1. You seem to be having trouble with the basic definition of what it means to be an interior point.

Since p is an interior point of A, there is an open set $O\subset A$ with $p\in O$. Similarly there is an open set $U\subseteq B$ with $p\in U$. Then $O\cap U\subseteq A\cap B$ is an open set with $p\in O\cap U$. This shows that $p\in (A\cap B)^o$.

2. @ alice8675309
Here is a way to think about both the interior of a set and the closure of a set.
The interior of a set is the ‘largest’ open subset of the set.
The closure of a set is the ‘smallest closed set containing the set as a subset.

Here is another suggestion. If you can find the small book Elementary Theory of Metric Spaces by Robert B Reisel, then that would be a great introduction for you. It is written as a self-study or Moore-style class.

3. Thanks for the help guys and Plato, thanks for suggesting the book. I'm obviously struggling with this stuff, but I HAVE to take it. So i appreciate all the patients and helpful hints/answers and i apologize for posting so much. Thanks

4. No need to apologize. Just keep trying things and posting your attempted solutions. Eventually after working hard to prove these things on your own things will click.

5. Originally Posted by alice8675309
What would the formal proof look like for this problem?

Prove that the interior of an intersection is the intersection of the interiors:

(A∩B)º =Aº∩Bº
Any point p in A^B must be a common point of A and B. There are two possibilities:
p is an interior point of both A and B, or
p is a boundary point of A and or B.

If p is an interior point of A & B, there must be a neighborhood of p containing only points common to A and B, and this neighborhood becomes the neighborhood of p in A^B and so is an interior point of A^B.

If p is a boundary point of, say, A, every neighborhood of p will contain points in A and not in A and so will have some points in common and not in common with the same point in B. Thus p is a boundary point of A^B.

Therefore the only interior points of A^B will be common interior points of A & B: Ao^Bo.

I cannot give a simpler explanation.

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# prove that intersection of interior is interior of intersection

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