Forcing homeomorphism

Question

I don't know much about forcing, only the basics.

But I know for instance that forcing can make two first-order structures isomorphic, by collapsing their cardinals to be $\aleph_0$. For this to work, it suffices that they were $\mathcal{L}_{\infty, \omega}$-elementarily equivalent in the "first" universe.

I've read some author say that if two such structures weren't isomorphic to begin with, it was for "a silly reason", i.e. cardinality.

But I wonder if this could work with topological spaces : can two topological spaces not be homeomorphic "for silly reasons" ? More precisely, in what way can forcing to change cardinals modify the answer to "Are $X$ and $Y$ homemorphic ?" ? More generally, how can forcing change the answer ?

And what conditions on $X$ and $Y$ can we find so that they are homemorphic in a forcing extension ? (A condition like the one for first-order structures)

The answers to these questions need not be extremely technical since, as I said, I only know the very basics of forcing theory.

Answer 1

FURTHER EDIT: 35093731895230467514051 raises a valuable point in the comments - do we want our space to "gain points" when we move to a generic extension?

The ur-example of this is the usual space of real numbers. If $V\subseteq W$ are models of set theory with the same $\omega$, then there are two ways to think of "$W$'s version of $\mathbb{R}^V$" - namely, as $\mathbb{R}^V$ with the topology induced by the set $\{$open sets of reals$\}^V$, or as $\mathbb{R}^W$ with the usual topology.

My answer below focuses on the former. There are certainly interesting results to be gotten in this direction; see e.g. this paper by Kunen. However, one might also be interested in the latter (and indeed the latter approach seems probably more interesting in general). Some interesting difficulties can crop here. If I have a definition of a space, $\varphi$, then I can talk about $\varphi^W$ without a problem. However, two different definitions of the same space in $V$ might not define the same space in $W$, so this ultimately is really about the definitions rather than the specific spaces themselves. Somehow, we want a way to take a (reasonable) space and get a "canonical definition" (or something like that) for it, which we can then use to get a notion of what that space should be in a different model.

For this direction, I suggest this paper of Zapletal. (That paper isn't really about generating canonical definitions in any sense, so you might view my focus on definitions above as a bit of a red herring; still, I think that at some level there is a notion of "canonical definition" kicking around, it's just that that notion is much more structural - how the space in question fits into an appropriate category of spaces.)

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EDIT: an important distinction between the spaces and structures contexts is: while a structure remains a structure after forcing, a space need not remain a space. This is because forcing may introduce new families of open sets, whose union was not accounted for in the ground model.

Of course, the old topology will form a subbase for a new topology. For that reason it's probably better to talk about spaces in terms of their subbases; I'll just ignore this point below, since it won't affect anything.

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An even simpler example of potential but not actual homeomorphism: if $X, Y$ are infinite sets with different cardinalities, then their discrete (or indiscrete) topologies are potentially homeomorphic but not actually homeomorphic. This exactly parallels the non-topological situation.

The interesting part, of course, is characterizing potential homeomorphism. You mention $\mathcal{L}_{\infty\omega}$-equivalence of structures as sufficient for isomorphism; it is in fact necessary as well, so we have a good characterization of potential isomorphism. Another characterization is given by the infinite-length back-and-forth game. So it's reasonable to hope for a good characterization(s) of potential homeomorphism.

It turns out that we can find such characterizations, which parallel the structures case:

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The first is a game characterization. It's ultimately subsumed by the next section, but I think it's interesting to think about on its own. Given spaces $\mathcal{X}=(X, \tau)$ and $\mathcal{Y}=(Y, \sigma)$ we let $G_{ph}(\mathcal{X},\mathcal{Y})$ be the following game:

• On their move, a given player will play either an element of $X$, an element of $\tau$, an element of $Y$, or an element of $\sigma$.

• What sort of thing player $2$ plays is determined by what sort of thing player $1$ plays:

  • If player $1$ plays an element of $X$ (resp. $Y$) then player $2$'s next move will be an element of $Y$ (resp. $X$).

  • If player $1$ plays an element of $\tau$ (resp. $\sigma$) then player $2$'s next move will be an element of $\sigma$ (resp. $\tau$).

• Player $2$ must play "bijectively" on points. E.g. suppose Player $2$ plays $y\in Y$ in response to Player $1$'s playing $x\in X$; then if Player $1$ later plays $x$ again Player $2$ has to respond with $y$, and if Player $1$ later plays $y$ then Player $2$ has to respond with $x$.

• Player $2$'s "set moves" restrict their "point moves": e.g. if Player $2$ plays $V\in\sigma$ in response to player $1$'s play of $U\in\tau$, then they've committed to responding with a point in $V$ whenever Player $1$plays a point in $U$.

(I'm leaving it as an exercise to fully write out the details of this game, but the idea should be clear.)

Say that Player $1$ wins a play of the game if Player $2$ eventually makes an illegal move, and Player $2$ wins otherwise. It's now easy to show that for countable spaces (EDIT: that is, countable set of points and countable set of opens), Player $2$ winning is equivalent to homeomorphism, and this same line of thought translates to:

$\mathcal{X}$ and $\mathcal{Y}$ are potentially homeomorphic iff Player $2$ has a winning strategy in $G_{ph}(\mathcal{X},\mathcal{Y})$.

(Note that $G_{ph}$-games are open games, hence always determined in ZFC, so there's no determinacy issue here. We are using choice, though; if we want to work in ZF alone we need to talk about quasistrategies.)

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The second characterization of potential homeomorphism - or rather, class of characterizations - can be given by translating directly into the language of structures. Any topological space $\mathcal{X}=(X,\tau)$ has an associated first-order structure $S_\mathcal{X}$:

• $S_\mathcal{X}$ has two sorts, a "points" sort and an "opens" sort; the elements of the points sort are exactly the elements of $X$, and the elements of the opens sort are exactly the elements of $\tau$.

• The language of $S_\mathcal{X}$, besides a binary relation, include the binary relation symbol "$\in$" which is interpreted in the obvious way.

It's easy to check that $\mathcal{X}$ and $\mathcal{Y}$ are homeomorphic iff $S_\mathcal{X}$ and $S_\mathcal{Y}$ are isomorphic; so we can now "port over" the characterizations of potential isomorphism of structures. In particular, note that the game characterization above is the "pullback" of the usual game characterization of potential isomorphism along the construction $\mathcal{X}\mapsto S_\mathcal{X}$.

This sort of translation is an instance of a useful general phenomenon: while first-order theories are quite weak, first-order structures are quite expressive. Note that of course the class of structures of the form $S_\mathcal{X}$ for some topological space $\mathcal{X}$ is definitely not first-order definable! This old question of mine contains a similar use of this idea.

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EDIT: It's worth mentioning that in the structures case, a lot of interesting stuff happens when you restrict the forcing notions you're using. The Levy collapse $Col(\omega, A\cup B)$ is universal for determining whether $A$ and $B$ are potentially isomorphic: making everything relevant countable solves all possible problems (I have that slogan on a t-shirt). However, we can have stronger notions of potential isomorphism like "become isomorphic in a c.c.c. forcing extension," and here the situation is more complex.

This was studied in particular from a model-theoretic perspective by Baldwin, Laskowski, and Shelah in their papers "Forcing isomorphism" and (sans Baldwin) "Forcing isomorphism II".

I have no idea how similar variations work in the context of topological spaces, but it seems potentially quite interesting. The $\mathcal{X}\mapsto S_\mathcal{X}$ construction suggests that it won't yield a picture which is too new from that of the structures context, but there might still be cool stuff.

Answer 2

Let me answer this question in the context where after forcing one does not just add open sets to the topology but one adds points to the topology as well. I am going to answer this question in the context of point-free topology and where all spaces considered satisfy higher separation axioms (regular and above), but don't worry, for every regular space $X$, after collapsing enough cardinals to $\omega$, you are spatial again. This interpretation of what it means to extend a topological space to a forcing extension by adding open sets and points takes some work to set up, so bear with me. I could have made this answer much shorter and more direct, but a greater background is usually a good thing.

This process that I am going to outline of extending topological spaces works well for forcing extensions, but I am unsure of how well it will work for extensions of the universe that are not forcing extensions. Throughout this post, I will use the Boolean-valued model approach to forcing.

Background and set-up

A frame is a complete lattice $L$ that satisfies the distributivity law $x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i}).$ Frames are typically thought of as point-free topological spaces since if $(X,\mathcal{T})$ is a topological space, then $\mathcal{T}$ is a frame; furthermore, nearly all of the notions from general topology seamlessly generalize to frames (frames actually behave much better than topological spaces). Yes. All the information in a Hausdorff space $(X,\mathcal{T})$ is contained in the frame $\mathcal{T}$ since two Hausdorff spaces are homeomorphic if and only if their frames of open sets are homeomorphic. As always, I refer you to the book Frames and Locales: Topology without Points by Picado and Pultr for more information on point-free topology.

Observe that every complete Boolean algebra is a frame. Complete Boolean algebras satisfy high separation axioms; complete Boolean algebras are always regular, normal, paracompact, ultraparacompact, etc. Since complete Boolean algebras are frames and complete Boolean algebras are the objects that people use to produce forcing extensions, one should ask whether one can use complete Boolean algebras to say anything about frames in forcing extensions; and the answer to this question is yes, complete Boolean algebras give a characterization of frames in forcing extensions and they give one a method of transferring spaces from a ground model to a forcing extension.

If $(X,\mathcal{T})$ is a Hausdorff space, then the set $X$ is in a one-to-one correspondence with the set of all frame homomorphisms $\phi:\mathcal{T}\rightarrow 2$ by the mapping $x\mapsto \phi_{x}$ where $\phi_{x}(U)=1$ if and only if $x\in U$. Therefore, in point-free topology, we define a point in a frame $L$ to be a frame homomorphism $\phi:L\rightarrow 2$. The notion of a point in point-free topology gives us a way to define a $B$-valued point. If $L$ is a frame and $B$ is a complete Boolean algebra, then we define a $B$-valued point to be a frame homomorphism $\phi:L\rightarrow B$. The set $\mathrm{Sp}_{B}(L)$ of $B$-valued points becomes a $B$-valued set closed under mixing where we define $\|\phi=\theta\|$ to be the largest $b\in B$ such that $\phi(x)\wedge b=\theta(x)\wedge b$ for each $b\in B$. Therefore, since $\mathrm{Sp}_{B}(L)$ is a $B$-valued set, we may immerse the structure $\mathrm{Sp}_{B}(L)$ into $V^{B}$, so we shall consider $\mathrm{Sp}_{B}(L)\in V^{B}$. The frame $L$ allows us to topologize $\mathrm{Sp}_{B}(L)$ in $V^{B}$. For each $x\in L$, let $\dot{x}:\mathrm{Sp}_{B}(L)\rightarrow B$ be the function where $\dot{x}(\phi)=\phi(x)$. Then $V^{B}\models``\text{$\{\dot{x}|x\in L\}$ is a basis for a topology on $\mathrm{Sp}_{B}(L)$.}"$ Therefore, every frame $L$ produces a topological space in the Boolean-valued model $V^{B}$; frames should therefore be thought of as virtual topological spaces whose points do not live in $V$ but live in $V^{B}$. If $(X,\mathcal{T})$ is a topological space, then for each $x\in X$, we have $\phi_{x}\in\mathrm{Sp}_{B}(L)$, but in general $\mathrm{Sp}_{B}(L)$ contains points which are not simply mixes of homomorphisms $\phi_{x}$, so this process adds points to Hausdorff spaces in forcing extensions.

This process for adding points to spaces in forcing extensions has been studied by Zapletal [1] and Fremlin [2].

One can lift frames in $V$ to frames in $V^{B}$ in a purely point-free manner, but these two methods of lifting frames in $V$ to frames in $V^{B}$ are equivalent in sufficiently large forcing extensions where the resulting frame lifted in a point-free manner is spatial.

Suppose that $L$ is a frame and $B$ is a complete Boolean algebra that is a subframe of $L$. Then define $\|x=y\|$ to be the largest $b\in B$ such that $x\wedge b=y\wedge b$ and $\|x\leq y\|$ to be the largest $b\in B$ such that $x\wedge b\leq y\wedge b$. Then $L$ becomes a Boolean-valued partial ordering and $V^{B}\models``\text{$L$ is a frame.}"$ Furthermore, every frame in $V^{B}$ is isomorphic to an object of this form.

Suppose that $L,M$ are frames. Then $L,M$ are both subframes of the frame coproduct $L\oplus M$. In particular, if $L$ is a frame and $B$ is a complete Boolean algebra, then since $B$ is a subframe of $L\oplus B$, $V^{B}\models``\text{$L\oplus B$ is a frame}."$ We shall therefore consider, $L\oplus B$ to be canonical extension of $L$ to a $B$-valued frame.

These two methods of extending frames to forcing extensions are in some sense one and the same $V^{B}\models``\text{The space $\mathrm{Sp}_{B}(L)$ is canonically homeomorphic to the space of all points in $L\oplus B$.}"$ Furthermore, if $L$ is regular and $B$ collapses $|L|$ to $\omega$, then $V^{B}\models\text{$L\oplus B$ is spatial}$, and therefore after collapsing cardinals these two approaches are equivalent. I however prefer the completely point-free approach since in $V^{B}$, the space $\mathrm{Sp}_{B}(L)$ typically does not contain as much information as the $B$-valued frame $L\oplus B$.

Why this construction?

The construction of lifting $L$ to the $B$-valued frame $L\oplus B$ behaves much better than you would expect and much better than if one lifts spaces to spaces instead of frames; Zapletal needed to restrict his attention to the rather narrow class of regular Hausdorff images continuous open images of $G_{\delta}$-subsets of compact regular spaces in order for everything to work correctly because Zapletal was working in the context of topological spaces instead of frames.

Theorem: Suppose that $L$ is a frame and a complete Boolean algebra $B$ is a subframe of $L$. Then suppose that $P$ is one of the following properties: regularity, complete regularity, paracompactness, zero-dimensionality, ultraparacompactness. Then $L$ satisfies $P$ if and only if $V^{B}\models``\text{$L$ satisfies $P$.}"$ This construction is also functorial; if $L,M$ are frames that both contain the complete Boolean algebra $B$ as a subalgebra, then if $f:L\rightarrow M$ is a frame homomorphism such that $f(b)=b$ for each $b\in B$, then $V^{B}\models``\text{$f$ is a frame homomorphism from $L$ to $M$}"$ and all frame homomorphisms in $V^{B}$ can be produced this way.

Theorem: Suppose now that $L$ is a frame and $B$ is some complete Boolean algebra. Suppose $P$ is one of the following properties: compactness, regularity, local connected connectedness, compact connectedness. Then $L$ satisfies $P$ if and only if $V^{B}\models``\text{$L\oplus B$ satisfies $P$.}"$

Theorem: Suppose that $L$ is a frame and $B$ is a complete Boolean algebra. Let $P$ be one of the following properties: complete regularity, paracompactness, complete metrizability, second countability, zero-dimensionality, ultraparacompactness, has dense set of points. Then $L$ satisfies property $P$ if and only if $V^{B}\models\text{$L\oplus B$ satisfies property $P$.}$

Forcing takes bad spaces and eventually turns them into very good spaces.

Theorem: Suppose that $L$ is a regular frame. Then there is a complete Boolean algebra such that $V^{B}\models``\text{the frame $L\oplus B$ is a Polish space.}"$

Forcing isomorphisms: the zero-dimensional case

Collapsing cardinals often takes zero-dimensional frames and turns these frames into well-known topological spaces like the Baire space or the Cantor space.

Theorem: Suppose that $L$ is a zero-dimensional frame without isolated points where there does not exist a compact complemented element $x\in L\setminus\{0\}$. Then there exists a cardinal $\lambda$ so that if $B$ is a complete Boolean algebra that collapses $\lambda$ to $\omega$, then $V^{B}\models``\text{$L\oplus B$ is homeomorphic to the Baire space.}$"

Theorem: Suppose that $L$ is a compact zero-dimensional frame without any isolated points. Then there exists a cardinal $\lambda$ such that if $B$ collapses $\lambda$ to $\omega$, then $V^{B}\models``\text{$L\oplus B$ is homeomorphic to the Cantor space.}"$

Observation: Suppose $L$ is a compact zero-dimensional frame. Then if $x\in L\oplus B$, then $V^{B}\models``\text{$x$ is complemented}"$ if and only if there is some partition $p$ of $B$ along with complemented elements $x_{a}\in L$ such that $x=\bigvee_{a\in p}(x_{a}\wedge a).$

Therefore, our method of extending a space preserves the Stone duality between Boolean algebras and compact zero-dimensional spaces, so by Stone-duality the problem of whether two compact zero-dimensional spaces becomes equivalent to the problem of whether two Boolean algebras become isomorphic in forcing extensions. Luckily, the countable Boolean algebras have been classified up to isomorphism by Ketonen [3], and one can probably extend this classification to obtain complete invariants that determine precisely when two Boolean algebras become isomorphic in a forcing extension (I have not found any classification for the zero-dimensional Polish spaces similar to Ketonen's classification of countable Boolean algebras).

The connected case and general case.

Since forcing preserves locally connected connectedness and compact connectedness, one should not hope to obtain an easy charaterization of when two regular frames are homeomorphic in a forcing extension unless one has a characterization of when Polish spaces are homeomorphic in general (and I do not know of any such characterization).

I conjecture that one can characterize the compact regular spaces which become homeomorphic in forcing extension using forcing extensions using nerves of finite covers. I also conjecture that we can characterize when compact regular spaces become homeomorphic in forcing extensions using some sort of logic similar to $\mathcal{L}_{\infty,\omega}.$

Let me give a characterization of when two regular spaces become homeomorphic in some forcing extension.

Theorem: Suppose that $\mathcal{S},\mathcal{T}$ are regular frames. Then the following are equivalent:

1. There is a complete Boolean algebra $B$ so that for every complete Boolean algebra $C$ with $B\subseteq C$, we have $V^{C}\models``\text{The spaces $\mathrm{Sp}_{C}(\mathcal{S})$ and $\mathrm{Sp}_{C}(\mathcal{T})$ are homeomorphic.}"$

2. There is a regular frame $N$ and a frame isomorphism $i:\mathcal{S}\oplus N\rightarrow\mathcal{T}\oplus N$ such that $i(n)=n$ for each $n\in N$.

There is no spatial version of the above theorem since the isomorphism $i$ is analogous to a homeomorphism $f:X_{1}\times Y\rightarrow X_{2}\times Y$ where for all $y\in Y$ there is some homeomorphism $f_{y}:X_{1}\rightarrow X_{2}$ with $f(x,y)=(f_{y}(x),y)$ for all $x,y$.

[1] https://people.clas.ufl.edu/zapletal/files/interpretations2.pdf

[2] David H. Fremlin. Topological spaces after forcing.

[3] J. Ketonen, The structure of countable Boolean algebras, Ann. of Math. (2) 106 (1978), 41-89.