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.

  • $\begingroup$ I think your proof is correct $\endgroup$ – chan kifung Jul 11 '17 at 23:46
  • 1
    $\begingroup$ you can just notice that a convergent sequence is a cauchy sequence, then you ar done $\endgroup$ – Jens Renders Jul 11 '17 at 23:49
  • $\begingroup$ @JensRenders ok, it's clear, thanks! $\endgroup$ – 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 $.

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