in Set Theory

Some Forcing Notions

In this post I will give a few examples of forcing notions and explore their combinatorial properties. Most of this is directly taken from “combinatorial set theory with a gentle introduction to forcing” by Halbeisen and condensed down a bit.

Ultrafilter Forcing \mathbb{U} := ([\omega]^\omega, \supseteq^*): 

What makes this one useful, is that a \mathbb{U}-generic filter G over \textbf{V} is also a Ramsey Ultrafilter in the extension, different from all ultrafilters in the ground model. Hence, when forcing with this notion, one adds at least one Ramsey Ultrafilter to the ground model. There are a few things one needs to check to be able to conclude:

  • RUF’s \math are characterised by the fact that for each coloring

        \[ \pi : [\omega]^2 \rightarrow 2\]

    there is a x \in \mathcal{U} such that \pi|_{[x]^2} is constant, i.e. for each coloring, \mathcal{U} contains a monochromatic set. We will conclude that the new filter G has such a set for each coloring in the ground model. However one also needs to make sure that no new colorings get added, so that indeed the property holds true for all colorings in \textbf{V}[G]. For this one checks that \mathbb{U} is \omega_1-closed, hence preserves all cardinals \le \omega_1, and in particular all real numbers r : \omega \rightarrow \omega. Since the reals are in one-to one correspondence with the two colorings of [\omega]^2, there are no new colorings appearing in \textbf{V}[G]
  • G is in fact an ultrafilter which is different from every ultrafilter in the ground model. For the first fact consider the open dense set: 

        \[ \mathcal{D}_x = \{ y \in [\omega]^\omega : x \supseteq^* y \vee \omega \setminus x \supseteq^* y\}\]

    G being \mathbb{U}-generic means it intersects this set, and hence either x \in G or \omega \setminus x \in G. For the second fact let \mathcal{V} be an ultrafilter in the ground model and consider the following open dense set:

        \[ \mathcal{D}_{\mathcal{V}} = \{ y \in [\omega]^\omega : y \not\in \mathcal{V}\}\]

    Here, G being \mathbb{U}-generic implies, that for every ultrafilter \mathcal{V} in the ground model, there is a set in G that is not contained in \mathcal{V}, hence in particular \mathcal{V} \ne G.

Cohen Forcing \mathbb{C}_\kappa = (\text{Fn}(\omega \times \kappa, 2), \subset): 

This forcing notion can be used to show that \neg \textbf{CH} is consistent with \textbf{ZFC}. Suppose \textbf{V} was a model of \textbf{ZFC}+\neg \textbf{CH}. Then forcing with \mathbb{C}_\kappa adds \kappa distinct new reals to the model, the so called Cohen reals,  and preserves all cardinals. In fact if G is \mathbb{C}_\kappa-generic over \textbf{V}, we get that \bigcup G is a function \omega \times \kappa \rightarrow 2. This can be concluded by looking at the open dense set:

    \[ \mathcal{D}_\lambda = \{ p \in \text{Fn}(\omega \times \kappa, 2) : \alpha \in \text{dom}(p)\}\]

which shows that \bigcup G is defined for all \lambda \in \kappa. Since \mathfrak{c} counts exactly the number of reals, if we choose \kappa > \omega_1 we get: 

    \[ \mathfrac{c} \ge \kappa > \omega_1 .\]

Hence we see that the continuum hypothesis no longer holds. However there are a few things one needs to check given a \mathbb{C}_\kappa-generic filter G over \textbf{V}:

  • The added reals are indeed are all distinct and different from all existing reals. First, for \lambda \in \kappa, we define the new reals by:

        \[ r_\lambda = \left( \bigcup G \right)|_{\omega \times \{ \lambda\}}\]

    Then you look at: 

        \[ \mathcal{D}_{\lambda, \lambda'} = \{ p \in \text{Fn}(\omega \times \kappa, 2): \exists n \in \omega (p(\lambda, n) \ne p(\lambda', n))\}\]

    which exactly tells you that no two r_\lambda‘s are the same, and consider a similar open dense set to show that they also differ from all reals in the ground model.
  • The forcing doesn’t collapse any cardinals, since otherwise \kappa might shrink down to \omega_1 again and we wouldn’t have proved anything. This is shown using the fact that \mathbb{C}_\kappa is ccc by a consequence of the \Delta-system lemma, and hence preserves all infinite cardinals. 

Collapsing Forcing \mathbb{K}_{\kappa, \omega_1}:

The conditions are given by partial functions \omega_1 \rightarrow \kappa with countable domain, and the order is the inclusion.  As the name suggests, this forcing can be used to collapse a cardinal \kappa to \omega_1. In particular for \kappa = \mathfrak{c} we have that for a \mathbb{K}_{\kappa, \omega_1}-generic filter over \mathbf{V}

    \[ \textbf{V}[G] \vDash \textbf{CH} \]

One needs to show a couple of things for that:

  • No new reals get added. This can be assured by verifying the \sigma-closedness of \mathbb{K}_{\kappa, \omega_1}.
  • \mathbb{K}_{\kappa, \omega_1} adds a surjective function f : \omega_1 \twoheadrightarrow \kappa, so that

        \[ \textbf{V}[G] \vDash \kappa \le \omega_1 \]

Sacks Forcing \mathbb{S}:

Here the conditions are given by:

    \[ S = \{ T \subset {}^{<\omega}2 : T \text{ is a perfect tree}\}\]

and \mathbb{S} = (S, \supset). A tree T \subset {}^{<\omega}2  = \bigcup_{n \in \omega} {}^n2 is a subset that is closed under taking initial segments, i.e. if p \in T, then p|_k \in T for all k \in \omega. We also say that t extends s, written s \prec t, if s is an initial segment of t. A perfect tree, is a tree that splits indefinitely, i.e. \forall t \in T \exists t' \ne t''\in T:  t \prec t' \wedge t \prec t''. Sacks forcing allows us to build a model \textbf{V}[G], such that there is no “intermediate model” between \textbf{V} and \textbf{V}[G], meaning if \textbf{V} \subset \textbf{W} \subset \textbf{V}[G] is a model of \mathrm{ZFC} then either \textbf{V} =  \textbf{W} or \textbf{W} = \textbf{V}[G].  Forcing with perfect trees was used by Gerald Enoch Sacks to produce a real a with minimal degree of constructibility.

Sacks Forcing adds neither splitting nor unbounded reals

Miller Forcing \mathbb{M}:

Here the conditions are given by:

    \[ M = \{ T \subset \text{seq}(\omega) : T \text{ is a superperfect tree} \} \]

and \mathbb{M} = (M, \supset). Here superperfect means that for s \in T:

    \[ \text{next}(t) := \{ t\in T: \text{dom}(t) = \text{dom}(s)+1 \wedge s \prec t\}\]

is infinite for all t \in T.

It adds unbounded reals but no splitting reals.

Mathias Forcing \mathbb{M}_{\mathcal{U}}:

where \mathcal{U} is a fixed Ramsey family. The conditions are given by: 

    \[ M_{\mathcal{U}} = \{ (s,x) : s\in \text{fin}(\omega) \land x \in \mathcal{U} <!-- /wp:paragraph --> <!-- wp:paragraph --> \land \text{max}(s) < \text{min}(x)\}\]

and the preorder is given by: 

    \[ (s,x) \le (t, y) \Leftrightarrow s \subset t \land y \subset x \land t \setminus s \subset x \]

Mathias forcing can be used to construct a pseudointersection of an ultrafilter.

Restricted Laver Forcing \mathbb{L}_{\mathcal{U}}: 

Here \mathcal{U} \subset [\omega]^\omega  is a fixed ultrafilter. Now a Laver tree is a tree T \subset \text{seq}(\omega) if it has a special element s \in T, called the stem, such that for any t \in T we have t \prec s \vee s \prec t, and for every  t \in T with s \prec t, we have that \text{next}(t) is infinite. A   Laver tree restricted to \mathcal{U} is a Laver tree such that \text{next}(t) \in \mathcal{U}. The conditions of Laver forcing are exactly the Laver trees restricted to \mathcal{U}, with reverse inclusion.

This forcing is \sigma-centred and adds a real which is almost homogenous for all colorings in the ground model.

Silver Forcing \mathbb{S}_{\mathcal{E}}:

Here \mathcal{E} is a fixed P-family. Define: 

    \[ S_\mathcal{E} = \bigcup\{ {}^x2:  \omega \setminus x \in \mathcal{E}\}\]

and for p,q \in S_\mathcal{E}:

    \[ p \le q  \Leftrightarrow \text{dom}(p) \subset \text{dom}(q) \wedge q|_{\text{dom}(p)} = p\]

Then \mathbb{S}_{\mathcal{E}} = (S_\mathcal{E}, \le). For \mathcal{E} = [\omega]^\omega  write \mathbb{S} and just call it Silver forcing. Silver reals are given by: 

    \[ r = \bigcup G\]

In general, Silver filters cannot be reconstructed from Silver reals, since multiple Silver filters yield the same real. However, if \mathcal{E} is an ultrafilter we can always do the reconstruction.

Shelah’s Tree forcing \mathbb{T}: See my other post here.