# If the interior of the boundary of a set is nonempty, then the interior of that set is empty

I was reading the post " A set which the interior of its boundary is not empty ", and I conjectured the following:

Let $$(X, \tau )$$ be a topological space, and let $$A \subseteq X$$. If $$int(\partial A) \neq \phi$$ , then $$int(A)= \phi$$.

(Here $$\phi$$ represents the empty set).

How can I prove this?

• What if $X=\mathbb R$ and $A=(0,\infty)\cup\mathbb Q$? – bof Jun 14 at 3:23
• @bof , thank you. I originally wanted to prove that $cl(int(\partial A)) \subseteq cl(A \cap int(\partial A))$ . Anyhow, now that I see the conjecture is wrong, what should I do with this post? Should I delete it? – evaristegd Jun 14 at 20:39
• @bof , would you mind writing an answer so I can accept it? Thanks – evaristegd Jul 10 at 2:42
• Leave it up. Just because a conjecture is false does not mean it should be forgotten. – DanielWainfleet Jul 10 at 13:40

Your conjecture is false. Let $$X$$ be the real line with the usual topology, and let $$A=(0,\infty)\cup\mathbb Q$$. Then $$\partial A=(-\infty,0]$$, so $$\operatorname{int}\partial A=(-\infty,0)\ne\emptyset$$, but $$\operatorname{int}A=(0,\infty)\ne\emptyset$$.