Irreducible Hausdorff Space is Singleton

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space which is irreducible and Hausdorff.


Then $T$ contains only one point.


Proof

Suppose $T$ has at least two points $x, y \in S$.

Because $T$ is irreducible, there are no two disjoint open sets such that $x$ is in one and $y$ is in the other.

This contradicts the fact that $T$ is Hausdorff.

Thus $T$ has only one point.

$\blacksquare$