Let $\kappa$ be an inaccessible cardinal. Is there any forcing notion $P$ with the following properties:

1-$P$ preserves GCH and the strong inaccessibility of $\kappa$,

2-$P$ adds a subset of $\kappa$ of size $\kappa$,

3-$P$ is the $< \aleph_1-$support product of some forcing notions $P_{\alpha}, \alpha < \kappa.$

4- The generic filter for $P$ can be reconstructed from the subset of $\kappa$ added in 2.

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    $\begingroup$ In 2 you want that no new subset of a smaller cardinal is added, right? (Otherwise adding a single Cohen real and somehow extending it to a set of size $\kappa$ would do.) $\endgroup$ Jun 8, 2011 at 6:17
  • $\begingroup$ This is still not the right question. Let $P_0$ be Cohen forcing, and let $P_\alpha$ be trivial for each $\alpha>0$. You get essentially Cohen forcing, but formally a countable support product. Or, let each $P_\alpha$ be (countable) Cohen forcing. Their countable support product will be a forcing which is equivalent to "countable partial functions from $\kappa$ to 2", which has $\aleph_2$-cc. $\endgroup$
    – Goldstern
    Jun 9, 2011 at 11:12
  • $\begingroup$ Martin, your latter example kills the inacessibility of $\kappa$. $\endgroup$ Jun 9, 2011 at 12:38

1 Answer 1


Here is my answer to the updated question:

If you allow the forcing $P_\alpha$ to be trivial, then the Cohen real forcing example still works: add a Cohen real, and then perform $\kappa$ many stages of trivial forcing, with countable support. Overall, this is the same as just adding a Cohen real, since the later stages don't do anything, and it will still have the properties you request as in the original question.

But if you insist that every $P_\alpha$ is nontrivial as a forcing notion, then the inaccessibility of $\kappa$ will necessarily be destroyed. To see this, suppose that each $P_\alpha$ has incompatible conditions. Pick a condition $p_\alpha$ in $P_\alpha$ such that there are other conditions in $P_\alpha$ incompatible with $p_\alpha$. Consider any block $B$ of $\omega_1$ many coordinates $\alpha$ at which such nontrivial forcing is to be done. Let $G$ be $V$-generic for the iteration. For each such $\alpha\in B$, let $s_B(\alpha)$ be $0$ or $1$ accordingly as to whether $p_\alpha\in G$ or not. Thus, for each block $B$, the binary pattern of $s_B$ determines a subset of $\omega_1$. Since your iteration uses only countable support, it is dense that the binary pattern of $s_B$ is different from that of $s_{D}$, if $B$ is disjoint from $D$, since we may introduce such a difference by moving beyond the support of any given condition. Since we may divide $\kappa$ into $\kappa$ many disjoint blocks $B$, it follows that in $V[G]$ we have $\kappa$ many subsets of $\omega_1$, and so $\kappa$ is no longer inaccessible, contrary to 1.

Here is my answer to the original question:

Stefan's comment provides an easy example if you don't intend that the forcing should not add bounded sets to $\kappa$. Just add a Cohen real. This of course preserves GCH, has countable conditions, and adds a subset to $\kappa$ (since every subset of $\omega$ is already a subset of $\kappa$); if you throw in all infinite ordinals as well, then you've got a new subset of $\kappa$ of size $\kappa$.

But even if you don't want to add bounded sets to $\kappa$, there are some easy examples showing that the question is probably not quite what you want to ask. This is because the size of conditions is not a fundamental feature of a forcing notion. For any partial order $\mathbb{P}$, let's make an isomorphic copy $\mathbb{P}'$ by replacing each condition $p$ with its singleton $p'=\{p\}$. We order $p'\leq q'$ just in case $p\leq q$. So now we have a partial order all of whose conditions have size $1$, and hence are countable. So any forcing notion is isomorphic and hence forcing equivalent to a forcing notion with countable conditions. Thus, this property is probably not really what you want to ask about.

  • $\begingroup$ The fact that this partial order is not hereditarily countable is not a fix either. Indeed, many arguments in forcing assume the partial order lives on an ordinal or other fixed set via some abstract isomorphism. $\endgroup$ Jun 8, 2011 at 11:40
  • $\begingroup$ Yes, Juris. Clearly, if you insist that conditions are hereditarily countable, then the forcing must be contained in $H_{\omega_1}$, which is too small to add a new subset to $\kappa$ if no bounded sets are to be added. $\endgroup$ Jun 8, 2011 at 12:47
  • $\begingroup$ @Joel, in your answer to the updated question, do you mean the $P_\alpha$ to be the $\alpha^{\mathrm{th}}$ iterand (what you often see denoted as $\dot{Q}_\alpha$ or do you mean the iteration up to the first $\alpha$ iterands? If it's the former, then I don't follow your definition of $s_B$ and if it's the latter, then I don't see how it's dense that $s_B$ and $s_D$ have different patterns - in particular Cohen forcing followed by trivial iterations is such that each $P_\alpha$ (in the sense of being the iteration of the first $\alpha$ iterands) is non-trivial and has incomparable elements. $\endgroup$ Jun 9, 2011 at 16:41
  • $\begingroup$ Amit, I meant that $P_\alpha$ was the forcing appearing at coordinate $\alpha$ (the OP is using product forcing, not iterated forcing, although my argument works in either case). Thus, the decision as to whether $p_\alpha$ is included in $G$ or not can be made independently for different $\alpha$. I am using $p_\alpha$ just to have an easy way of having the coordinate $\alpha$ part of the generic filter give me a binary digit, and $s_B(\alpha)$ is that digit, for $\alpha$ in $B$. For example, among the first $\omega_1$ many stages of forcing, some choose $p_\alpha$ and some do not, and this... $\endgroup$ Jun 9, 2011 at 17:15
  • $\begingroup$ ...determines a subset of $\omega_1$. And the same for each subseqeunce block of size $\omega_1$. These patterns must be different by a density argument, so we get $\kappa$ many subsets of $\omega_1$ in the extension $V[G]$. $\endgroup$ Jun 9, 2011 at 17:16

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