Proof: Interior of two intersecting sets equals the intersecting interiors of the sets seperately Prove that the following is true:
$$\operatorname{int}A∩\operatorname{int}B=\operatorname{int}(A∩B)$$
I'm in general really bad at writing proofs. I do not know where to start even when looking at the definitions given in the textbook. I am well aware of the definition of the interior of sets, but I can't apply it in a meaningful way. My attempt at writing that proof is laughable since I'm doing the proof backwards.
We know that $int(A)\subset A$, thus we have:
$$\operatorname{int}(A∩B)=\operatorname{int}A∩\operatorname{int}(B)\\
\operatorname{int}(A\cap B)\subseteq A \cap B$$
The latter is per definition the interior of the set $A \cap B$.
Q.E.D.
 A: I can't make any sense of your proof - it seems like you use what you are trying to prove in the first line.
Here are some hints:

*

*You will have to use the definition of interior. Since this looks like an introductory proofs class, I assume that you are using the definition: $x$ is in the interior of $A$ if there exists an open ball centered at $x$ that is contained in $A$.


*The easiest way to show that two sets are equal is usually to show that the first is contained in the second, and then show that the second is contained in the first. So, you can first try to show $\operatorname{int}(A \cap B) \subseteq \operatorname{int} A \cap \operatorname{int}B$, and then show the opposite inclusion.
A: Doing a proof backwards does not have to be a bad idea. It's often a useful tool to use when trying to figure out how to prove it, only to work it out backwards later. In your case however, you use what you're trying to prove as a given.
Try proving that
$$\operatorname{int}(A∩B)\subseteq\operatorname{int}A∩\operatorname{int}B$$
and that
$$\operatorname{int}A∩\operatorname{int}B\subseteq\operatorname{int}(A∩B)$$
