Prove that S is closed IFF S contains all of its limit points

Question

I've seen the solution of this problem involving the closure of S; however, I solved it differently the first time through. Is this a valid approach?

Proof:

Assume S is closed, then $S^{c}$ must be open. Next, let y $\in$ X be a limit point of S such that y $\in S^{c}$. Thus, there exists a sequence ($S_n$) from S that converges to y. Furthermore, every $S_n$ = S(n) is an element of S and, for every $\epsilon$ > 0, there is an M $\in$ N so that d($s_n$,y) < $\epsilon$ for all n $\ge$ M. However, y $\in S^c$ so there exists some $\delta$ > 0 such that $N_{\delta}^{S^c}$(x) = {x $\in$ X : d(x,y) < $\delta$} $\subset S^c$.

Consider the case $\epsilon = \delta$. Then for some $S_n \in$ S with n $\ge$ M, d($S_n$,y) < $\epsilon = \delta$ and it follows $S_n \in N_{\delta}^{S^c}$(x) for some x $\in S^c$ (thus $S_n \in S^c$). This is a contradiction as $S_n \in S^c$ and $S_n \in S$ but $S^c \cap S = \emptyset$. Therefore, if S is closed and y is any limit point of S, y $\in$ S.

{The reverse statement still must be proved. I made a massive mistake}

Q.E.D.

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Thank you for the help!

Answer 1

i'll put this as an answer because it exceeds the character limit on comments. however it is only an observation.

the idea of your proof is sound, though its application is limited to metric spaces. more generally any open set $U$ is a neighbourhood of any of its points $x \in U$, and obviously $U \cap U^c = \emptyset$. thus there cannot be a sequence in $U^c$ which converges to $x \in U$, since a convergent sequence must eventually lie within any neighbourhood of its limit ($x$). for this reason the apparatus of real arithmetic is not required, and it is advisable always to seek to economize on assumptions. indeed this a one of the main drives for theoretical development in mathematics. on the other hand a proof is a proof! however you should explicitly mention metric spaces as the context. btw a small typo - you have written $s_n$ in place of $s_m$ (line 3 of paragraph 2)