Topology, Int(A) is an open set

by rourky
Tags: inta, topology
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,348 By the way, there are a number of different ways of approaching topology. You appear (since you say "there exists r > 0 such that B(x, r) is a subset of Int(A)") to be assuming a topology defined by a metric. I notice Dick refers to |x-y| rather than d(x,y), so he may be assuming you are in R, the real numbers with the usual topology. In more general topology an "open set" is simply a member of the "topology" and the "interior" of a set, A, is defined as the union of all open subsets of A. It would help if you would say explicitely what kind of topology you have, what definitions you are using. In any case, since you are definitely in a metric space, you can use the "triangle" inequality: that $d(x,y)\le d(x,z)+ d(z,y)$ (or $|x-y|\le |x-z|+ |z- y|$. That should be useful.