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$F \subseteq X$, where $X$ is a metric space.

Then $F$ is closed if and only if every convergent sequence in F converges to a limit in $F$.

Attempt

$ \implies$ If $F$ is closed, $F= \overline F$, but $\overline F=F \cup F'$ where $F'$ is the set of all accumulation points of $F$.

Thus, $F' \subseteq F$ which means that $F$ contains all of its limit points which means that every sequence in F converges to a limit that is in F.

Is this a valid proof for this direction?

Edit After reading the comments, I should have written every convergent sequence in F converges to a limit in F.

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  • $\begingroup$ You may find your answer here: math.stackexchange.com/questions/882876/… $\endgroup$ – Aniruddha Deshmukh Nov 29 '18 at 4:20
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    $\begingroup$ If F is the reals, your conclusion that every sequence within F converges to a limit of F, is false. $\endgroup$ – William Elliot Nov 29 '18 at 4:26
  • $\begingroup$ So if F contains all its limit points, it does not imply that every sequence in F converges to a limit in F? $\endgroup$ – Snop D. Nov 29 '18 at 4:31
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    $\begingroup$ @SnopD.: You can say for every point in $F'$ there exist a sequence in $F$ such that it converges to that point. $\endgroup$ – Yadati Kiran Nov 29 '18 at 4:36
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That's a valid proof (but you wrote every sequence where you should have wrote every convergent sequence).

For the converse: If every convergent sequence in $A$ converges to a point in $A$, then $A'\subset A$. Thus A is closed.

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