Let (X,d) be a metric space and assme that AX. 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 inthat 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?