Closure and convergence of sequences

Question

I have a question about closure and convergence of sequences.

The theorem is:

(a) For any set $S$ $\subseteq$ $R^n$, a point $a$ in $R^n$ belongs to $cl(S)$ if and only if $a$ is the limit of a sequence {$x_k$} in $S$.

(b) A set $S$ $\subseteq$ $R^n$ is closed if and only if every convergent sequence of points in $S$ has its limit in $S$.

I don't understand the proof from "if $a$ is the limit of a sequence {$x_k$} in $S$" to "a point $a$ in $R^n$ belongs to $cl(S)$". The proof reads:

Assume that $a =\lim_{x\to\infty}x_k$ for some sequence {$x_k$} in $S$. We claim that $a \in cl(S)$. Indeed, for any $r>0$ we know that $x_k \in B(a;r)$ for all large enough $k$. Since $x_k$ also belongs to $S$, it follows that $B(a;r) \cap S \neq \emptyset$. Hence $a \in S$.

I know that $a$ is a boundary point and therefore in $cl(S)$ if $B(a;r) \cap S \neq \emptyset$. But I thought this condition must be true for all $B(a;r)$. In this case, it is only true when $k$ is sufficiently large so that $x_k \in B(a;r)$. Could someone let me know where I am going wrong?

Thank you for your help.

Answer 1

$$a\in cl(S)\iff$$ $$( \forall r>0) \;\; S\cap B(a,r)\ne \emptyset$$

$$\iff (\forall n>0) \;\; (\exists s_n\in S) \; : d(a,s_n)<\frac 1n$$

$$\iff a=\lim_{n\to+\infty}s_n$$

Answer 2

Fact (a) is about individual points of $\operatorname{cl}(S)$: how can we tell from sequences convergence that some point $x$ lies in $\operatorname{cl}(S)$? The answer is: precisely when we can find a sequence $(x_n)_n$, where all $x_n \in S$ and such that $x_n \to x$.

One direction (the one you ask about) is about the "easy" (i.e. not even specific for metric spaces or subsets of $\mathbb{R}^n$) direction. We have in our hand a sequence $(x_n)$ of points of $S$ that converges to $x$. We take an open ball $B$ (or open set, in a general space) containing $x$ and by the definition of convergence we know that there is some $N$ such that for all $n \ge N$ we have $x_n \in B$. But then $x_N$ (but also $x_{N+1}$ or $x_{2N}$ etc. etc.) all lie in $B \cap S$, so every open ball around $x$ intersects $S$, which means that $x \in \operatorname{cl}(S)$.

(In your proof you use $a$ instead of $x$, no problem, but it also states at the end that $a \in S$, which need not be; we just know that $a \in \operatorname{cl}(S)$, which was also the initial claim).

Every ball $B(a,r)$ contains at least one $x_n$ (different $n$ for different $r$) such that $x_n \in B(a,r) \cap S \neq \emptyset$ and that's enough to show $a$ (or in my case $x$) to be in the closure of $S$.