8
$\begingroup$

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

$\endgroup$
  • 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 at 13:11
  • $\begingroup$ Thanks so much for the reference! $\endgroup$ – Corey Switzer Aug 17 at 18:29
8
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.