# A set F is closed if and only if every convergent sequence in F converges to a limit in F

$$F \subseteq X$$, 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.

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

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