Prove that $A^o \cap B^o = (A \cap B)^o$ Prove that $A^o \cap B^o = (A \cap B)^o$.
I am able to prove that if $x \in A^o \cap B^o$, then $x \in (A \cap B)^o$. But this only implies that $A^o \cap B^o \subset (A \cap B)^o$. To show that $A^o \cap B^o = (A \cap B)^o$, I must now show that $A^o \cap B^o \supset (A \cap B)^o$. In other words, let $x \in (A \cap B)^o$, show that $x \in A^o \cap B^o$.
To me, however, it seems that once I prove it one way, then to prove it the other way, I simply "unwind" the previous proof. There seems to be nothing novel about it, which makes me feel that I am doing something wrong. Is that the way these proofs typically go?
Here's what I have:
Let $x \in A^o \cap B^o$.
Then $x \in A^o$ and $x \in B^o$.
Then $\exists$ $\alpha > 0$ and $\beta >0$ s.t. $B_{\alpha}(x) \subset A$ and $B_{\beta}(x) \subset B$. Take $r = \text{min}\{\alpha,\beta\}$. Then $B_r(x) \subset A$ and $B_r(x) \subset B$. Therefore, $B_r(x) \subset A \cap B$. Thus, $x \in (A \cap B)^o$.
Again, I see nothing novel about the reverse direction. 
 A: Well, you wouldn't strictly reverse the steps because in the first direction, you took the minimum of two radii, and how can you reverse a "taking minimum" step?
But you are right, the proof in the reverse direction is just as easy and uses most of the same steps.  Still, you should write down these steps to convince yourself.
If $x \in (A \cap B)^{o}$, then there is some $\epsilon > 0$ such that $B_{\epsilon}(x) \subseteq A \cap B$.  But that means $B_{\epsilon}(x) \subseteq A$ and $B_{\epsilon}(x) \subseteq B$ since $A\cap B \subseteq A$ and $A \cap B \subseteq B$.  Thus, $x \in A^{o}$ and $x \in B^{o}$, so $x \in A^{o} \cap B^{o}$, as desired.
A: I think that's a general theme, for most "basic" results in most topics. We cut our teeth doing a lot of symbol-pushing, to build up a reserve of little lemmas and theorems that will help us more elegantly achieve more results. Once we have an arsenal of lemmas/theorems, we can start using them to prove more intermediate things, and don't need to resort to fighting in the trenches quite as much.
For example, proving De Morgan's laws in set theory is pretty mundane, but once we have it, we can use it, avoiding looking at elements of sets. In group theory, proving very basic results is again a lot of symbol-chasing, but as we progress, we chain these results together, to pay less attention to the group elements themselves.
So, in my opinion, the fact that "the intersection of interiors is the interior of the intersection" is a very basic result, and unsurprisingly has a rather uninspired proof. However, it will open doors in the future, to more abstract and elegant proofs, perhaps not using open balls, etc, quite as much. 
A: Let $x∈(A∩B)^0$. So there exists a neighborhood $N_r(x)$ around $x$ which is a subset of $(A∩B)$. So $N_r(x)$ is a subset of both $A$ and $B$. So $x∈A^0$ and $x∈B^0$. This implies $x∈A^0∩B^0$. Hence Proved.
A: Let x∈(A∩B)0. So there exists a neighborhood Nr(x) around x which is a subset of (A∩B). So Nr(x) is a subset of both A and B. So x∈A0 and x∈B0. This implies x∈A0∩B0. Hence Proved.
