Noetherian Hausdorff Topological Spaces

Question

What are the Noetherian Hausdorff topological spaces? (We know that Zariski topology is not Hausdorff, so we should look for another topological spaces).

I would be thankful for any comments.

Answer 1

Rolands idea: A Noetherian space $X$ can be (uniquely) written as $$X = \bigcup_{i=1}^n C_i$$ where each $C_i$ is an irreducible component of $X$ and $N$ is some finite number.

So each $C_i$ is a Hausdorff irreducible space and we saw here that this means that $|C_i| = 1$ and so $|X| = N$. So $X$ is finite and discrete (a finite Hausdorff space is discrete as all singletons are closed, so all subsets are closed so all subsets are open).

Another proof can go by contradiction: if $X$ is an infinite Hausdorff space then $X$ has an infinite discrete subspace $D$, and if $X$ were Noetherian, so would its subspace $D$ be and this is clearly false.

Answer 2

You might know that in Noetherian space $X$ every subset is compact. So if $a \in X$ then $X-\{a\}$ is compact. For every $y \in X-\{a\}$ choose $y\in U_y$ and $a \in V_y$ open subsets of $X$ such that $U_y \cap V_y = \emptyset$ by Hausdorffness. By compactness of $X-\{a\}$ finite $U_{y_i}$s cover $X-\{a\}$. Consider $\cap V_{y_i} $- this is a open set such that $a \in \cap V_{y_i}$ and $\cap V_{y_i} \bigcap (X-\{a\})= \emptyset$. So $\{a\}$ is a open set. The result then follows from compactness of $X$.

Answer 3

Every subspace of a Noetherian space is Noetherian and hence compact. In a Hausdorff space, all compact subspaces are closed. Thus every subspace is closed and hence the topology is discrete. By compactness, the space is also finite.