Irreducible and Hausdorff implies one-point space

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Suppose X is a non-empty topological space that is both an irreducible space and a Hausdorff space. Then, X must be a one-point space.

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We prove a somewhat modified form: we prove that an Hausdorff space containing two distinct points cannot be irreducible.

Given: A non-empty topological space X that is Hausdorff and has at least two points. Say x,y are two distinct points of X.

To prove: X is not irreducible, i.e., it contains two non-empty open subsets with empty intersection.

Proof: Use the definition of Hausdorff to find non-empty open subsets U \ni x, V \ni y that have empty intersection. This completes the proof.