Question

X is a limit point iff every open ball around x contains infinitely many points of a

Answer 1

I assume that "x is a limit point of S" means that:

Every open ball that is centered at x contains some point of S other than x.

For metric spaces.

Suppose there is an ϵ>0 such that B(x,ϵ)∩S−{x}={s1,…,sn} is finite. Let δi=d(x,si) be the distance from x so si for each i. Now let δ=12min(δ1,…,δn). Then B(x,δ)∩S−{x}=∅, which means that x cannot be a limit point of S.

By contrapositive, if x is a limit point of S, then every open ball centered at x contains infinitely many points of S.

For topological spaces.

Modifying the proposition by replacing "open ball containing x" with "open set containing x" (i.e., "open neighborhood of x"), the result is false in general but true in T1 (or better) spaces. A space is T1 if and only if for every x and y, x≠y, there are open subsets U and V such that x∈U−V and y∈V−U.

One argues as above: suppose there is an open set U containing x such that U∩S−{x}={s1,…,sn} is finite. For each i, let Ui be an open set containing x but not containing si (the existence of Ui is guaranteed by the separation property). Then V=U∩U1∩⋯∩Un is a finite intersection of open sets, hence open, it contains x, and V∩S−{x}=∅, so x is not a limit point of S.

However, the proposition fails if X is not T1. Let x and y be elements that witness the fact that X is not T1. So either all open sets that contains x also contain y, or all open sets that contain y also contain x. Assume without loss of generality that all open sets that contain x also contain y. Then S={y} has x as a limit point, but every open set that contains x intersects S at a single point, not infinitely many.