Collapsing cardinal $\lambda$ to $\omega$ by Cohen foricng

Question

Does this forcing notion: $$\mathbb{P}=\{f:f \text{ is a finite function from }\lambda \text{ to }\{0,1\}\}$$ collapse $\lambda$ to $\omega$ in $V[G]$?

Where can I consult this fact in Kunen's book IIP? How can this $\mathbb{P}$ collapse $\lambda$ to $\omega$ if $$\Vdash_P"2^{\aleph_0}=\lambda"$$ (see the snippet below) which implies $$\Vdash_P"\lambda > \omega"$$

Answer 1

Expanding some discussion in the comments about the likely source of OP's confusion into an answer...

There are two distinct things that can happen in the forcing extension that can potentially impact the value of $\mathfrak c.$

1. The cardinality of the reals in $V$ can collapse. In other words, $\mathfrak c^V$, defined as the unique least ordinal such that there is an $f\in V$ that is a bijection $\mathfrak c^V\to P(\omega)^V$ (where $P(\omega)^V$ is the set of all subsets of $\mathbb N$ that are in $V$), can cease to be a cardinal in $V[G]$. The way this happens is that $V[G]$ has a bijection between $\mathfrak c^V$ and a smaller ordinal (one that obviously can't be in $V$ since $\mathfrak c^V$ is a cardinal in $V$). So in other words, we have $|\mathfrak c^V|^{V[G]} < \mathfrak c^V.$
2. There can be reals in $V[G]$ that are not in $V.$ In other words, $P(\omega)^V\subsetneq P(\omega)^{V[G]}.$ In this case it is perfectly possible that $P(\omega)^{V[G]}$ has a larger cardinality than $P(\omega)^V$ in $V[G].$ (Note that even though $P(\omega)^V$ is defined in reference to $V$, it is a set in $V[G]$ simply because $V\subset V[G]$, so it makes perfect sense to talk about its cardinality in $V[G].$)

A key distinction is that while $\mathfrak c$ is defined as the cardinality of a set meeting some definition, by contrast, $\aleph_1$ and $\aleph_{17}$ (for example) are defined simply by their position in the cardinal hierarchy (one can either think of $1$ and $17$ as stand-ins for fixed ordinals or note that $1$ and $17$ are absolute so it doesn't matter if we think of them as defined). Thus these latter two are only subject to the first kind of disturbance, not the second kind.

In a c.c.c. forcing like your example, it can be shown that the first kind of disturbance cannot occur. For any $\kappa$ that is a cardinal in $V$ there are no new bijections in $V[G]$ between $\kappa$ and a smaller ordinal. In other words, a cardinal in $V$ remains a cardinal in $V[G].$ This means we do not have to distinguish between $|\cdot|^V$ and $|\cdot|^{V[G]},$ nor do we have to distinguish between $\aleph_1^V$ and $\aleph_1^{V[G]}.$ So the change the value of $\mathfrak c$ is entirely driven by the fact that there are more subsets of $\mathbb N$ in the forcing extension. Assuming CH holds in the ground model and $|\lambda|>\aleph_1,$ we have $$|P(\omega)^{V[G]}|\ge \lambda > \omega_1= |P(\omega)^V|.$$

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An instructive contrasting example is to force with countable partial bijections between $P(\omega)^V$ and $\aleph_1^V.$ This forcing notion is not c.c.c., but is $\omega$-closed, which is another nice property with very different implications; for instance, it implies that no new maps $\omega\to V$ exist in $V[G].$ This means that $P(\omega)^V=P(\omega)^{V[G]}$ since new reals would mean new maps $\omega\to 2.$ It also means $\aleph_1^{V} = \aleph_1^{V[G]}$, but for a very different reason than in the previous example: a collapsing bijection $\omega\to \aleph_1^V$ would be a new map $\omega\to V.$ Note also that this same reasoning doesn't extend to $\aleph_{17}$: it won't be collapsed to be countable, but it could be collapsed to some smaller uncountable cardinal.

In fact, provided our ground model violates the CH, a cardinal will collapse in this forcing, namely the cardinality of the reals in the ground model. The generic set gives a bijection between $P(\omega)$ and $\aleph_1$ (no superscript needed on either of these per the last paragraph), so CH holds in $V[G].$ Thus we have $$\mathfrak c^{V[G]}=|P(\omega)^{V[G]}|^{V[G]} = \aleph_1 < |P(\omega)^V|^V =\mathfrak c^V.$$

(I left the superscripts on the $P(\omega)$ just for emphasis... as mentioned before, they are the same set in this particular case, in contrast to the previous example of Cohen forcing where there are new reals in the forcing extension.)

So for instance if we had $\mathfrak c=\aleph_{17}$ in $V$, then $|\aleph_{17}^V|^{V[G]} = \aleph_1^{V[G]},$ so $\aleph_{17}$ in $V$ has collapsed to have cardinality $\aleph_1$ in $V[G].$

Answer 2

No, Cohen forcing does not collapse any cardinals. Specifically, no c.c.c. forcing will ever collapse cardinals. A forcing $\mathbb{P}$ is c.c.c. iff $\mathbb{P}$ has no uncountable antichains (remember that "antichain" in the context of forcing means a set of incompatible rather than merely incomparable elements). This is a standard result, which appears in Kunen explicitly (I don't have my copy on my so I can't give a reference number at the moment). The c.c.c.-ness of Cohen forcing is a consequence of the $\Delta$-system lemma, and both this lemma and its application appear in Cohen explicitly as well.

Meanwhile, it's worth noting that there are two ways that the lower bound $(2^{\aleph_0})^{V[G]}\ge\lambda$ could fail to be sharp:

• We could have $(2^{\aleph_0})^V>\lambda$ already at the beginning. Nothing says that we have to start with a model of CH (although when first introducing forcing we generally do, since all we know at that point is the relative consistency of ZFC+CH per Godel). Since c.c.c. forcings don't collapse cardinals, this will prevent $(2^{\aleph_0})^{V[G]}=\lambda$.

• We could wind up with $cf(\lambda)^{V[G]}=\omega$ at the end. Since ZFC proves $cf(2^{\aleph_0})>\omega$, this would mean that in $V[G]$ we can't possibly have $2^{\aleph_0}=\lambda$. Note that a priori to tell if this situation holds we need to look at cofinalities in $V[G]$, but it turns out that in this particular case this situation will arise exactly when $cf(\lambda)^V=\omega$: any c.c.c. forcing preserves cofinalities as well as cardinalities, by basically the same argument.

In light of the above - especially the second bulletpoint - you should be even more interested in the question of exactly computing $2^{\aleph_0}$ in $V[G]$. It turns out this is something that we can do without too much effort, the relevant term being "nice names" (and Kunen will treat them later on); the exact calculation is $(2^{\aleph_0})^{V[G]}=\vert\lambda^\omega\vert^V$, as DanielWainfleet commented.

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Incidentally, finiteness of domain is crucial here. In general, if we force with partial functions from $\kappa$ to $\lambda$ with domain of cardinality $<\mu$, we may indeed collapse cardinals: e.g. forcing with countable partial functions from $\omega_1$ to $\omega_2$ will preserve $\omega_1$ but collapse $\omega_2$ (preservation of $\omega_1$ follows from countable closure, and collapsing $\omega_2$ is a good exercise - I believe as above this all appears explicitly in Kunen).

There is an important principle at work here, which is often counterintuitive at first:

Forcing with "larger" conditions generally adds fewer "small" sets!

In particular, forcing with finite partial functions adds reals while forcing with countable partial functions won't. This same point will come back with a vengeance when you look at iterated forcing in the context of the set theory of the reals, where finite support iterations turn out to be nasty while countable and full support iterations are quite tame.