We all know that
$(A\cap B)^\mathrm{o}=A^\mathrm{o} \cap B^\mathrm{o}$, where $A,B \subset X$ which is a metric space.
The proof is not also difficult, but actually I cannot visualize or feel intuitively that this should happen.
Can someone provide me with an intuition that this is obvious even intuitively?
Some diagramatic representation is also welcome.
Intuitively: This is intuitive so you can't quote or use it.
A set has an inside and an edge. And the inside of a set is all the set minus the edge.
So $A\cap B$ is 1)Everything in both the inside of $A$ and then inside of $B$ plus 2) Everything in the inside of $A$ and the edge of $B$ and 3) Everything on the edge of $A$ and the inside of $B$ plus 4) everything on the edge of $A$ and on the edge of $B$.
Now if $x$ is on the edge of $A$ then it's right next to things not in $A$ so it is right next to things not in $A\cap B$. So $x$ is either on the edge of $A\cap B$ or not in $A\cap B$ at all; it certainly can't be that $x$ is on the inside of $A\cap B$.
And the same goes for $y$ on the edge of $B$.
In other words if any point is on the edge of $A$ or $B$ it's not on the inside of $A\cap B$.
So $(A\cap B)^\circ$, which is $A\cap B$ minus the edges of $A\cap B$ must strip away the edges of $A$ and $B$.
So $(A\cap B)^\circ$ can contain 1)Everything in both the inside of $A$ and then inside of $B$ but not 2) Everything in the inside of $A$ and the edge of $B$ nor 3) Everything on the edge of $A$ and the inside of $B$ and not 4) everything on the edge of $A$ and on the edge of $B$.
So $(A\cap B)^\circ$ can only contain points in $A^\circ \cap B^\circ$. But does it contain all of them? If $y \in A^\circ \cap B^\circ$ then everything immediately around it is in $A$ and everything immediately around it is in $B$. So everything immediately around it is in both $A$ and in $B$.
So $(A\cap B)^\circ$ consists only of and all of the points in $A^\circ \cap B^\circ$.