# Does fast function forcing really have $\kappa$-Knaster property?

I ran into a claim concerning Woodin's fast function forcing in the following paper of Apter and Cummings which sounds no right to me:

A. Apter, J. Cummings, Blowing up the power set of the least measurable, Journal of Symbolic Logic 67 (2002), no. 3, 915--923.

Definition: Woodin's fast function forcing on $\kappa$ consists of partial functions $p$ from $\kappa$ to $\kappa$ ordered by inclusion such that:

• The domain of $p$ consists of inaccessible cardinals $\lambda <\kappa$ which are closed under $p$, namely for every $\lambda, \theta\in dom(p)$ if $\theta<\lambda$ then $p(\theta)<\lambda$.

• For every $\lambda\in dom(p)$ we have $|dom(p)\cap \lambda|<\lambda$

The claim is as follows:

Claim: For Mahlo $\kappa$ the fast function forcing satisfies $\kappa$-Knaster property.

I have seen no such claims discussing the chain condition of fast function forcing in other related papers.

Also I think the following counterexample works for refuting the $\kappa$-cc property of the fast function forcing as defined above.

.Counterexample: Fix an inaccessible cardinal $\lambda <\kappa$. Consider the set of $\kappa$-many incompatible conditons as follows: $p_{\alpha}:=\{\langle\lambda,\alpha\rangle\}$ for $\alpha<\kappa$.

Finally the questions are:

Question 1: Am I missing something?! Does fast function forcing really have $\kappa$-Knaster property?

Question 2: If $f$ is the fast function added by fast function forcing, is the following statement true?

For all functions $g:\kappa\rightarrow\kappa$ in the forcing extension there is a function $h:\kappa\rightarrow\kappa$ such that $\forall \alpha\in \kappa ~~~ g(\alpha)<f(h(\alpha))$.

## 2 Answers

I agree with you; I think your antichain example shows that fast function forcing is not $\kappa$-c.c.

Since I am a little surprised to hear that there has been a mistake about this in the literature, I wonder whether they were talking about some modified version of the fast function forcing? For example, I believe that if one modifies the forcing to insist that $p(\gamma)<\gamma$, then I believe that one could run a $\Delta$-system argument and show that it has the $\kappa$-c.c. In this case, however, the generic function would not be so fast, and so the "fast function" moniker would no longer be apt.

Meanwhile, let me also mention that there are several different versions of fast function forcing mentioned in the literature. The definition you give is one that I also used in an early paper, but I came eventually to realize that it is not as good with factoring as the definition that requires the smallness property $|p\upharpoonright\lambda|<\lambda$ at every inaccessible cardinal $\lambda$, rather than just at $\lambda\in\text{dom}(p)$. I now always use this latter version of fast function forcing. The two forcing notions are different, because one could have a condition $p$ with $p\upharpoonright\lambda$ having domain unbounded in an inaccessible cardinal $\lambda$, provided only that $\lambda\notin\text{dom}(p)$. This can cause some irritating issues with factoring the forcing, and it is more natural to avoid it.

But for both these kinds of fast-function forcing, the same simple antichain that you mention shows it is not $\kappa$-c.c. Indeed, to my way of thinking, the main point of fast function forcing is to arrange that $j(f)(\kappa)$ can be anything you want up to $j(\kappa)$, and the conditions forcing these various values of course form a large antichain; so the lack of the $\kappa$-c.c. seems to be inherent in fast function forcing.

As for question 2, the answer is clearly yes, if you allow $h$ in $V[f]$, since the range of $f$ is unbounded in $\kappa$, and so you can just let $h(\alpha)$ be an ordinal $\beta$ for which $g(\alpha)<f(\beta)$. But do you want $h\in V$? In this case, the answer is negative, since one can take $g=f$ itself, and there there will be no $h\in V$ with $f(\alpha)<f(h(\alpha))$, since if a condition $p$ forced that $h$ worked, we could extend $p$ be specifying it at $\alpha$ and $h(\alpha)$ in such a way that violated the requirement.

• A very nice and informative answer! Thanks. – Morteza Azad Jun 15 '17 at 3:42

As some colleagues asked about the validity of the main result in the following paper in the absence of $\kappa$-cc property for fast function forcing, (with the permission of both authors) I would like to post an answer provided by Arthur in order to clarify the situation:

A. Apter, J. Cummings, Blowing up the power set of the least measurable, Journal of Symbolic Logic 67 (2002), no. 3, 915--923.

The counterexample certainly shows that fast function forcing for $\kappa$ a Mahlo cardinal isn’t $\kappa$-c.c. (let alone $\kappa$-Knaster). However, as best as I can tell, it doesn’t pose a problem for the remainder of the arguments of the paper.

The reason is that we use this false fact only in two places. These occur on page 920, when we infer that $\kappa$ isn’t measurable in $M[H]$, and when we infer that $M[H]$ is closed under $\kappa$ sequences inside $V[G]$.

To infer that $\kappa$ isn’t measurable in $M[H]$, simply introduce a gap at $\aleph_1$ into the fast function forcing (e.g., begin by adding one Cohen real and then define the fast function forcing as we did earlier). We may then infer by Hamkins’ Gap Forcing Theorem that the fast function forcing doesn’t make $\kappa$ measurable in $M[G]$. This is since $G$ is $M$-generic over a partial ordering containing a very low gap. Consequently, since $\kappa$ wasn’t measurable in $M$, it can’t be measurable in $M[G]$. It then follows, as we point out in our paper, that $\kappa$ isn’t measurable in $M[H]$. Note that Joel himself mentions this idea of introducing a very low gap on page 107 of his lottery preparation paper (Remark on Gap Forcing 1.2).

To infer that $M[H]$ is closed under $\kappa$-sequences inside $V[G]$, as we mentioned, it suffices to show that $V[G] \models ON^\kappa \subseteq M[G]$. However, even though fast function forcing isn’t $\kappa$-c.c., because $\kappa$ is Mahlo and hence inaccessible, it is trivially $\kappa^+$-c.c. This is since as Joel mentions on page 106 of his lottery preparation paper, we assume under these circumstances that a condition has the size less than $\kappa$. Thus, fast function forcing has cardinality $\kappa$. It then follows that because $V \models M^\kappa \subseteq M$ and we are forcing with a $\kappa^+$-c.c. poset, $V[G] \models ON^\kappa \subseteq M[G]$.

Arthur Apter