# Proper topological spaces

Recall that a topological space is ccc, or has the countable chain condition, if every family of pairwise disjoint open sets is countable.

But equivalently, we can say that the forcing defined with the non-empty open sets is a ccc forcing.

This lends itself to many other definitions that can be translated from forcing terms to topological spaces. But whereas some of them are "natural" enough to recast in topological terms, others might not be as clear.

Definition. A topological space $(X,\tau)$ is a proper space, if the forcing $(\tau\setminus\{\varnothing\},\subseteq)$ is proper.

Question. Is there a reasonable topological characterization of proper spaces?

• Just to mention that there's a widely used notion of proper metric space (this means that closed bounded subsets are compact), and that this is a priori unrelated. – YCor Mar 24 '18 at 12:32
• I think you want to force via $\subseteq$ rather than $\supseteq$. That is, $p\Vdash q$ if and only if $p\subseteq q$. – Joel David Hamkins Mar 24 '18 at 14:34
• @Joel: The smaller open set is stronger. – Asaf Karagila Mar 24 '18 at 14:35
• Yes, that is what I wrote. So why did you reverse the order? This seems wrong even in light of the Isreali convention on $\geq$ vs. $\leq$ for forcing. – Joel David Hamkins Mar 24 '18 at 14:44
• @Joel: It's fine. As I wrote in my edit summary. Forcing ordering can be confusing sometimes. – Asaf Karagila Mar 24 '18 at 14:51

Here's the following combinatorial characterization of a proper topological space in terms of a topological game.

We say a topological space $(X,\tau)$ is a proper space if player $II$ has a winning strategy in the two player topological game $\Gamma(X,\tau)$ defined as follows. Player $I$ begins at the $0^{th}$ move by choosing an open set $O_0\in \tau \setminus\{\emptyset\}$ and a dense open set $D_0$ of $X$. Player $II$ responds by choosing a countable subset $Q_0^0$ of $O_0$. Next player $I$ chooses another dense open set $D_1$ of $X$ and player $II$ responds by choosing two countable subsets $Q^1_0\subseteq D_0$ and $Q^1_1\subseteq D_1$. At the $n^{th}$ move, for $n>0$ player $I$ chooses a dense open set $D_n$ and player $II$ chooses countable subsets $Q^n_k\subseteq D_k$ for all $k\leq n$. This topological game has $\omega$ many moves and the payoff condition is given by the following: player $II$ wins the game $\Gamma(X,\tau)$ if and only if there exists a open set $P\subseteq O_0$ such that for every $k$ if we define $Q_k=\bigcup_{k\leq n}Q_k^n$, then $Q_k$ intersects every open subset of $P$.

This characterization holds because $(\tau\setminus\{\emptyset\},\subseteq)$ is a proper partial order if and only if player $II$ has a winning strategy in the game $\Gamma((\tau\setminus\{\emptyset\},\subseteq))$ defined just as above (I will let you show this).

Reasonable? Not really.

Moreover, if I'm not mistaken, being proper isn't preserved by continuous surjections; for example assuming $\mathsf{CH}$ there is a continuous map from the space $\omega^\ast$ (which is proper) onto the space associated with the po-set which shoots a club through a stationary/co-stationary subset of $\omega_1$ (contrast this with the $c.c.c.$ which is preserved by continuous surjections.) This makes the idea that there is some useful topological characterization of proper unlikely.

That said, the topological interpretation of forcing has had an influence on how proper forcings are cooked up. The relevant topological observation being that the only forcings worth playing with, come from spaces which are completely regular, making them comfortably embedded as a subspace of some product space $[0,1]^S$ (with $S$ some infinite set; normally taken to be an ordinal.) This is desirable, since in this case, the base for the product topology admits a convenient coding using elements of $\mathsf{Fn}(S, \cal I)$ (where $\cal I = \{ (a,b)\cap [0,1]: a< b\in \mathbb{Q}\}$.) Moreover, the exploitation of such codings/embeddings/etc.. is implicit in topological analysis of forcing extensions and motivates axioms like the various flavors of $\mathsf{OCA}$.

Anecdotally, I find that this type of perspective/coding makes tradmark tools of the trade, like side-condition style forcings (i.e. $\in$-chains of models with countable/finite working part), promises, creatures, souslin-c.c.c./proper, ... easier to digest.

• “Moreover, if I'm not mistaken, being proper isn't preserved by continuous surjections […] This makes the idea that there is some useful topological characterization of proper unlikely.” I don’t follow this point: plenty of nice topological properties (e.g. Hausdorffness, for a start) are not preserved by continuous surjections. – Peter LeFanu Lumsdaine Apr 9 '18 at 11:58
• @PeterLeFanuLumsdaine I was coming at this from the perspective that a useful purely topological characterization of proper should provide some potential reduction in work.Especially since any such characterization would potentially obfuscate the benefits of being proper from the perspective of forcing. For example, it doesn't readily follow from the game characterization given in the first answer, that in the forcing extension generated by a space with an appropriate winning strategy that stationary subsets of $[H(\theta)]^{\omega}$ from the ground-model remain stationary. – Not Mike Apr 9 '18 at 13:39