$. Obviously, $$\lim\limits_{n\rightarrow\infty}p_n=p$$. Moreover, for any $$\epsilon>0$$, $$d(p_m, p_n) \leq d(p_m, p)+d(p_n, p) < \frac{1}{m}+\frac{1}{n} < \epsilon$$ whenever $$m,n>N=\frac{2}{\epsilon}$$. Thus, $$\{p_n\}$$ is a Cauchy sequence. Since $$(Y, d)$$ is complete, we have $$p \in Y$$. Therefore, $$Y$$ is closed, for every limit point of $$Y$$ is in $$Y$$.