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

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

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    $\begingroup$ 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

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

enter image description here

  • 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

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


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


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