About product of Baire spaces and forcing

Question

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:

7. 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.

Answer 1

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 $n$.