# Topology in which every open set is compact: Noetherian and, if Hausdorff, discrete

Question:

Let $$X$$ be a topological space in which every open set is compact. Show that:

a. Every ascending chain of open sets in $$X$$ stabilizes: If $$U_1\subseteq U_2\subseteq U_3\subseteq\dots$$, $$\exists n\in \mathrm{N} : U_n=U_m, \forall m>n$$.

b. If $$X$$ is Hausdorff, then it has a discrete topology. (Recall that compact sets in a Hausdorff space are closed)

What I tried:

For the first item, consider $$\cup_{i=1}^{\infty} U_i$$. It's open, thus compact. If the chain didn't stabilize, there would be no finite subcover, contradicting its compactness.

For the second item, I know that I must show that every subset of $$X$$ is open. I haven't studied irreducibility yet, but this seems close to what I need to show. I understand that a finite set that is Hausdorff has the discrete topology, so I wonder if there is way to not use the irreducibility related stuff in the answer in the link above and show $$X$$ has to be finite.

If, as in the first item, we consider $$\cup_{i=1}^{\infty} U_i$$, we can write it in terms of disjoint open sets: $$\cup_{i=1}^{n} U_i= \cup_{i=1}^{n} U_i - \cup_{j=0}^{i-1} U_j$$. But I'm not sure if this is useful.

Let $$x \in X$$. Since $$X$$ is Hausdorff, $$\{x\}$$ is closed. Hence its complement is open. By hypothesis the complement is compact. In a Hausdorff space compact sets are closed. So the complement of $$\{x\}$$ is closed. This means $$\{x\}$$ is open. Since every singleton set is open the space has discrete topology.
• +1....So if X is Hausdorff it is discrete and also finite, otherwise the open set X would not be compact....An infinite set with the co-finite topology is an example of an infinite non-Hausdorff space in which $every$ subset is compact. Mar 8, 2020 at 20:10