Recent content by bobsagat57

Maybe I understand now: So, given Q\inN(P,ε), let d(P,Q)< r. Choose z s.t. d(Q,z) < ε - r. So, d(P,z) ≤ d(P,Q) + d(Q,z) < r + (ε - r) = ε. Therefore N(Q,ε-r) \subset N(P,ε) and every ε-neighborhood in a metric space is an open set. Hopefully this is somewhat the right idea.

So, d(P,z) < ε by the triangle inequality, so z is in N(P,ε), so N(Q,ε) is contained in the metric space as well? Not sure if that is correct, and I'm not sure if I'm seeing how this will help to prove the original problem.

I guess I'm still not understanding what I'm supposed to do, I'm sorry.

Sorry, yes I meant for it to say a subset of T And if it helps our definition of an epsilon-neighborhood is: An ε-neighborhood of a point P is N(P,ε) = {Q\inS : d(P,Q) less than ε

An set is open if every point in the set is an interior point. P in T in an interior point if there exists some ε greater than 0 such that N(P,ε) is an element of T And yes N(P,ε) is an open set

Homework Statement In any metric space, every ε-neighborhood N(P,ε) is an open set Homework Equations The Attempt at a Solution I'm completely lost on how to start this proof. I considered assuming that there existed a neighborhood that was closed and show by contradiction but...