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:
- $\nexists U_1, U_2 \in \tau: U_1 \cap U_2 = \O, x \in U_1, y \in U_2$
This contradicts the fact that $T$ is Hausdorff.
Thus $S$ has only one element.
$\blacksquare$