Do spaces where all singletons are closed have a name? I know for example that $\mathbb R$ is one of these spaces since the complement of a singleton $\{x\}$ is $(-\infty,x)\cup (x,\infty)$ which is open. I know also that a space where all singletons are open is a discrete space since if every singleton is open in $X$ then this would imply that every subset of $X$ is open in $X$. Thank you for your help!!
[Math] Spaces where all singletons are closed
general-topology
Related Solutions
Let $F$ be an arbitrary subset of $X$ where $X$ is equipped with discrete topology.
As you said: in a discrete topological space all singletons are open.
As you said: arbitrary unions of singletons are open so $F^c=\bigcup_{x\in F^c}\{x\}$ is an open set.
(You don't even need this subroute: in a discrete space all sets are open by definition)
Then its complement $F$ is a closed set.
A set is irreducible if it cannot be written as the union of two proper closed subsets. The only proper subset of a singleton is the empty set, so a singleton cannot be written as the union of two proper closed subsets and is therefore irreducible.
Suppose that $X$ is Hausdorff, and let $S\subseteq X$ contain at least two points, $x$ and $y$. Since $X$ is Hausdorff, there are open sets $U$ and $V$ in $X$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$. But then $S\setminus U$ and $S\setminus V$ are relatively closed proper subsets of $S$ whose union is $S$, so $S$ is not irreducible. Thus, the only irreducible subsets of $X$ are its singletons, which are therefore also the irreducible components of $X$.
You can see how the hypothesis that $X$ is Hausdorff is used in the proof. It is not a necessary condition, however: the converse of the theorem is false, because there are non-Hausdorff spaces in which the irreducible components are the singletons. For instance, let $p$ and $q$ be distinct points not in $S=\left\{\frac1n:n\in\Bbb Z^+\right\}$, and let $X=\{p,q\}\cup S$, topologized so that the points $\frac1n$ are isolated, the open nbhds of $p$ are the sets $\{p\}\cup(S\setminus F)$ for finite $F\subseteq S$, and the open nbhds of $q$ are the sets $\{q\}\cup(S\setminus F)$ for finite $F\subseteq S$. (You can think of $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ as a sequence converging to both $p$ and $q$.) The points $p$ and $q$ do not have disjoint open nbhds, so $X$ is not Hausdorff. The closed subsets of $X$ are the finite subsets and the infinite subsets that contain both $p$ and $q$. Suppose that $A\subseteq X$ is not a singleton. If $A$ is finite, it is easy to write $A$ as the union of two proper subsets that are finite and therefore closed. If $A$ is infinite, let $x\in A\cap S$; then $A=\{x\}\cup(A\setminus\{x\})$, where $\{x\}$ and $A\setminus\{x\}$ are both relatively closed subsets of $A$, so $A$ is reducible.
Best Answer
They are called $T_1$-spaces.