Closed sets as compact sets

Question

While playing around with some basic general topology, I have thought of some problems whose solutions are not so obvious (at least to me), and surprisingly I do not remember having seen these anywhere.

Disclaimer #1 : I have already posted those problems elsewhere, but I was unlucky and received no answer.

Disclaimer #2 : The French and English notions of compact sets are notoriously different. I will try to write it right (and it might be actually easier in English).

So, let us take any topological space $(X, T)$. As a matter of prophylaxy, I will assume that $X$ is not empty. A subset $K$ of $X$ is said to be compact if, from any covering of $K$ by open sets one can extract a finite sub-cover.

My first question is: When are the compact subsets for the topology $T$ the closed subsets for another topology $T^*$ (modulo the whole space $X$)?

Trivially, any finite union of compact subsets is compact. If any intersection of compact subsets is compact, then the set of compact subsets plus $X$ are the closed subsets for some topology $T^*$ on $X$. One can see the open subsets of $(X,T^*)$ as the open neighborhoods of infinity for the Alexandroff compactification of $(X, T)$, without said infinity. If $(X,T)$ is Hausdorff then any intersection of compact sets is compact (this is because the compact sets are then closed). However, this condition is not necessary: with the coarse or cofinite topology, any subset is compact, so that any intersection of compact subsets is compact. Do the topological spaces for which this property holds have a name, or a characterization?

My second question is: Assuming that the property discussed above is satisfied, is there anything interesting to be said about the operation $T \to T^*$?

For instance, if $T$ has the property that any intersection of compact subsets is compact, then does $T^*$ have the same property? This would be important as it would allow us to iterate the operation $T \to T^*$. In addition, it looks like $T^*$ can be seen, in some cases, as a dual topology on $X$. If $T$ is the coarse topology on $X$, then $T^*$ is the discrete topology, and $T^{**}$ is the cofinite topology. Then the sequence stabilizes, as $T^{***}$ is again the discrete topology. In a similar way, If $T$ is the usual topology on $\mathbb{R}^d$, I think (I have not written down the argument properly) that $T=T^{**}$. Are those special cases, or is there a more general pattern?

Answer 1

I don’t have an answer to the first question, but I can show that the desired property isn’t necessarily preserved by the $\tau\mapsto\tau^*$ operation. (Barring mental hiccups, anyway: keeping track of these topologies is a bit of a pain.)

Here’s an example of a non-Hausdorff KC-space $\langle X,\tau\rangle$ such that $\langle X,\tau^*\rangle$ is also a non-Hausdorff KC-space, but $\langle X,\tau^{**}\rangle$ does not have the desired property: $\langle X,\tau^{**}\rangle$ is the space obtained by splitting the point $\omega_1$ in two in the compact Hausdorff space $\omega_1+1$ (with the order topology), as is done in constructing the line with two origins. The space $\langle X,\tau\rangle$ was originally constructed by Bill Fleissner as an example of a KC-space whose topology contains no minimal KC-topology.

Let $p$ and $q$ be two points not in $\omega_1$, and let $X=\omega_1\cup\{p,q\}$. Let $\tau_0$ be the order topology on $\omega_1$. Let $I=\{0\}\cup\{\xi+1:\xi\in\omega_1\}$, the set of isolated points of $\omega_1$. For $\alpha<\omega_1$ let $$B_\alpha(p)=\{p\}\cup I\setminus\alpha\text{ and }B_\alpha(q)=\{q\}\cup I\setminus\alpha\;,$$ and let $\tau$ be the topology on $X$ generated by the base $$\tau_0\cup\{B_\alpha(p):\alpha<\omega_1\}\cup\{B_\alpha(q):\alpha<\omega_1\}\;.$$ Let $\mathscr{K}_0$ be the set of $\tau_0$-compact subsets of $\omega_1$; $\mathscr{K}_0$ is simply the set of countable $\tau_0$-closed subsets of $\omega_1$. Let $\mathscr{K}\;$ be the set of $\tau$-compact subsets of $X$; clearly $$\mathscr{K}=\big\{K\cup F:K\in\mathscr{K}_0\land F\subseteq \{p,q\}\big\}\;,$$ which is precisely the set of countable $\tau$-closed subsets of $X$, and it follows that $$\tau^*\triangleq\{X\setminus K:K\in\mathscr{K}\;\;\}\cup\{\varnothing\}$$ is a topology on $X$. Let $\tau_0^*=\tau^*\cap\wp(\omega_1)\setminus\{\varnothing\}$; $\tau_0^*$ is the set of co-countable $\tau_0$-open subsets of $\omega_1$, and $$\tau^*=\big\{V\cup F:V\in\tau_0^*\land F\subseteq \{p,q\}\big\}\cup\{\varnothing\}\;.$$

Let $\mathscr{K}\;^*$ be the set of $\tau^*$-compact subsets of $X$; $$\mathscr{K}\;^*=\mathscr{K}\cup\big\{C\cup F:C\subseteq\omega_1\text{ is a }\tau_0\text{-cub} \land F\subseteq \{p,q\}\big\}\;.$$

Clearly $\tau^{**}\triangleq \{X\setminus K:K\in\mathscr{K}\;^*\}$ is a topology on $X$, and it’s not hard to see that $$\tau^{**}=\big\{V\cup F:V\in\tau_0\land F\subseteq\{p,q\}\big\}\subsetneqq\tau\;.$$

The subspaces $\omega_1\cup\{p\}$ and $\omega_1\cup\{q\}$ of $\langle X,\tau^{**}\rangle$ are homeomorphic in an obvious way to the compact Hausdorff space $\omega_1+1$ (with the order topology), and $\langle X,\tau^*\rangle$ is therefore $\omega_1+1$ with the point $\omega_1$ split into $p$ and $q$, as mentioned above.

Finally, let $\mathscr{K}\;^{**}$ be the set of $\tau^{**}$-compact subsets of $X$; $$\mathscr{K}\;^{**}=\mathscr{K}\cup\big\{C\cup F:C\subseteq\omega_1\text{ is a }\tau_0\text{-cub} \land\varnothing\ne F\subseteq \{p,q\}\big\}$$ and is not closed under intersections, since $X\setminus\{p\},X\setminus\{q\}\in\mathscr{K}\;^{**}$, but $X\setminus\{p,q\}\notin\mathscr{K}\;^{**}$.