(Dis)proving $(\operatorname{Int}(A))'=A'$.

Question

Let $A'$ denote the set of the accumulation points of a set $A\subseteq\Bbb R^n$ and let $\operatorname{Int}(A)$ denote the interior of $A$, that is $$(\operatorname{Int}(A))'=\bigcup_{U\subset A,\\\text{U is open}}U.$$ Does the following hold: $(\operatorname{Int}(A))'=A'?$

I think it does not hold.

I considered $n=1$ and took $A=(-1,-1/2)\cup\left\{\frac1m:m\in\Bbb N\right\}.$

Then, $\operatorname{Int}(A)=\left(-1,-1/2\right),(\operatorname{Int}(A))'=[-1,-1/2],$ while $A'=[-1,-1/2]\cup\{0\}.$

Therefore, I believe $(\operatorname{Int}(A))'\subseteq A'$

May I ask for verification?

Answer 1

Yes. it is correct. But it is much simpler to take $A=\left\{\frac1n\mid n\in\Bbb N\right\}$. Then $A'=\{0\}$ and $(\operatorname{Int} A)'=\emptyset'=\emptyset$.

Answer 2

Looks OK, but purely technically, it's a little awkward.

You didn't prove that $(\mathrm{Int}(A))'\subseteq A'$. What you proved is that $(\mathrm{Int}(A))'\neq A'$.

It is also true that $(\mathrm{Int}(A))'\subseteq A'$, but that's not part of your task.