# Is $\overline F \setminus A$ perfect where $A$ is the set of isolated points of $F$?

Let $X$ be a metric space and $F\subset X$. I have proved that $\overline F \setminus A$ is closed.

I'm having trouble with showing that $\forall x\in \overline F \setminus A$, $x$ is a limit point.

Plus, how do i show $\overline {F \setminus A} = \overline F \setminus A$?

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Apologies. My answer outlined an argument to show that if $x$ is a limit point of $F$, then $x\in\overline{F}\smallsetminus A$. Of course that's not what you were looking for.... I have deleted it. – Cameron Buie Sep 14 '12 at 23:40
I just noticed I didn't mention of which $x$ is a limit point. It's my mistake. Thanks ! – Katlus Sep 16 '12 at 0:34
If I'd paid closer attention to the title, I'd have figured that out by context. Alas, I did not! – Cameron Buie Sep 16 '12 at 2:04

You can’t show that every $x\in(\operatorname{cl}F)\setminus A$ is a limit point of $(\operatorname{cl}F)\setminus A$, because it isn’t true.

Let $$A=\left\{\left\langle n,\frac1m\right\rangle\in\Bbb R^2:n\in\Bbb Z\text{ and }m\in\Bbb Z^+\right\}\;,$$ and let $F=A\cup\{\langle n,0\rangle:n\in\Bbb Z\}$. Then $F$ is a closed subset of $\Bbb R^2$, and $A$ is its set of isolated points, so $(\operatorname{cl}F)\setminus A=F\setminus A=\{\langle n,0\rangle:n\in\Bbb Z\}$. This is a closed, discrete set in $\Bbb R^2$: it has no limit points.

Corrected:

Did you want to show that every $x\in(\operatorname{cl}F)\setminus A$ is a limit point of $F$? That is true. Suppose that some point $x\in(\operatorname{cl}F)\setminus A$ is not a limit point of $F$; then $x$ has an open neighborhood $U$ such that $U\cap F\subseteq\{x\}$. $U\cap F\ne\varnothing$, since $x\in\operatorname{cl}F$, so $U\cap F=\{x\}$. But then $x$ is an isolated point of $F$, i.e., $x\in A$, contradicting the choice of $x\in(\operatorname{cl}F)\setminus A$.

It’s not necessarily true that $\operatorname{cl}(F\setminus A)=(\operatorname{cl}F)\setminus A$; we might have $F=A=\{1/n:n\in\Bbb Z^+\}$ in $\Bbb R$, for example, in which case $F\setminus A=\varnothing$, but $(\operatorname{cl}F)\setminus A=\{0\}$.

It is true, however, that $\operatorname{cl}(F\setminus A)\subseteq(\operatorname{cl}F)\setminus A$, since $(\operatorname{cl}F)\setminus A$ is a closed set containing $F\setminus A$. It is also true that $(\operatorname{cl}F)\setminus A=\operatorname{Lim}F$, the set of limit points of $F$. We saw above, it is true that $(\operatorname{cl}F)\setminus A\subseteq\operatorname{Lim}F$, and the opposite inclusion is clear.

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I think the last equality holds only when $F\setminus A ≠ \emptyset$. – Katlus Sep 16 '12 at 3:57
For example, $\{1/m | m \in \mathbb{Z}^+ \}$. I don't understand where in your argument implies this constraint. Is it because "$x\notin \overline {F\setminus A}$" is vacuously true? – Katlus Sep 16 '12 at 4:07
@Katlus: You’re right; I was too hasty at the end. I’ve corrected it and slightly improved the exposition. – Brian M. Scott Sep 16 '12 at 6:26