Prove that $\overline{int(Fr(A))}=\overline{A\cap int(Fr(A))}$

Solution 1:

In order to show $\subseteq$, you just need to show that every point $x\in\operatorname{int}(\operatorname{Fr}(A))$ belongs to the closure of $A\cap\operatorname{int}(\operatorname{Fr}(A))$.

This is equivalent to requiring that every open neighborhood of $x$ intersects $A\cap\operatorname{int}(\operatorname{Fr}(A))$.

Take an open neighborhood $U$ of $x$ and consider $U'=U\cap\operatorname{int}(\operatorname{Fr}(A))$. Then $U'$ consists of points of $\operatorname{Fr}(A)$ and $U'$ is an open set so it must contain a point $a\in A$.

Therefore $a\in A\cap U'\subseteq A\cap\operatorname{int}(\operatorname{Fr}(A))$, but also $a\in U$ and so $U\cap(A\cap\operatorname{int}(\operatorname{Fr}(A))\ne\emptyset$ as required.

Thus we proved that $\operatorname{int}(\operatorname{Fr}(A))\subseteq\overline{A\cap\operatorname{int}(\operatorname{Fr}(A))}$ and we're done.