# Effect of adding one Hechler real versus adding two on the meager ideal

In "The Kunen-Miller Chart (Lebesgue Measure, The Baire Property, Laver Reals and Preservation Theorems for Forcing)" by Haim Judah and Saharon Shelah JSL Vol. 55, No. 3 (Sep., 1990), pp. 909-927 ([JdSh308] in Shelah's numbering) the authors remark on the final page that adding a pair of Hechler reals makes the union of the meager sets coded by the ground model meager. My questions are:

1. How does one prove this? (and/or is there a citation with the proof, in the paper none is given that I can find)

2. Does this not happen when one adds only one Hechler real?

I suppose more generally my question is as follows: clearly in the Hechler model $$\mathrm{add}(\mathcal M) = \mathfrak{c}$$ by Miller-Truss theorem that $$\mathrm{add}(\mathcal M) = \min\{\mathfrak{b}, \mathrm{cov}(\mathcal M)\}$$ since both dominating reals and Cohen reals are added but I don't quite see how this proliferates down to the finite steps. Is there a more direct way to see how Hechler forcing effects $$\mathrm{add}(\mathcal M)$$?

• That one Hechler real does not suffice is Theorem 3.5.4 in Bartoszynski-Judah book. The argument appears before Corollary 1.4 in J. Pawlikowski, Why Solovay real produces Cohen real, JSL, Vol. 51 No. 4, Dec. 1986 Aug 1, 2019 at 13:11
• Thanks so much for the reference! Aug 17, 2019 at 18:29

If $$c$$ is Cohen over $$V$$, and $$d$$ is dominating over $$V[c]$$ (not necessarily Hechler-generic), then in $$V[c][d]$$ there is a meager set covering all meager sets from $$V$$. Hence 2 successive Hechler reals make the union of all old meager sets meager.

(I suspect that this is not true if you just add one Hechler real.)

Proof: (This is implicit in Bartoszynski-Judah 2.2, and implicit or perhaps even explicit in other papers, such as Miller or Truss. I seem to remember that Andreas Blass invented a notion of "composition" of Galois-Tukey relations, which gives a general framework for arguments such as the one I give below.)

For each meager set $$M\subseteq 2^\omega$$ coded in $$V$$ there are functions $$f:\omega\to \omega$$ and $$x\in 2^\omega$$ (again in $$V$$) such that $$M$$ is contained in $$M_{f,x}:= \{ y\in 2^\omega\mid \forall^ \infty n: x\restriction I_n\not= y\restriction I_n\}$$, where $$I_n:=[f(n), f(n+1))$$.

In $$V[c]$$ there are infinitely many $$n$$ such that $$c\restriction I_n = x\restriction I_n$$. (As $$c$$ is Cohen). So there is an increasing sequence $$(n_k)$$ such that $$x\restriction I_{n_k} = c\restriction I_{n_k}$$ for all $$k$$. (The sequence $$(n_k:k\in \omega)$$ depends on $$x$$, of course. But all we will use about it in the next paragraph is that it is an element of $$V[c]$$.)

In $$V[c][d]$$ we may wlog assume that $$d$$ strongly dominates $$V[c]$$, e.g. by first assuming $$d(k)>k$$ for all $$k$$, and then replacing $$d$$ with $$k\mapsto d^{(k)}(0)$$ ($$k$$-th iterate). Let $$J_\ell$$ be the interval $$[ d(\ell), d(\ell+1))$$. As these intervals are very long (compared with anything from $$V[c]$$), almost every $$J_\ell$$ contains at least one intervall of the form $$I_{n_k}$$.

I claim that $$M_{f,x} \subseteq M_{d,c}$$. So let $$y\in M_{f,x}$$ (in $$V[c][d]$$). For notational simplicity assume that $$\forall n: y\restriction I_n \not= x\restriction I_n$$. Since $$x$$ and $$c$$ agree on all the $$I_{n_k}$$, we also have $$y\restriction I_{n_k}\not= c\restriction I_{n_k}$$. For any (sufficiently large) $$\ell$$ we can find $$k$$ such that $$J_\ell$$ contains $$I_{n_k}$$; so we also have $$y\restriction J_\ell\not= c\restriction J_\ell$$. Hence $$y\in M_{d,c}$$.

• Thank you so much! That's perfect. Aug 17, 2019 at 18:28