The Irreducible components of a Hausdorff space are the singletons. I'm interested in understanding that in a Hausdorff space the only irreducible components are the singletons. Take the example of $\mathbb R$. For any subset $X$, take any real number $a$ and we can write $$X=(X\cap (-\infty, a])\cup (X\cap [a,\infty))$$ so $X$ is written as the union of two closed subsets of $X$ since $(-\infty,a]$ and $[a,\infty]$ are closed subsets of $\mathbb R$. Now my two questions are :

*

*Where did we use the Hausdorff property here?


*Why we exclude the singleton from this construction, for example we can write $$\{2\}=(\{2\}\cap (-\infty,3])\cup (\{2\}\cap [3,\infty))$$
 A: 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.
A: Let $Y$ be a Hausdorff space. If $X \subseteq Y$ is a subset with a least two elements, say $x$ and $y$, then there are open of $Y$ $U \ni x$ and $V \ni y$ whose intersection is empty. Call $F=Y \setminus U$ and $F^\prime=Y \setminus V$. Then $X= (X \cap F) \cup (X \cap F^\prime)$ and neither $X \cap F$ nor $X \cap F^\prime$ is equal to $X$.
In the case of $\mathbb{R}$, if $X$ has two distinct elements $x<y$ you can choose some $x<a<y$ and then $X=\left( X \cap(-\infty,a] \right) \cup \left( X \cap [a,+\infty) \right)$. Both $X \cap(-\infty,a]$ and $X \cap [a,+\infty)$ are closed in $X$ and neither is equal to $X$. In the case of $\mathbb{R}$ the Hausdorff assumption is used in the fact that $(-\infty,a)$ and $(a,+\infty)$ are opens containing $x$ and $y$ respectively but have empty intersection.
