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?

  • 6
    $\begingroup$ What if $X=\mathbb R$ and $A=(0,\infty)\cup\mathbb Q$? $\endgroup$ – bof Jun 14 at 3:23
  • $\begingroup$ @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? $\endgroup$ – evaristegd Jun 14 at 20:39
  • $\begingroup$ @bof , would you mind writing an answer so I can accept it? Thanks $\endgroup$ – evaristegd Jul 10 at 2:42
  • $\begingroup$ Leave it up. Just because a conjecture is false does not mean it should be forgotten. $\endgroup$ – 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 $\operatorname{int}A=(0,\infty)\ne\emptyset$.


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