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The statement is the following:

Let X be a topological space and let A be a subset of X. Then $ cl(int(\partial A)) \subseteq cl(A \cap int (\partial A) ) $

Notation: $cl$ means closure, $\partial $ means boundary of a set, and $int $ means interior of a set.
These are the posts I have read so far:
A set which the interior of its boundary is not empty
When is the closure of an intersection equal to the intersection of closures?
If the interior of the boundary of a set is nonempty, then the interior of that set is empty

However I have not been able to prove it yet.

How can I prove that statement?

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  • $\begingroup$ What definitions are you using for closure, interior, and boundary? Since you're working in a general topological space, I assume that the closure of $A$ is defined to be the intersection of all closed sets containing $A$, the interior of $A$ is defined to be the union of all open sets contained in $A$, and the boundary of $A$ is the set of points in the closure but not in the interior of $A$. $\endgroup$
    – Robert Shore
    Jun 14, 2019 at 23:03
  • $\begingroup$ @RobertShore thank you for the interest. Yes, those definitions would work fine. $\endgroup$
    – evaristegd
    Jun 14, 2019 at 23:15
  • $\begingroup$ Unless you show some work you've done on the question, you may soon learn a new meaning of the word 'closure'! Surprisingly, however, I can't find a statement of this requirement in the Help Center. I only see a Meta post, How to ask a homework question?, which reads in part, "Show your work. You should definitely include any partial work you have done." I could have sworn there was more definite guidance on this, especially in view of the severe actions that are sometimes taken (e.g. deletion of questions that people have worked hard to answer). $\endgroup$
    – Calum Gilhooley
    Jun 14, 2019 at 23:31
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    $\begingroup$ @CalumGilhooley , Thank you for the heads up. I will see how can I edit my question. $\endgroup$
    – evaristegd
    Jun 14, 2019 at 23:33

1 Answer 1

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To clarify:

  • A neighbourhood of $a$ is an open set containing $a$.
  • The interior of a set $A$ is the set of points which have a neighbourhood $B$ which is a subset of $A$.
  • The closure of $A$ is the set of points for which all neighbourhoods intersect $A$.

The boundary is the set of points whose neighbourhoods always intersect $A$, but do not lie in the interior of $A$.

Its interior is the set of points who have a neighbourhood which lies entirely in the boundary. As boundary points lie in the closure, this neighbourhood must also intersect $A$.

The closure of the interior of the boundary are the points for which every neighbourhood intersects the interior of the boundary.

Now assume $a \in cl(int(\partial A ))$. Let $U_a$ be one of its neighbourhoods; it intersects $int(\partial A)$ in at least one point $y$. Let $V_y$ be a neighbourhood of $y$ which lies completely in the interior boundary. The intersection $W_y := U_a \cap V_y$ is also a neighbourhood of $y$ which lies completely in the interior boundary; furthermore, it is a subset of $U_a$. As argued before, $W_y$ must contain a point $z$ of $A$. This point $z$ is also a point of the interior boundary. Therefore, $z \in A \cap int(\partial A)$.

We have just proven that every neighbourhood of $a$ intersects a point $z \in A \cap int(\partial A)$. By definition it then follows that $$cl(int(\partial A )) \subset cl( A \cap int(\partial A) )$$

which concludes the proof.

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    $\begingroup$ That was impressive, thank you. One question, when you say "interior boundary", you mean "interior of the boundary", right? $\endgroup$
    – evaristegd
    Jun 14, 2019 at 23:59
  • $\begingroup$ Yes, I was using shorthand! $\endgroup$
    – Alexander Geldhof
    Jun 15, 2019 at 0:05
  • $\begingroup$ Is that a widely used shorthand? Can I use it in other places too? $\endgroup$
    – evaristegd
    Jun 15, 2019 at 0:11
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    $\begingroup$ I don't think it's a convention, although most people will understand what it means. $\endgroup$
    – Alexander Geldhof
    Jun 15, 2019 at 0:15
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    $\begingroup$ To the proposer: Since $U\subset V$ implies $cl(U)\subset cl(V),$ a corollary here is that $cl(int (\partial A))=cl(int (A\cap \partial A))$... BTW another notation for $\partial A$ or $\partial (A)$ is Fr($A$)... (for Frontier). $\endgroup$
    – DanielWainfleet
    Jun 15, 2019 at 0:22

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