# Interpreting a space in Baire space: how many facts do I need to understand the whole thing?

Below I'm working in ZF+DC+AD or similar; I want enough choice that things don't explode, but I also want the Wadge hierarchy to be well-behaved everywhere. Since this question is a bit long, I've put the actual question in red.

EDIT in response to a comment by მამუკა ჯიბლაძე: the only reason I'm focusing on Baire space is that it has a descriptive set theory that we understand well with many nice properties (and nothing would change if we replaced it with another uncountable Polish space). So describing things in terms of how complicated they are to "represent" via Baire space seems non-entirely-silly.

## Background

I'm interested in using Baire space to give a presentation of a given space - or rather, of the algebra of open sets of such a space:

(I'm really just thinking about structures with operations of infinite arity, but I'm focusing on algebras of open sets of spaces for now, partly for concreteness and partly because they're what I'm most interested in at the moment.)

Definition: A partial map $$\rho:$$ $$\subseteq\omega^\omega\rightarrow\tau$$ is a generator of a topological space $$(X,\tau)$$ (or rather, just of $$\tau$$) if its range is a base for $$\tau$$.

For the rest of this question, by "space" I mean "$$T_0$$ space which has a generator," and by "base" I mean "base which is the image of some generator." (In the presence of choice I'd say "base of size at most $$\mathfrak{c}$$," but this phrasing is less ambiguous here.)

Having fixed a generator, the rest of the structure of $$\tau$$ is determined by some (infinitary) relations, namely the covering relations: statements of the form $$\bigcup_{f\in A\cap dom(\rho)}\rho(f)\supseteq\rho(g)$$ or their negations. (If the statement above holds, say "$$A$$ covers $$g$$.") I'm thinking of facts of this type as forming the "atomic diagram" of the presentation of $$\tau$$ given by $$\rho$$, or some significant fragment thereof.

It's now reasonable to ask how simple this bundle of data can be, or can be made to be. The problem is on the $$A$$ side. Since arbitrary-complexity index sets are allowed, the amount of possible complexity here seems to be "all of it;" I'm interested in when things don't stink quite so badly.

## Question

My overall interest is in understanding when some "bounded complexity" amount of information about a topology, phrased in the language of a generator, gives me a complete picture of the topology. Specifically, and using terms defined below:

1. Is there a non-BCP$$_0^{+,uni}$$ space, at all?

2. Is there a BCP$$_0^{-,uni}$$ space such that there is no generator $$\rho$$ with $$\{f: x\in\rho(f)\}$$ open in $$dom(\rho)$$?

Here's what all that (and a bit more) means. Since we're thinking about sets of reals here, the right first stab at a complexity notion seems to be Wadge reducibility. Conveniently, every Wadge degree (other than the top degree $$\mathcal{P}(\omega^\omega)$$) has size at most continuum, and clearly every family of sets of size at most continuum is contained in a Wadge degree strictly below $$\mathcal{P}(\omega^\omega)$$, so we can phrase things just in terms of cardinality which is a bit easier. A couple weak notions of when a space has a bounded complexity presentation appear particularly natural here, specifically with respect to how hard it is to determine "atomic" facts about covering or non-covering; the below includes a couple not relevant to the specific question above,but I think they help flesh out the picture:

• A space is BCP$$_0^+$$ if for some generator $$\rho$$ there is a family $$\mathfrak{S}\subseteq\mathcal{P}(\omega^\omega)$$ of size at most continuum such that whenever $$A$$ covers $$g$$ there is some $$B\subseteq A$$ with $$B\in\mathfrak{S}$$ which also covers $$g$$.

• A space is BCP$$_0^-$$ if for some generator $$\rho$$ there is a family $$\mathfrak{S}\subseteq\mathcal{P}(\omega^\omega)$$ of size at most continuum such that whenever $$A$$ does not cover $$g$$ there is some $$B\supseteq A$$ with $$B\in\mathfrak{S}$$ which also doesn't cover $$g$$.

• A space is BCP$$_0^{+,uni}$$ (respectively, BCP$$_0^{-,uni}$$) if it is uniformly BCP$$_0^+$$ (resp. BCP$$_0^-$$); that is, if there is a specific function $$A\mapsto B$$ witnessing the relevant property when we take $$\mathfrak{S}$$ to be the range of this function. (Since we're not assuming choice, this isn't a trivial addition.)

Finally, there is an asymmetry between $$+$$ and $$-$$ here, regarding topological equivalence. Given a generator $$\rho$$, let $$A_{ext(\rho)}=\{f:\exists h\in A(\rho(f)=\rho(h))\}$$ be the "extensional closure" of $$A\subseteq\omega^\omega$$. Then if $$A$$ fails to cover $$g$$, so will $$A_{ext(\rho)}$$, and clearly $$A_{ext(\rho)}\supseteq A$$; thus, "extensionality is free" as far as BCP$$_0^-$$ is concerned. However, this is a serious issue for BCP$$_0^+$$: since we want to wind up with a subcover as opposed to a supernoncover, we can't freely pass from $$A$$ to $$A_{ext(\rho)}$$, and so the problem of finding a simple subcover of $$g$$ could be much harder for $$A$$ than for $$A_{ext(\rho)}$$. This seems somewhat artificial; to get around this, it may be more natural to consider the following:

• A space is BCP$$_0^{+/ext}$$ (respectively, BCP$$_0^{+/ext,uni}$$) if it has a generator $$\rho$$ witnessing BCP$$_0^+$$ (resp., BCP$$_0^{+,uni}$$) restricted to sets satisfying $$A=A_{ext(\rho)}$$.

Of course, we could apply the same restriction to the "$$-$$-notions," but as noted above this wouldn't change anything. Meanwhile, I'm keeping the non-extensional notions in mind since it's not actually clear to me that the "noise" we throw away by restricting attention to extensionally-closed covers is actually uninteresting.

The broad question I'm trying to answer here is, which spaces have these various properties? On reflection I think this is far too broad, hence the more specific focus above.

## Coda

Let me end by mentioning a couple facts motivating said specific focus:

First of all, it's easy to see that any second-countable space is BCP$$_0^+$$, and that any space with a base of compact sets is BCP$$^+_0$$. However, to get BCP$$_0^+$$ the latter doesn't obviously suffice, and we need instead a well-orderable base of compact sets. It's also, interestingly, not clear to me that if one generator witnesses BCP$$_0^+$$ then every generator witnesses BCP$$_0^+$$. By contrast, BCP$$_0^-$$ is generator-invariant (as is BCP$$_0^{-,uni}$$): if $$\rho$$ witnesses BCP$$_0^-$$, $$\rho'$$ is another generator, and $$A$$ covers $$g$$ in the sense of $$\rho$$, then $$t(A):=\{f: \rho'(f)\subseteq \bigcup_{g\in A}\rho(g)\}$$ also, and the map $$t$$ being nicely definable we also get that $$\rho'$$ witnesses BCP$$_0^{-,uni}$$ if $$\rho$$ does.

BCP$$_0^-$$ alone seems a very weak property: it is implied by the existence of a base $$\rho$$ such that $$\{\{f: x\in\rho(f)\}: x\in X\}$$ has bounded Wadge degree (more simply: has cardinality at most continuum), and is equivalent to that under the further assumption of $$T_1$$. (But $$T_1$$-ness is necessary for this equivalence: consider the topology on $$\mathcal{P}(\omega^\omega)$$ generated by $$\{\{S\subseteq\omega^\omega\}:x\in\omega^\omega\}$$.)

The interesting example is BCP$$_0^{-,uni}$$. Say that a generator is open-pointed if for each point $$x$$ the set $$\{f\in dom(\rho): x\in\rho(f)\}$$ is open in $$dom(\rho)$$. Then any s pace with an open-pointed generator has BCP$$_0^{-,uni}$$: $$A$$ covers $$g$$ iff (WLOG assuming $$A\subseteq dom(\rho)$$) $$cl(A)\cap dom(\rho)$$ covers $$g$$. To prove this we just need to show that any $$f\in cl(A)\cap dom(\rho)\setminus A$$ "adds no new points," and this follows immediately from open-pointedness property: if $$x\in\rho(f)$$, then $$x\in\rho(g)$$ for some $$g\in A$$ since $$f$$ is in the closure of $$A$$. Note that we can't obviously replace "open" with "Borel." So BCP$$_0^{-,uni}$$ actually looks at the

So the situation we're left with is that BCP$$_0^+$$ is odd enough that I really have no idea what it looks like in any nontrivial sense, and BCP$$_0^{-,uni}$$ actually has something apparently interesting going on but it seems unlikely to give a full characterization.

• Incidentally, there is (to me) a flavor similarity between this question and this old question of mine about the complexity of the Banach game, in terms of how we try to measure the complexity of "high-order" structures. – Noah Schweber Aug 11 '18 at 23:00
• Can this second example be extended to show that as long as $\{f \in dom(\rho) : x \in \rho(f) \}$ are Borel with uniformly bounded rank for $x\in X$, then $(X, \tau) \in BCP_0^-$? – James Aug 12 '18 at 1:17
• Personally, I would understand the question much better if you could single out what kind of structure on the Baire space you need to formulate everything. Is it possible to name some abstract structure (like a topological space with certain properties + some notion of degree) which would allow to define these bcp conditions and pose the same questions? How much of it does actually depend on the Baire space per se? – მამუკა ჯიბლაძე Aug 12 '18 at 5:33
• @მამუკაჯიბლაძე See my edit. – Noah Schweber Aug 12 '18 at 16:18
• – Federico Poloni Oct 18 '18 at 12:52

This is just a comment, but it is too long for the comment section.

Given a "generator" $\rho$, of the topology $\tau$, as Noah defined it, one could define $\varphi: \mathcal{P}(\omega^{\omega}) \rightarrow \tau$ by $\varphi(A) = \bigcup_{f \in A} \rho(f)$. $\varphi$ is a surjective $\subseteq$-homomorphism. One could then take the quotient of $\mathcal{P}(\omega^{\omega})$ by $A \sim B$ iff $\phi(A) = \phi(B)$. We can completely recover $\tau$ (considered as a pointless topology) from this quotient together with the covering relation Noah defined. We can also recover it completely from just the $\sim$-invariant binary relation $\mathcal{R}$ on $\mathcal{P}(\omega^{\omega})$ defined by $A\mathcal{R}B$ iff $\phi(A)\supseteq \phi(B)$

This allows any topology which has a "generator" to be presented as a binary covering relation $\mathcal{R} \subseteq \mathcal{P}(\omega^{\omega}) \times \mathcal{P}(\omega^{\omega})$. So we are in the realm of higher-order descriptive set theory. It wouldn't be hard to axiomatize exactly which such $\mathcal{R}$ yield a topology, but I won't do this because I would probably leave something out by accident.

A presentation of $\tau$ in this form witnesses the $BCP_0^+$ condition if: $\exists \alpha < \Theta$ $\forall g \in \omega^{\omega}$ the upward $\subseteq$ closure of $\{A : A\mathcal{R}\{g\} \text{ and } A \text{ has Wadge degree } < \alpha \}$ is $\{A : A \mathcal{R} \{g\}\}$

A presentation of $\tau$ in this form witnesses the $BCP_0^+$ condition if: $\exists \alpha < \Theta$ $\forall g \in \omega^{\omega}$ the downward $\subseteq$ closure of $\{A : \neg A\mathcal{R}\{g\} \text{ and } A \text{ has Wadge degree } < \alpha \}$ is $\{A : \neg A \mathcal{R} \{g\}\}$

To motivate Noah's question a bit for those who don't want to read between the lines: usually a relation on $\mathcal{P}(\omega^{\omega})$ is hard to get your hands around. Noah is interested in cases where you can recover $\tau$ completely just from the restriction of $\mathcal{R}$ to a nice, small (meaning, size $\leq 2^{\aleph_0}$) subclass of $\mathcal{P}(\omega^{\omega}) \times \mathcal{P}(\omega^{\omega})$. Then you can hope to represent $\mathcal{R}$ as a relation on $\omega^{\omega}$, and use all your tools from classical descriptive set theory.

Noah's first observation is that if a topological space has base of compact sets of size at most $2^{\aleph_0}$, there is a representation in which you can recover $\tau$ completely from the restriction of $\mathcal{R}$ to the finite powerset $\mathcal{P}_{Fin}(\omega^{\omega})$

His second observation is that under a certain condition, you can recover $\tau$ completely from the restriction of $\mathcal{R}$ to the $\Pi^0_1$ powerset $\mathcal{P}_{\Pi^0_1}(\omega^{\omega})$.

These conditions in $BCP_0^+$ and $BCP_0^-$ are conditions to ensure a (relatively) easy recovery process.

• This seems a good explanation to me! However, it's worth pointing out that you've rephrased my goal: "Noah is interested in cases where you can recover $\tau$ completely just from the restriction of $\mathcal{R}$ to a nice, small (meaning, size $\le 2^{\aleph_0}$) subclass" uses the fact that a family of continuum-many sets is bounded in the Wadge hierarchy, even though the Wadge hierarchy has length only $\Theta$; this fact is pretty trivial, since we can code continuum-many sets into a single set appropriately, but I want to point out the change in phrasing even though it's easy. – Noah Schweber Aug 12 '18 at 16:24