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?

  • 6
    $\begingroup$ 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 $B (x,r) \cap F \subset A \Longrightarrow x \in A$). $\endgroup$ – user642796 Sep 16 '12 at 5:02
  • $\begingroup$ 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. $\endgroup$ – Alex Becker Sep 16 '12 at 5:07
  • $\begingroup$ @Alex Yes. Edited $\endgroup$ – Katlus Sep 16 '12 at 5:12
  • $\begingroup$ The title is not so much ambiguous, but unspecific. $\endgroup$ – joriki Sep 16 '12 at 5:14
  • $\begingroup$ "a set of all isolated points of $F$" is ungrammatical. There is only one such set, so the article has to be definite. $\endgroup$ – joriki Sep 16 '12 at 5:16

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,


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