Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\overline {F \setminus A} = \overline F \setminus A$?

share|cite|improve this question
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
up vote 1 down vote accepted

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.


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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.