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

Question

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?

@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 '19 at 20:39
@bof , would you mind writing an answer so I can accept it? Thanks — evaristegd, Jul 10 '19 at 02:42
Leave it up. Just because a conjecture is false does not mean it should be forgotten. — DanielWainfleet, Jul 10 '19 at 13:40

score 3 · Accepted Answer · edited Apr 27 '23 at 06:31

3

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 $\operatorname{int}A=(0,\infty)\ne\emptyset$.

edited Apr 27 '23 at 06:31

Greg Martin

78,820

answered Jul 10 '19 at 03:01

bof

78,265

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

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