1
$\begingroup$

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

$\endgroup$
  • 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.