Isolated points in subsets of a topological space For a topological space $X$, is there an equivalent condition under which every subset of $X$ has finitely many isolated points?
 A: Every Noetherian space has this property, and the converse holds if you assume the space is $T_1$.
First, suppose some subset of $X$ has infinitely many isolated points.  Taking a countably infinite subset of those isolated points, we get a subspace $D=\{x_n\}_{n\in\mathbb{N}}\subseteq X$ which is discrete.  Now for each $m$, let $A_m=\overline{\{x_n:n\geq m\}}$.  Note that if $n<m$, then $x_n\not\in A_m$, since there is an open set containing $x_n$ and no other element of $D$.  It follows that the sets $A_m$ are all distinct, and form a strictly decreasing infinite sequence of closed subsets of $X$.  Thus $X$ is not Noetherian.
Conversely, suppose $X$ is $T_1$ and not Noetherian.  Let $A_1\supset A_2\supset\dots$ be a strictly decreasing infinite sequence of closed subsets of $X$.  For each $n$, choose a point $x_n\in A_n\setminus A_{n-1}$.  Then I claim the subset $D=\{x_n\}_{n\in\mathbb{N}}\subseteq X$ is discrete (and in particular has infinitely many isolated points).  Indeed, for each $n$, the set $\{x_1,\dots,x_n\}$ is open in $D$, being the intersection of $D$ with the open set $X\setminus A_{n+1}$.  Since $X$ is $T_1$, it follows that every singleton is open in $D$.
The $T_1$ hypothesis is necessary here.  For instance, let $X=\mathbb{N}$, topologized by saying $\{m\in\mathbb{N}:m\leq n\}$ is open for each $n$.  Then $X$ is not Noetherian, but no subset of $X$ has more than one isolated point.
