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 $\kappa$.


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.


3 Answers 3


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.


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.

  • $\begingroup$ Very nice. Thank you for this, it resolves several mysteries for me. $\endgroup$ Feb 17, 2015 at 7:22
  • $\begingroup$ This is a really nice result! $\endgroup$
    – David Roberts
    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.

  • $\begingroup$ 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. $\endgroup$ Sep 14, 2013 at 11:13
  • $\begingroup$ 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. $\endgroup$ Sep 14, 2013 at 13:05
  • $\begingroup$ Dear Prof. Hamkins, You are right; I edited my answer. $\endgroup$ Sep 14, 2013 at 13:38
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    $\begingroup$ 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. $\endgroup$ Sep 14, 2013 at 16:22
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    $\begingroup$ 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 $P_U$. $\endgroup$ 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.)

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

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