# The closure of $S$ is the set of all limits of sequences of points in $S$ that converge in $E$

My book asks the reader to prove that the closure of $S$ (the intersection of all closed sets in $E$ that contain $S$) is the set of all limits of sequences of points in $S$ that converge in a metric space $E$. I believe I have a proof for this but am not sure of its formality:

① Take some sequence in $S$, $~~ a_1, a_2, a_3, \ldots ~$ that converges to some $a \in E$.

② That is, for any $~\epsilon > 0$, $~~\exists N \in \mathbb{N} \mathrm{~~~s.t.~~} \forall n > N$, $~~ \mathrm{d}(a,a_n) < \epsilon$.

③ So by definition 4 of closed sets on Wolfram MathWorld, for $a$ to be outside of any closed set containing $S$ there would have to exist some ball of center $a$ that is disjoint from $S$.

④ Since $a$ is arbitrarily close to $S$ by ②, such a ball cannot exist, so $a \in \mathrm{closure}(S)$.

⑤ Conversely, any $b \in \mathrm{closure}(S)$ is the limit in $E$ of some convergent sequence in $S$.

Q.E.D.?

Thanks!

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I think you can write ④ clearer. You also need to write that a point in the closure of $S$ is the limit of a sequence of points in $S$. –  Makoto Kato Dec 25 '12 at 19:38
This shows that the set of limits is a subset of the closure. To complete the proof, you also need to show that the closure is a subset of the limit points. –  copper.hat Dec 25 '12 at 19:40
You need to prove ⑤. –  Makoto Kato Dec 25 '12 at 19:41
Alright, thanks! I'll do it –  enthdegree Dec 25 '12 at 19:41

$\newcommand{\cl}{\operatorname{cl}}$You’ve written it in a rather confusing way, but you have essentially half of the proof.
You have a metric space $\langle E,d\rangle$ and a set $S\subseteq E$, and you’re to prove that $\cl S$, the closure of $S$, is the set of limits of convergent sequences in $S$. If $L$ is the set of limits of convergent sequences in $S$, you must show that $\cl S=L$. This typically requires you to show two things: that $\cl S\subseteq L$, and that $L\subseteq\cl S$.
You’ve essentially shown that $L\subseteq\cl S$. To restate your argument a bit more clearly, if $x\notin\cl S$, then there is some $r>0$ such that $B(x,r)\cap S=\varnothing$, where $B(x,r)$ is the open ball of radius $r$ centred at $x$. If $\langle x_n:n\in\Bbb N\rangle$ is any sequence of points of $S$, $B(x,r)$ is clearly an open nbhd of $x$ that contains no point of the sequence, so the sequence does not converge to $x$. That is, no sequence in $S$ converges to $x$, so certainly $x\notin L$.
You still have to show that $\cl S\subseteq L$: if $x$ is in the closure of $S$, then some sequence $\langle x_n:n\in\Bbb N\rangle$ of points of $S$ converges to $x$. If $x\in S$ this is trivial: just let $x_n=x$ for each $n\in\Bbb N$. Thus, you can focus on the case in which $x\in(\cl S)\setminus S$. In this case we know that $B(x,r)\cap S\ne\varnothing$ for each $r>0$. In particular, for each $n\in\Bbb Z^+$ there is a point $x_n\in B\left(x,\frac1n\right)\cap S$; does the sequence $\langle x_n:n\in\Bbb Z^+\rangle$ converge to $x$?
Oh, I see. In your final paragraph you have taken a point $x$ in $\mathrm{cl}S$ and constructed a sequence in $S$ that converges to it. The sequence does indeed converge in $E$, since for $\epsilon > 0 \exists N \in \mathbb{N}$ s.t. $\forall n > N, \mathrm{d}(x,x_n) < \epsilon$. I thought I was able to simply write `⑤ Conversely, [...]' and be done. Thank you for the answer! –  enthdegree Dec 25 '12 at 20:13