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First of all, sorry for the unspecific title, but can't think of a better one..

Here's an argument.

Let $F$ be a set in a metric space $X$ and $A$ be the set of all isolated points of $F$. Then, $\overline F \setminus A$ is closed and is the set of all limit points of $F$. Since $F\setminus A \subset \overline F \setminus A$, $\overline {F\setminus A} \subset \overline F \setminus A$.

Now, fix $x\in \overline F \setminus A$. Suppose $x\notin \overline {F\setminus A}$. Then there exists $r>0$ such that $B(x,r)\cap (F\setminus A)\setminus \{x\} = \emptyset$. Since $x\notin F\setminus A$, $B(x,r)\cap (F\setminus A) = \emptyset$.

Since $x\in \overline F \setminus A$, $x$ is a limit point of $F$. Thus, $B(x,r)\cap F ≠ \emptyset$. Thus, $B(x,r)\cap F \subset A$ hence $x\in A$, which is a contradiction. Thus $\overline F \setminus A \subset \overline {F\setminus A}$.

Consequently, $\overline F \setminus A = \overline {F\setminus A}$. Q.E.D.

Consider a set $F=\{1/m \in \mathbb{R} | m \in \mathbb{N}\}$. This shows equality does not hold. Where in the argument is wrong and what kind of constraint should I mention?

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    I'll just find the first statement which is false for $x = 0$ and $F = A = \{ 1/m : m \in \mathbb{N} \}$: I believe this occurs when you claim that $x \in A$. Upon inspection, I think you make this inference by erroneously thinking that $x \in F$ (the step $B (x,r) \cap F \subset A \Longrightarrow x \in A$).2012-09-16
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    Do you mean that $A$ is the set of *all* isolated points of $F$? Otherwise the line "Since $x\in \overline F\setminus A$, $x$ is a limit point of $F$" is false.2012-09-16
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    @Alex Yes. Edited2012-09-16
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    The title is not so much ambiguous, but unspecific.2012-09-16
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    "a set of all isolated points of $F$" is ungrammatical. There is only one such set, so the article has to be definite.2012-09-16
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    @Arthur You are correct. However, does the equality still not hold when $F\setminus A ≠ \emptyset$?. I think it is likely to hold.2012-09-16
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    @joriki Edited!2012-09-16
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    @Katlus: Take $F = \{ 1/m : m \in \mathbb{N} \} \cup [4842,5623]$. Then $A = \{ 1/m : m \in \mathbb{N} \}$, and we have that $\overline{F} \setminus A = \{ 0 \} \cup [4842,5632]$, while $\overline{F \setminus A } = [4842,5632]$. To get something close to the equality you desire I think you might need to ensure that the set $A$ is closed discrete (meaning that every point has a neighbourhood containing at most one element of $A$) but this seems to trivialise things.2012-09-16
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    @Arthur Great. Thank you!2012-09-16
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    "Now fix $x \in \overline F \backslash A$. Suppose $x \notin \overline{F \backslash A}$. Then there exists $r > 0$ such that $B(x,r) \cap (F \backslash A) = \varnothing$." In other words, you don't need to remove $x$ from there. I don't know why you do it in two steps. The set $\overline E$ is often defined as $\{ x \in X \, | \, \forall r > 0, \quad B(x,r) \cap E \neq \varnothing \}$, so the negation of that is not a two-step thing. But if you have chosen another definition it's fine too.2012-09-16

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Your argument goes wrong at

"Thus, $B(x,r) \cap F \subset A$, hence $x \in A$, which is a contradiction."

Why would $x$ need to belong to $A$? Because it belongs to $B(x,r) \cap F$? I don't see why it would. Sure, $x$ belongs to the ball, but you've only shown that the intersection $B(x,r) \cap F$ was non-empty, not that it contained $x$. It is precisely what goes wrong in your example with $\{ 1/m \}$ ; you assume in the proof that $x \in \overline F \backslash A$, but that does not imply anywhere that $x \in F$. $0$ is precisely one of those limit points that are not in $F$.

If you don't see my point, notice that $0 \notin A$ in your example, but all the other assumptions you make (beginning with $x = 0$) work.

Hope that helps,