Proposition 1

Let $X$ be a non-empty topological space. The following conditions are equivalent: (a) $X$ is irreducible; (b) every non-empty open set of $X$ is dense in $X$; (c) every open set of $X$ is connected.

By definition, a set which is dense in $X$ is a set which meets every non-empty open set, hence (a) and (b) are equivalent. It is immediate that (c) implies (a), for if $U_1$ and $U_2$ are disjoint non-empty open sets, $U_1 \cup U_2$ is a disconnected open set. Finally let us show that (a) implies (c): if $U$ is a disconnected open set, it is the union of two disjoint non-empty sets $U', U''$ which are open in $U$ and hence also open in $X$, which implies that $X$ is not irreducible.

A Hausdorff space is irreducible only if it consists of a single point.

A subset $E$ of a topological space $X$ is called an irreducible set if the subspace $E$ of $X$ is irreducible. For this to be so, it is necessary and sufficient that, for every pair of sets $U, V$ which are open in $X$ and meet $E$, $U \cap V$ also meet $E$, or (what amounts to the same thing) that, for every pair of sets $F, G$ which are closed in $X$ and satisfy $E \subset F \cup G$, $E \subset F$ or $E \subset G$. By induction on $n$, we deduce that, if $(F_i)_{1 \leq i \leq n}$ is a finite family of closed sets of $X$ such that $E \subset \bigcup_{i=1}^n F_i$, there exists an index $i$ such that $E \subset F_i$.