# About product of Baire spaces and forcing

Let $$\mathbb{P}=\langle P, \leq \rangle$$ be a p.o.

• Two elements $$p$$ and $$q$$ of it are called compatible if there is an $$r \in \mathbb{P}$$ such that $$r\leq p$$ and $$r \leq q$$; otherwise they are called incompatible.
• A subset $$D$$ of $$\mathbb{P}$$ is said to be dense in $$\mathbb{P}$$ if for each $$p\in \mathbb{P}$$ there is a $$d\in D$$ such that $$d\leq p$$.
• A partial ordering $$\leq$$ is said to be separative if for any two elements $$p$$ and $$q$$ of $$\mathbb{P}$$ either $$q\leq p$$ or there is an $$r\leq q$$ that is not compatible with $$p$$.

• We define on $$P$$ a topology $$\tau_{\leq}$$ by declaring each set $$\{q : q\leq p\}$$ to be open. Note that if the space is derived from a p.o. set as above, then any such countable intersection of open sets is necessarily open.

Now let $$\mathcal{M}$$ be any model and $$\mathbb{P}$$ any p.o. set in $$\mathcal{M}$$, let $$G$$ be an $$\mathcal{M}$$-generic subset of $$\mathbb{P}$$, and $$\mathcal{M}[G]$$ the corresponding generic extension of $$\mathcal{M}$$.

The elements of the p.o. set $$\mathbb{P}$$ are often called conditions. We say that a condition $$p$$ forces a sentence $$A$$ (to be true in the model $$\mathcal{M}[G]$$) if $$A$$ holds in $$\mathcal{M}[G]$$ whenever $$G$$ contains $$p$$. In symbols this is written $$p \Vdash A$$.

Fundamental theorem of forcing

A sentence $$A$$ is satisfied in $$\mathcal{M}[G]$$ if and only if there is a condition $$p\in G$$ such that $$p\Vdash A$$.

From a properties of generic subsets and the fundamental theorem of forcing it follows that to prove that $$A$$ holds in $$\mathcal{M}[G]$$ it suffices to prove that $$\{p:p\Vdash A \}$$ is a dense subset of $$\mathbb{P}$$.

Proposition 1 The basic properties of the forcing relation are as follows.

1. $$p\Vdash \neg A$$ if and only if no $$q\leq p$$ forces $$A$$; We note that $$p\Vdash \neg \neg A$$ is equivalent to $$p\Vdash A$$, therefore,

2. $$p\Vdash A$$ if and only if no $$q\leq p$$ forces $$\neg A$$,

3. $$p \Vdash A \wedge B$$ if and only if $$p\Vdash A$$ and $$p\Vdash B$$;

4. $$p \Vdash A \vee B$$ if and only if $$(\forall q\leq p)(\exists r\leq q)[r\Vdash A \hspace{0.1cm}\text{or}\hspace{0.1cm} r\Vdash B]$$;

5. $$p\Vdash \forall x A(x)$$ if and only if $$(\forall x \in \mathcal{M}^{\mathbb{P}})[p \Vdash A(x)]$$;

6. $$p\Vdash \exists x A(x)$$ if and only if $$(\forall q\leq p)(\exists r\leq q)(\exists x \in \mathcal{M}^{\mathbb{P}})[r\Vdash A(x)]$$.

An important property of the forcing relation is the following:

1. for any sentence $$A$$ and any $$p\in \mathbb{P}$$ $$(\exists q\leq p)[q\Vdash A \hspace{0.1cm} \text{or}\hspace{0.1cm} q\Vdash \neg A]$$

The most important connection between forcing and topology is as follows:

Lemma 1. Suppose that $$P$$ is a separable p.o. Then $$(P, \tau_{\leq})$$ is a Baire space if and only if for every $$\mathcal{M}$$-generic subset $$G$$ of $$\mathbb{P}$$ no new $$\omega$$-sequences of ordinals occur in $$\mathcal{M}[G]$$.

Proof

First, suppose that $$(P, \tau_{\leq})$$ is a Baire space, and let $$f\in \mathcal{M}[G]$$ with $$dom f=\omega$$, whose values are ordinals, as the formula $$f: \omega\to \text{Ord}$$ is a function in $$\mathcal{M}[G]$$ is satisfied, then by Fundamental Theorem of forcing, there exists $$p^{\prime}\in G$$ such that $$p^{\prime}\Vdash f:\omega \to \text{Ord}$$.

For every $$n\in \omega$$ consider the set $$D_{n}=\{p \in P : (\exists \alpha\in\text{Ord})(p \Vdash f(\check{n})=\check{\alpha}” ) \}$$. Note that $$D_{n}\not=\emptyset$$, because there is $$q\leq p^{\prime}$$ such that $$q\in D_{n}$$.

We claim that for each $$n\in\omega$$, $$D_{n}$$ is open and dense. For this, let $$p\in P$$, if $$q \leq p$$ we are done. So suppose that there is $$r\leq q$$ that is not compatible with $$p$$, by Proposition 1.(6), there is $$s\leq p$$ such that $$s\Vdash f(\check{n})=\check{\alpha}”$$ or $$s\Vdash f(\check{n})\not=\check{\alpha}”$$, for some $$\alpha \in \text{Ord}$$. If $$s\Vdash f(\check{n})=\check{\alpha}”$$ we are done, so suppose that $$s \leq p$$ and $$s\Vdash f(\check{n})\not=\check{\alpha}”$$. My question is, in the latter case, how can I conclude that $$D_{n}$$ is dense?

Thanks a lot.

• It is usually called "separative" not "separable". – Joel David Hamkins Dec 11 '19 at 17:20
• It's good that you write a lot. But please for next time, make your actual question prominent. At the very least, in its own separate paragraph. – Asaf Karagila Dec 12 '19 at 10:17

The way you set this up, it might not be dense, since you only have that $$p'$$ forces that $$f$$ is a function from $$\omega$$ to the ordinals. Perhaps other incompatible conditions force that $$f$$ is not a function, or empty, or is whatever, in such a way that $$f(\check n)$$ is not meaningful.
But you can fix things by arguing differently. First, you don't really need the dense sets to be dense, but just dense below a condition that you know is in the filter. So it suffices to work below $$p'$$.
And now, the basic observation to make is that if a condition $$p'$$ forces that $$\dot f$$ is a function from $$\omega$$ to the ordinals, then for every $$n\in\omega$$ it will be dense below $$p'$$ to decide what the value of $$\dot f(\check n)$$ is. The reason is that if $$p'$$ forces that $$\dot f(\check n)$$ has a value, then what this means is that there is a dense set of conditions that force a particular value, which is what it means for $$D_n$$ to be dense below $$p'$$. If this wasn't dense, then we could find a condition $$p''$$ below $$p'$$, which could not be extended to a condition forcing a particular value of $$\dot f(\check n)$$, and then any generic filter containing $$p''$$ would not make $$\dot f$$ into a function defined at $$n$$.
• I understood, in addition to the proof it is enough that they are dense below $p^{\prime}$, since an open subspace of a Baire remains Baire. – Gabriel Medina Dec 11 '19 at 17:41
• I have one last question, I can prove that $D_ {n}$ is open and dense at $\{q: q\leq p^{\prime} \}$, then $\bigcap _{n\in \omega}D_{n}$ is dense in $\{q: q\leq p^{\prime} \}$, but I need that $G$ intersects $\bigcap _{n\in \omega}D_{n}$, and this will happen if $\bigcap _{n\in \omega}D_{n}$ is dense. How can I get that $\bigcap _{n\in \omega}D_{n}$ is dense in $P$? – Gabriel Medina Dec 11 '19 at 20:30
• For general partial orders, this isn't true, but it is exactly equivalent to the partial order being $(\omega,\infty)$-distributive. I assumed that this is the property you mean by saying the space was `Baire'. So it follows from that assumption. The usual way of stating your lemma is that a forcing notion adds no new $\omega$-sequences over the ground model if and only if it is $\omega$-distributive, which is equivalent to the assertion that the countable intersection of open dense sets is dense. – Joel David Hamkins Dec 12 '19 at 10:24