# Singularizing forcing of "small" cardinality?

Can there be a large cardinal $\kappa$ and a forcing of size $\kappa$ that makes $\kappa$ a singular cardinal? The motivation is that the standard Prikry forcing does not have a dense set of size $\kappa$.

Edit:

In response to some attempts at a positive answer, let me explain something that does not work. If $\mathbb{P}$ is the Prikry forcing and $\mathbb{Q}$ is something like $Coll(\kappa,2^\kappa)$, one may expect under suitable indestructibility hypotheses, $\mathbb{P}$ works in $V^\mathbb{Q}$. But this never works.

The following lemma is based on an exercise in Kunen's book: Suppose $\kappa$ is a singular cardinal and $\mathbb{R} = \{ f : f$ is a partial function from $\kappa$ to $2$ with domain bounded below $\kappa \}$, ordered by extension. Then $\mathbb{R}$ collapses $\kappa$ to $cf(\kappa)$.

Proof: Suppose for simplicity $cf(\kappa) = \omega$, and let $\langle \kappa_n : n \in \omega \rangle$ be an increasing cofinal sequence. If $G \subseteq \mathbb{R}$ is generic, define in $V[G]$ the function $f : \omega \to \kappa$ by $f(n) = \beta$ where $\beta < \kappa_n$ and for some $\delta$, the ordinal $\kappa_n \cdot \delta + \beta$ is the $\kappa_n$-th element of $\{ \alpha : \bigcup G(\alpha) = 1 \}$. A simple density argument shows that $f$ is surjective.

Now let $\kappa$ be our large cardinal. It follows from a general folklore fact that there is a dense embedding $e : Add(\kappa,1) \times Coll(\kappa,2^\kappa) \to Coll(\kappa,2^\kappa)$. After forcing with $\mathbb{P}$, the $Add(\kappa,1)$ of the ground model becomes the forcing with bounded functions from the lemma, and the map $e$ is still a dense embedding. So if $G \times H$ is $\mathbb{P} \times \mathbb{Q}$-generic, then by the lemma, $\kappa$ is collapsed to $\omega$. Therefore in $V^\mathbb{Q}$, $\mathbb{P}$ collapses $\kappa$ to $\omega$.

I suspect that if a positive answer is possible, the forcing must be significantly different from the standard Prikry forcing or some combination of it with simple forcings.

The answer to your question is no. We have the following theorem.

Theorem. Suppose $\kappa$ is a regular uncountable cardinal and $|P|=\kappa.$ Then $\Vdash_P cf(\kappa)=|\kappa|.$

Proof. Let $\tau$ be a name of an unbounded subset of $\kappa.$ We show there is $f\in V, f:\kappa\to\kappa$ such that $\Vdash_P f''[\tau]=\kappa.$ The result will follow immediately.

Let $(p_i: i<\kappa)$ enumerate $P$. Define by induction, for each pair $(i, j)\in \kappa\times \kappa$ an ordinal $\alpha_{ij}<\kappa$ and a condition $q_{ij}$ such that:

(1) $(i', j') < (i,j) \implies \alpha_{i'j'} < \alpha_{ij},$ (where $<$ denote any well ordering of $\kappa\times\kappa$ of order type $\kappa,$ say like Godel ordering)

(2) $q_{ij}\leq p_i$ and $q_{ij}\Vdash \alpha_{ij}\in \tau.$

Then $f$ defined by $f(\alpha_{ij})=j$ is as required as can be proved easily.

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Note on 23/07/2015: I realized that the result with the same proof has appeared in Hiroshi Sakai's paper Semiproper ideals'' as Fact 2.2.

• Very nice. Thank you for this, it resolves several mysteries for me. Commented Feb 17, 2015 at 7:22
• This is a really nice result! Commented Jul 23, 2015 at 7:13

It is consistent that the answer is no:

Let $V=L[U],$ where $U$ is a normal measure on a measurable cardinal $\kappa.$ First note that we can apply Prikry forcing over $V$ to change the cofinality of $\kappa$ to $\omega.$

Now we show that there is no forcing of size $\kappa$ changing the cofinality of $\kappa.$ Suppose not. Let $P$ be such a forcing notion and let $G$ be $P-$generic over $V$. By Dodd-Jensen covering theorem for $L[U],$ there exists an $\omega-$sequence $C\in V[G]$ cofinal in $\kappa$ which is a Prikry sequence for the classical Prikry forcing $P_U$ in $V$ and $V[G]$ is covered by $V[C]$. Then $V[C]\subset V[G],$ and we have $P_U$ is a projection of $P$, so $P$ has size $>\kappa.$

Remark. In fact the above proof shows that if there is such a forcing notion, then $0^\dagger$ exists.

• Very nice! In your yes' part, you need the GCH in your argument, but you could instead use $\text{Coll}(\kappa,2^{2^\kappa})$ and avoid this. Commented Sep 14, 2013 at 11:13
• Your argument in the yes' part is not right, since no embedding $j:V\to M$ can lift to $j^*:V[G]\to M[j(G)]$, if $G$ adds a new subset to $\kappa$ but no bounded sets to $\kappa$, since those new sets would have to be in $M$ and hence in $V$, contradicting the fact that they are new. With the Laver preparation, you need to do forcing also below $\kappa$. So you might modify your forcing to explicitly include the Laver preparation into $P$, and carry out a similar argument. Commented Sep 14, 2013 at 13:05
• Dear Prof. Hamkins, You are right; I edited my answer. Commented Sep 14, 2013 at 13:38
• Mohammad, I agree that $H$ is $P_U$-generic over $V$, this follows from Adrian Mathias's theorem about Prikry sequences. However, you have not argued, and I don't think it is the case, that $H$ is $P_U^V$-generic over $V[G][g]$. Indeed, this impossible because the Prikry sequence $C$ determines $H$ and $H^*$, depending on the model it's computed from, so if $H$ were $P_U^V$-generic over $V[G][g]$, then we'd have that $P_{U^*}$ has a dense subforcing of size $\kappa$, which is false. Commented Sep 14, 2013 at 16:22
• Yes, @Monroe, that is precisely what I was thinking. One can see that $H$ is not $V[G][g]$ directly as follows: in $V[G][g]$ there is a single set $A\in U^*$ such that $U$ is contained on the tail filter of $A$. But it is dense in $P_U$ that the generic sequence contain infinitely many elements not in $A$, since every set in $U$ has such points. Meanwhile, a tail of the Prikry sequence of $H^*$ is contained in $A$. So $H$ cannot be $V[G][g]$-generic for $P_U$. Commented Sep 14, 2013 at 17:40

What about the following. Let $V$ be a model of GCH and let $\kappa$ be Laver-prepared. Force with $P\oplus Q$ where $P$ is the Prikry forcing making $\mathrm{cf}(\kappa)=\omega$ and $Q=\mathrm{Coll}(\kappa,\kappa^+)$. Notice that $P\oplus Q$ has the Prikry property: if $\varphi$ is a sentence in the forcing language $((s,A),q)\in P\oplus Q$, then let $\psi$ be: there is $q'\leq q$ such that $q'$ forces $\varphi$ and apply the Prikry property to $\psi$. This implies that $P\oplus Q$ does not add new elements to $V_\kappa$. Clearly, $P\oplus Q$ changes the cofinality of $\kappa$ to $\omega$ ($P$ already does). As $P\oplus Q=Q\oplus P$, $V^Q$ is as required. (shame on me to write this as an answer, only it was too long for a comment.)

• By $\oplus$ you mean cartesian product, right? Commented Sep 15, 2013 at 17:04
• If so, see the edit of the OP. Commented Sep 15, 2013 at 17:41
• I agree with Monroe that this does not actually work. $P\times Q$ will collapse $\kappa$ to $\omega$. Commented Sep 15, 2013 at 18:02