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
    $\begingroup$ +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. $\endgroup$ – DanielWainfleet Mar 8 '20 at 20:10

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