# Is $\DeclareMathOperator{int}{int}\int(A\cup B)=\int(A)\cup \int(B)$?

I have posted a counterexample given in a solution below, but when I attempted the problem I did something different.

My Attempt

Take complements of both sides

$$\int(A\cup B)\stackrel{?}{=}\int(A)\cup \int(B) \\ \big(\int(A\cup B)\big)^c\stackrel{?}{=}\big(\int(A)\cup \int(B)\big)^c \\ \overline{ (A\cup B)^c}\stackrel{?}{=}\overline{(A)^c}\cap \overline{(B)^c}$$ I got this step from this answer Prove that the closure of complement, is the complement of the interior $$\overline{ (A)^c\cap (B)^c}\stackrel{?}{=}\overline{(A)^c}\cap \overline{(B)^c}$$

My question

Could I prove from where I stopped that the two sides are not equal? I tried to sketch a ven diagram but both sides seem to give the same intersected set.

Counterexample

• From your example, you can get a counterexample to $\overline{A\cap B}=\overline A\cap\overline B$ too. – Lord Shark the Unknown Jul 28 '18 at 11:48
• Let $A=(0,1)$ and $B=(1,2)$ thus $\overline{A\cap B}=\emptyset$ and $\overline A\cap\overline B=1$ – john fowles Jul 28 '18 at 12:01

int A$\cup$B = int A $\cup$ int B.
The correct answer to, "does $\operatorname{int}(A\cup B)$ equal $\operatorname{int}(A)\cup\operatorname{int}(B)$?" is:
Perhaps yes, perhaps no. That depends on which sets $A$ and $B$ are.
So you have no hope of starting out from nothing and then prove that $\operatorname{int}(A\cup B) \ne \operatorname{int}(A)\cup\operatorname{int}(B)$. That conclusion would be as wrong as it is to claim $\operatorname{int}(A\cup B) = \operatorname{int}(A)\cup\operatorname{int}(B)$.