# Choice of bound in the Laver property

$$\newcommand{\P}{\mathbb{P}}$$ Recall the following formulation of the Laver property (Page 459, Combinatorial Set Theory: With a Gentle Introduction to Forcing (Second Edition) by Halbeisen):

Laver Property. Let $$\mathcal{F}$$ be the set of all functions $$S : \omega \to \operatorname{fin}(\omega)$$ such that for every $$n < \omega$$, $$|S(n)| \leq \color{red}{2^n}$$. A forcing notion $$\P$$ has the Laver property iff for all $$f \in {}^\omega\omega \cap V$$ in the ground model and every $$\P$$-name $$\dot{g}$$ for a function in $${}^\omega\omega$$ such that $$1 \Vdash_\P \forall n < \omega(\dot{g}(n) \leq f(n))$$, we have $$1 \Vdash_\P \exists S \in \mathcal{F} \cap V \, \forall n < \omega(g(n) \in S(n))$$.

Is it possible to modify the bound $$\color{red}{2^n}$$ above? For instance, is the Laver property equivalent to the statement above with the bound replaced by $$r^n$$ for some $$r < \omega$$, or even by $$h(n)$$ where $$h \in {}^\omega\omega \cap V$$?

You can replace $$2^n$$ by any function $$h: \omega \to \omega \setminus \{0\}$$ in $$V$$ so that $$h(n)$$ diverges to infinity. For instance if you want to get the Laver property for $$h_0$$ from the Laver property for $$h_1$$, let $$i_n$$, for each $$n$$, be the least $$m$$ so that $$h_0(m') \geq h_1(n),$$ for every $$m' \geq m$$.
Instead of $$\dot g$$ simply consider a name for $$g'(n) = g \restriction [i_n, i_{n+1}).$$
The Laver property for $$h_1$$ gives $$h_1(n)$$ guesses for the values of $$g$$ on the interval $$[i_n, i_{n+1})$$ which is less than the number of guesses asked by $$h_0$$ on that interval (and in fact ever after $$i_n$$). The only additional guess we have to make is on $$[0, i_0)$$, but that's of course fine.
• Thanks Jonathan. Do you know if there's any known property that's equivalent to the Laver property, with the bound replaced by some constant (i.e. $h$ may be chosen to be uniformly bounded)? Commented Aug 12 at 5:10
• @ClementYung Yes, that would be equivalent to not adding any new real. For any $x \in 2^\omega$, let $f(n) = x \restriction n$ (which clearly gives rise to a bounded fundtion). If you could guess $c$ values for each $f(n)$, where $c$ is a constant, then also in the end there can only be at most $c$ possibilities for $x$ and you can compute these possibilities in $V$. Commented Aug 12 at 21:26
• @ClementYung cnt'd... More specifically, if $S \in V$ is your guessing function consider $\{ y \in 2^\omega : \forall n (y \restriction n \in S(n)) \}$. If it had $c+1$ many elements, then there must be a point from which all of their initial segments are pairwise different, which is not possible. This set is a finite Borel set coded in $V$, so it is contained in $V$. Commented Aug 12 at 21:29