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.



Singularizing forcing of "small" cardinality?
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](http://www2.kobe-u.ac.jp/~hsakai/Research/papers/semiproper_ideals.pdf)'' as Fact 2.2. 


