IDs 600-699

If it has a closed point… it has… a point.

A space is a subspace of itself.

Any set in any local basis is compact.

Let $x \sim y$ iff they are indistinguishable. The equivalent classes $[x]$ form a basis for a topology which must be finer than $X$ (if $U$ is a nbd of $x$, it must contain all $y \sim x$). It suffices to show each $[x]$ is an open set. But $Y = X \setminus [x]$ is compact, and any points of $Y$ are distinguishable from $x$, so it’s possible to find $x \in U$ and $Y \subseteq V$ nbds with $U \cap V = \emptyset$ (this is analogous to the result that for $T_2$ spaces, any point $x$ and a compact $K$ can be separated)