Irreducible Hausdorff Space is Singleton

Theorem
Let $T = \struct {S, \tau}$ be a non-empty topological space which is irreducible and Hausdorff.

Then $S$ is a singleton.

Proof
Suppose $S$ has exactly one element.

Then by definition $T = \struct {S, \tau}$ is the trivial topological space.

Hence, from Trivial Topological Space is Irreducible, $S$ is irreducible.

Suppose $S$ has at least two distinct elements:
 * $x, y \in S, x \ne y$

By definition of irreducible space:
 * there are no two disjoint open sets $T$ such that $x$ is in one and $y$ is in the other:


 * $\nexists U_1, U_2 \in \tau: U_1 \cap U_2 = \O, x \in U_1, y \in U_2$

This contradicts the fact that $T$ is Hausdorff.

Thus $S$ has only one element.