Show that the intersection of two non-empty open sets is open.
Proof:
Let X and Y be two nonempty open sets. Let p e XnY (n = intersection) We must show that p is an interior point of XnY. Since p e XnY, by def. of intersection, p e X and p e Y.
NOT SURE HOW TO CONTINUE... PLEASE HELP
Well, I don't know what you know or what kind of rigor this requires.
Either the two open sets are disjoint in which case their intersection is empty, one is a subset of the other in which case their intersection is themselves, or neither is true. In this case, this will be an ovaloid with no boundary, which is clearly geometrically "open".
That's all you get till I see some work!
So we are working, at least, in a metric space. (In a general topological space, the fact that the intersection of two open sets is open is part of the definition of "topology".)
If the intersection is empty it is, by definition, open.
If the intersection is not empty, let p be any point in the intersection. Then it is in both X and Y.
Since p is in X and X is open, there exist such that the open ball (neighborhood) centered on p with radius is a subset of X.
Since p is in Y and Y is open, there exist such that the open ball (neighborhood) centered on p with radius is a subset of Y
. Let r be the smaller of and and show that the neighborhood centered on p with radius r is a subset of .