Irreducible Hausdorff Space is Singleton

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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.

$\Box$


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.

$\blacksquare$