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 $\mathrm{add}(\mathcal M)$?

  • 3
    $\begingroup$ 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 $\endgroup$
    – Ashutosh
    Aug 1, 2019 at 13:11
  • $\begingroup$ Thanks so much for the reference! $\endgroup$ Aug 17, 2019 at 18:29

1 Answer 1


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 $y\in M_{d,c}$.

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

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