# Is completeness equivalent to closure in complete metric spaces?

Let $(X, d)$ be a complete metric space and consider a subset $A \subset X$:

• $A$ closed $\implies$ $A$ complete (I know that)
• $A$ complete $\stackrel{(?)}{\implies}$ $A$ closed

I was wondering about the truth of the second implication. If it's true does the following proof is correct?

Proof (Proof by contrapositive, $\neg A$ closed $\implies \neg A$ complete ):
Take $a \in A' \cap A^c$ (exists since A is not closed, closed sets contains their limit points) and consider the following sequence in $A$ such that

$$\large{(x_n)_{n\in\mathbb{N}} \in N_{1/n}(a) }$$

This is a Cauchy sequence in fact $\forall \epsilon>0$ I can set $\large{n_\epsilon := \lceil{\frac{1}{\epsilon}} \rceil}$ to have $d(x_n, x_m)<\epsilon \hspace{4pt} \forall n \ge n_\epsilon$.
But $(x_n)$ does not converge because $a \in A^c. \blacksquare$

Notations: with $A'$ I mean the derived set; with $N_r(p)$ I mean the neighbourhood of $p$ with radius r.

• I think your proof is correct – chan kifung Jul 11 '17 at 23:46
• you can just notice that a convergent sequence is a cauchy sequence, then you ar done – Jens Renders Jul 11 '17 at 23:49
• @JensRenders ok, it's clear, thanks! – Leonardo Vannini Jul 12 '17 at 0:04

Assume $A$ complete.
let prove that $\bar {A}\subset A$. take $a\in \bar {A}$.
then $a=\lim_{n\to+\infty}a_n$ with $a_n\in A$.
$(a_n)$ is Cauchy in $A$ since it converges.
but $A$ is complete then $a\in A$.