Hausdorff implies sober
This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., Hausdorff space) must also satisfy the second topological space property (i.e., sober space)
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Further information: Hausdorff space
A topological space is said to be Hausdorff if, for any two points in , there exist open subsets of such that and is empty.
Further information: sober space
A topological space is termed sober if the only irreducible closed subsets of are the closures of one-point subsets.
We execute this proof by showing that, for a Hausdorff space, the only irreducible closed subsets are the one-point subsets. Note that this in particular shows that the only irreducible closed subsets are the closures of one-point subsets.
Given: A topological space , an irreducible closed subset of .
To prove: is a one-point subset.
Proof: Since is irreducible, it is nonempty, so it has at least one point. We show that if has two points, we obtain a contradiction.
Suppose are two distinct points. By the definition of Hausdorffness, there exist open subsets of such that , and . Then, the sets and are both closed subsets, their union is , and they are both proper subsets of since and . Thus, we have written as a union of two proper closed subsets, contradicting the assumption of irreducibility.