Prove $b(int(A)) \subset b(A) $

where $b$ is boundary, $int$ is interior and $ext$ is exterior

if $x \notin b(A)$ then $ x \in int(A) \cup ext(A) $

if $x \in int(A) \to x \in int(int(A))$ because $int(int(A)) \subset int(A)$ namely $x \notin b(int(A))$

I don't known if it's right and complete. Thanks

  • 1
    $\begingroup$ What is the exterior of a space? And replacing arrows for english words such as 'then' makes it harder to read,. $\endgroup$ – bbnkttp May 28 '15 at 18:38
  • $\begingroup$ Excuse me, now it's OK and thank you very much! $\endgroup$ – Ryoma May 28 '15 at 19:48

$b(int(A))=cl(int(A))\setminus int(int(A))=cl(int(A))\setminus int(A)$ because $int(A)$ is open. Now $int(A)\subset A$, so $cl(int(A))\subset cl(A)$ and thus $$cl(int(A))\setminus int(A)=b(int(A))\subset cl(A)\setminus int(A)=b(A)$$


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