A Noetherian space satisfies that any non-empty set of open subsets must have a maximal element. But a Hausdorff space satisfies that, for $x\neq y$ two different points, there exists open neighborhoods $U,V$ containing $x,y$ respectively such that $U\cap V=\emptyset$.

My question is: when we consider the set of open subsets $$\{U,V\}$$ there is no maximal element in this set. So how can a Noetherian space be Hausdorff too? Thank you!

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    $\begingroup$ Easiest example are finite sets with the discrete topology. Those are both noetherian and Hausdorff. $\endgroup$ Aug 25, 2022 at 20:16
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    $\begingroup$ And finite discrete spaces are in fact the only examples of Noetherian Hausdorff spaces. $\endgroup$ Aug 25, 2022 at 21:06

1 Answer 1


Actually both $U$ and $V$ are maximal elements of $\{U,V\}$. It seems you’re confusing maximal elements with greatest elements. If we have a partially ordered set $(X,\leq)$, then an element $x$ is called a greatest element if we have $y\leq x$ for all $y$. It is called a maximal element if for all $y$, we have that $x\leq y$ implies $y=x$. The characterization of Noetherian spaces that you mention talks about maximal elements, not greatest elements.

  • $\begingroup$ Ah! It is such a trivial confusion. Thank you! $\endgroup$
    – Mizutsuki
    Aug 26, 2022 at 10:08

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