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)


2 Answers 2


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.

  • $\begingroup$ proof is not correct mathematically ;assume $x_n$ converges to some $y\in A$ then use uniqueness of limit $\endgroup$
    – Learnmore
    Apr 14, 2015 at 3:49
  • $\begingroup$ @learnmore: what about now? $\endgroup$
    – shalop
    Apr 14, 2015 at 16:27
  • $\begingroup$ @Shalop: Now it’s fine. $\endgroup$ Apr 14, 2015 at 16:51
  • $\begingroup$ its right now @Shalop $\endgroup$
    – Learnmore
    Apr 15, 2015 at 5:06

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 $Y$.


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