We know that for a Noetherian topological space $X$, the following are equivalent:

  1. $X$ is a Hausdorff space
  2. $X$ is finite and has discrete topology

Is this statement true if we replace Noetherian by Artinian?

Call a topological space $X$ Artinian if every nested sequence of closed sets $C_1 \supset C_2 \supset C_3 \supset \cdots$ is eventually constant. Also,a topological space $X$ is called Noetherian if it satisfies the descending chain condition for closed subsets

  • $\begingroup$ How are Artinian and Noetherian different? $\endgroup$ – AJY Sep 25 '16 at 18:02
  • $\begingroup$ @AJY: It’s Noetherian if every ascending sequence of closed sets is eventually constant. $\endgroup$ – Brian M. Scott Sep 25 '16 at 18:06
  • $\begingroup$ My apologies, indeed. $\endgroup$ – AJY Sep 25 '16 at 18:33

One direction is true for all $X$, Artinian or not. For the other direction, suppose that $X$ is Hausdorff and infinite. Clearly $X$ cannot have an infinite set of isolated points, so without loss of generality we may assume that $X$ has no isolated points.

Let $x_0$ and $x_1$ be distinct points of $X$; there are disjoint open sets $U_0$ and $V_1$ such that $x_0\in U_0$ and $x_1\in V_1$. Let $x_2\in V_1\setminus\{x_1\}$; there are disjoint open sets $U_1,V_2\subseteq V_1$ such that $x_1\in U_1$, $x_2\in V_2$. In general, given $x_n\in V_n$, let $x_{n+1}\in V_n\setminus\{x_n\}$; there are then disjoint open sets $U_n,V_{n+1}\subseteq V_n$ such that $x_n\in U_n$ and $x_{n+1}\in V_{n+1}$. In this way we recursively construct a family $\{U_n:n\in\Bbb N\}$ of pairwise disjoint open subsets of $X$. For $n\in\Bbb N$ let $F_n=X\setminus\bigcup_{k\le n}U_k$; then $\langle F_n:n\in\Bbb N\rangle$ is a strictly decreasing infinite sequence of closed sets, and $X$ is not Artinian. Thus, an Artinian Hausdorff space must be finite, and a finite Hausdorff space is discrete.


Yes, any Artinian Hausdorff space is finite. We use the following claim:

$(*)$ If $X$ is Hausdorff and infinite, and $A\subseteq X$ is open and $X\setminus A$ is infinite, then there is an open $B$ such that $B\not\subseteq A$ and $X\setminus (A\cup B)$ is infinite.

First, note that $(*)$ implies that no infinite Hausdorff space is Artinian: inductively build an increasing sequence of open sets $A_1\subsetneq A_2\subsetneq A_3\subsetneq . . .$, then their complements form a counterexample to Artinian-ness.

So how do we prove $(*)$? Well, let $x\not=y$ be elements of $X\setminus A$. By Hausdorffness, we get open $B_0\ni x, B_1\ni y$ with $B_0\cap B_1=\emptyset$. Now there are two cases:

  • Either $B_0\setminus A$ or $B_1\setminus A$ is finite. WLOG, assume $B_0\setminus A$ is finite. Then since $X\setminus A$ is infinite, we may take $B=B_0$.

  • Both $B_0\setminus A$ and $B_1\setminus A$ are infinite. Then since $B_0\cap B_1=\emptyset$, we may take $B=B_0$ or $B=B_1$.


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