Home » Problems for Fun » A Noetherian and Hausdorff space is finite

A Noetherian and Hausdorff space is finite

In this post, I will prove that a Noetherian and Hausdorff topological space is finite (and therefore has the discrete topology, being Hausdorff). The proof is very short and pleasant.

Proof: Let {X} be such a space, and suppose that it is infinite. Let {\Sigma} be the collection of infinite closed subsets of {X}. It is nonempty since {X \in \Sigma}, and therefore has a minimal member {Z} by the Noetherian assumption. Let {p,q} be points of {Z}, and {U,V} be disjoint open neighborhoods of {p} and {q} respectively (such {U} and {V} exist by the Hausdorff assumption). Then {X = (X-U) \cup (X-V)} since {U} and {V} are disjoint, so {Z = (Z \cap (X-U)) \cup (Z \cap (X-V))}. Now each of {Z \cap (X-U)} and {Z \cap (X-V)} is closed in {X}, and is properly contained in {Z} (the first one doesn’t contain {p}, and the second one doesn’t contain {q}). Therefore, by minimality of {Z}, each must be finite, and therefore {Z} is also finite, which is a contradiction. \Box

Corollary: in any infinite Hausdorff space, there exists a strictly descending infinite chain of closed subsets Z_1 \supset Z_2 \supset Z_3 \dots. The proof above can be easily adapted to construct such a sequence.

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