0
$\begingroup$

let $A,B$ be subsets of topological space $X$. Does it follow

$\partial (A \cap B) = \partial A \cap \partial B $ ?

$\endgroup$
  • 5
    $\begingroup$ No. There is absolutely no reason to think this would be the case. Work on your intuition. $\endgroup$ – TBrendle Oct 21 '13 at 18:44
  • $\begingroup$ Try to find counterexamples for both inclusions. $\endgroup$ – Stefan Hamcke Oct 21 '13 at 18:45
15
$\begingroup$

On these types of problems it's always best to draw picture when there's an opportunity to!

enter image description here

$\endgroup$
  • 1
    $\begingroup$ ms paint XD /// $\endgroup$ – user153330 Feb 20 '16 at 20:04
  • $\begingroup$ @user153330 yes, ms paint, I would use Paint 3D that comes with windows nowadays. $\endgroup$ – BananaCats Category Theory App Oct 29 '18 at 21:44
3
$\begingroup$

Let $X = \mathbb{R}$, $A = \mathbb{Q}$, and $B = \mathbb{R} \setminus \mathbb{Q}$ be the rationals and irrationals. What do you get?

$\endgroup$
2
$\begingroup$

Consider $X=\mathbb{R}$, with the usual topology. If $A=[-1,1]$ and $B=[-2,2]$, then $\partial(A\cap B)=\partial([-1,1])=\{-1,1\}$; $\partial A=\{-1,1\}$; and $\partial B=\{-2,2\}$.

But certainly, $\{-1,1\}\neq \{-1,1\}\cap\{-2,2\}$.

$\endgroup$

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.