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Definition) A point $x\in X$ is a limit point of S if every ball $B(x;r)$ contains infinitely many points from $S$.

A point $x\in X$ is called an isolated point of S if $\exists r > 0$ such that $B(x;r)\cap S = \{x\}$.

Problem) $S\subset X^{metric}$. Let $S_1$ be the set of limit points of S. Let $S_2$ be the set of isolated points of S.

Show that $\bar S = S_1\cup S_2$ and $S_1 \cap S_2 = \emptyset$.

Suppose x is not a limit point of S. Then $\exists \epsilon > 0$ such that $B(x;\epsilon)$ contains only finitely points of S. By shrinking $\epsilon$, we can assume that $B(x;\epsilon)$ contains no points of S other than possibly x itself. This means x is an isolated point. So $S_1\cap S_2 = \emptyset$. (Until now, is this right?)

Then I confused about limit point and closure. Originally I thought $\bar S=S_1$. But it is wrong and I lost my way.

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  • $\begingroup$ In your start of proof, you have not considered the possibility $x\notin S$. In that case, your argument shows that $x\notin \overline{S}$. If you look carefully, you see that $X = S_1 \cup S_2 \cup (X\setminus\overline{S})$ from your argument. Note also that $\overline{S} = S \cup S_1$. $\endgroup$ – Daniel Fischer Apr 16 '14 at 13:10
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To see that $S_1 \cap S_2 = \emptyset$ you only need to remark that an isolated point of $S$ (i.e. a point in $S_2$) is certainly not a limit point of $S$, i.e. a point of $S_1$.

Your other argument goes a long way in proving the equality $\overline{S} = S_1 \cup S_2$, though.

First, $S_1 \subset \overline{S}$, trivially, and $S_2 \subset S$, so certainly $S_2 \subset \overline{S}$, to take care of the inclusion $S_1 \cup S_2 \subset \overline{S}$.

Now (to see $\overline{S} \subset S_1 \cup S_2$): if $x \in \overline{S}$, then it can be a limit point of $S$, and we'd be done, or there is a ball $B(x,r)$ containing only finitely many point of $S$, say $s_1,\ldots, s_n$. If $x$ is not one of the $s_i$, we'd take $r'$ smaller than $r$ and all $d(x, s_i)$ and have a ball $B(x,r')$ around $x$ missing $S$, which cannot be as $x \in \overline{S}$. So $x = s_i$ for some $i$. But now take $r'$ smaller than $r$ and all $d(x, s_j)$, where $j \neq i$, and we have $B(x, r') \cap S = \{x\}$, so $x \in S_2$.

In short, your argument (slightly extended) shows that $x \in \overline{S}$ and $x \notin S_1$, then $x \in S_2$, which shows the required other inclusion.

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In a metric space $X$, the closure of $S$ is the set of all points $x \in X$ such that $x_n \to x$ for some sequence $\{x_n\}_n$ of points from $S$. Be careful, the constant sequence is allowed, and that's why you need isolated points (belonging to $S$, of course).

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