A nonempty subset $A$ of a topological space $X$ is called irreducible if, whenever we have open sets $U$ and $V$ such that $U \cap A$ and $V \cap A$ are nonempty, then we also have $U \cap V \cap A \neq \emptyset$. Show that the closure of an irreducible set is irreducible, and deduce that the closure of any singleton set $\{x\}$ is irreducible.

$X$ is said to be a sober topological space if, for any irreducible closed $A \subseteq X$, there is a unique $x \in X$ such that $A=\overline{\{x\}}$. Show that any Hausdorff space is sober, but that an infinite set with the cofinite topology is not sober.

Given an arbitrary topological space $(X, \mathcal{T})$, let $\widehat{X}$denote the set of all irreducible closed subsets of $X$, and for each $U \in \mathcal{T}$ let

$$\widehat{U}=\{A \in \widehat{X} \mid U \cap A \neq \emptyset\}$$

Show that the sets $\{\widehat{U} \mid U \in \mathcal{T}\}$ form a topology $\widehat{\mathcal{T}}$on $\widehat{X}$, and that the mapping $U \mapsto \widehat{U}$is a bijection from $\mathcal{T}$ to $\widehat{\mathcal{T}}$. Deduce that $(\widehat{X}, \widehat{\mathcal{T}}$) is sober. [Hint: consider the complement of an irreducible closed subset of $\widehat{X}$.]