Let (X,d) be a metric space and assme that A X. Prove that iff there exists a sequence in A converging to x.
Here's a rushed proof. you can streamline it.
Assume there is a sequence of points in that converges to . Then lies in any closed set containing . That is, .
Conversely, assume . Then the "ball" of radius centered at will contain points in . For each , pick one of these points and call it , then the sequence converges to
I assume you know how to describe a "ball" in a metric space. we derive this from the metric
gee, i don't know of a theorem that says we can do this. it just makes sense that we can based on our definitions.
what does it mean to be in ? Well, it means that if then any "open ball" centered at will contain points in . what does "open ball" mean in a metric space? the "open ball" centered at x is the set . so being in the closure means that this set is nonempty. if the set is nonempty, what's stopping us from picking elements in the set?