What does it mean that the exponential topology of a space is T$_1$? If the exponential topology $\exp(X)$ is $T_1$, does that mean that for every $x \in X$ we have that $\{x\}$ is closed in $\exp(X)$ or does that mean that $\{\{x\}\}$ is closed or does it mean something completely different? Unfortunately, this concept is rather confusing to me, so I would appreciate any kind of clarification.
 A: I'll use the definitions and notations from this answer.
Just recall two things:


*

*A topological space $Y$ is T1 if $\{ y \}$ is closed in $Y$ for all $y \in Y$; and

*The points of the space $\exp (X)$ are the closed subsets of $X$.


Therefore $\exp(X)$ is T1 if $\{ F \}$ is closed in $\exp(X)$ for each $F \in \exp(X)$, or rather, $\{ F \}$ is closed in $\exp(X)$ for each closed $F \subseteq X$.
If $X$ itself is T1, then $\exp(X)$ being T1 would imply that $\{ \{ x \} \}$ is closed in $\exp(X)$ for each $x \in X$.

Added. A fairly simply argument gives that $\exp(X)$ is T1 whenever $X$ is.

sketch. Assume $X$ is T1, and let $F,E \subseteq \exp(X)$ be distinct.  There are two cases:
  
  
*
  
*If $F \not\subseteq E$, then $\langle X , X \setminus E \rangle$ is an open neighbourhood of $F$ in $\exp(X)$ not containing $E$.
  
*If $E \not\subseteq F$, then pick $x \in E \setminus F$.  It follows that $\langle X \setminus \{ x \} \rangle$ is an open neighbourhood of $F$ in $\exp(X)$ not containing $E$.
  

