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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.

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    $\begingroup$ It is usually called "separative" not "separable". $\endgroup$ – Joel David Hamkins Dec 11 '19 at 17:20
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Dec 12 '19 at 10:17
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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 $n$.

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  • $\begingroup$ Thanks a lot @Joel David Hamkings. I'm new studying forcing I didn't know that last facts. $\endgroup$ – Gabriel Medina Dec 11 '19 at 17:35
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    $\begingroup$ If you have a name for an element of a set in the ground model, then it will always be dense (below any condition forcing that) to decide which particular element it is. The reason is that otherwise you can find a G containing some condition below the given condition, in which the interpretation of the name isn't any particular one of those elements, contrary to the assumption. $\endgroup$ – Joel David Hamkins Dec 11 '19 at 17:38
  • $\begingroup$ 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. $\endgroup$ – Gabriel Medina Dec 11 '19 at 17:41
  • $\begingroup$ 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 $P$? $\endgroup$ – Gabriel Medina Dec 11 '19 at 20:30
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    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Dec 12 '19 at 10:24

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