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In their paper "A barren extension", Henle, Mathias, and Woodin considered the forcing whose conditions are the sets of natural numbers ordered by inclusion up to finite error. If we force over $L(\mathbb{R})$ and assume that this is a model of determinacy, they show that this will add a free ultrafilter U on the natural numbers, and moreover in the extension there are no new (ordinal-indexed) sequences of elements of the ground model.

Let us order the functions $f\colon\omega\rightarrow\omega$ in $L(\mathbb{R})[U]$ by domination mod $U$. We obtain a linear order $(L,\leq)$.

(1) Is there a strictly increasing $\omega_1$-sequence in $(L,\leq)$?

(2) Is there a strictly increasing $\omega_1$-sequence of gaps in $(L,\leq)$?

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Does the argument that no sequences were added depends on the presence of AD in the ground model, or does it follow from $L(\Bbb R)$ being a Solovay model in general? –  Asaf Karagila Feb 19 '13 at 17:29
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@Asaf: they only use that the ground model satisfies ZF and the partition relation $\omega \to (\omega)^\omega$. –  Ramiro de la Vega Feb 19 '13 at 18:33
    
@Ramiro: Thanks! Taking this with the exercise from Jech's "The axiom of choice" (ch. 12 ex. 5) it means that it really holds in Solovay's model. –  Asaf Karagila Feb 19 '13 at 19:02
    
@Asaf: In "Happy families" Mathias showed that $\omega \to (\omega)^\omega$ is consistent with ZF+DC (provided that there are inaccessibles). According to Di Prisco (in "Partitions of the reals in models of ZF"), Mathias´ result actually "establishes that $\omega \to (\omega)^\omega$ holds in Solovay models". –  Ramiro de la Vega Feb 19 '13 at 21:25
    
Thank you for the comments, so the question also makes sense over a Solovay model. The difficulty is that the elements of L are not in the ground model. –  Philipp Schlicht Feb 25 '13 at 16:38

1 Answer 1

up vote 8 down vote accepted

The answer to the first question is no, at least if one is willing to assume large cardinals in $V$. I'd have to research what the optimal hypothesis is. It is shown in the two Neeman-Zapletal papers that if there are infinitely many Woodin cardinals below a measurable cardinal and above some cardinal $\kappa$, then forcing with a c.c.c. partial order of cardinality less than $\kappa$ cannot change the theory of $L(\mathbb{R})$ with ordinal parameters. Since the cardinal invariant $\mathfrak{t}$ (the tower number) can be made arbitrarily large by c.c.c. forcing, this shows from the same hypothesis that the intersection of any wellordered collection of dense open subsets of $\mathcal{P}(\omega)/\mathrm{Fin}$ in $L(\mathbb{R})$ is dense open. For this specific problem, we need to intersect only $\omega_{1}$ many dense sets. Since c.c.c. forcings don't change $\omega_{1}$, we can do with the ostensibly weaker hypothesis that c.c.c. forcing doesn't change the theory of $L(\mathbb{R})$ (not allowing for ordinal parameters), which gives that intersections of $\omega_{1}$ many dense open subsets in $\mathcal{P}(\omega)/\mathrm{Fin}$ are dense open. I don't know if this can be proved just assuming $\mathrm{AD}$ + $V = L(\mathbb{R})$, or even whether one can deduce it from the Henle-Mathias-Woodin result.

Suppose now that $\langle \tau_{\alpha} : \alpha < \omega_{1} \rangle$ is a sequence of $\mathcal{P}(\omega)/\mathrm{Fin}$ names for nonempty sets of reals. By the assumptions of the previous paragraph, there is an infinite $a \subseteq \omega$ such that for each $\alpha < \omega_{1}$, the set $B_{\alpha}$, consisting of those reals forced by $a$ to be in $\tau_{\alpha}$, is nonempty. The sequence $\langle B_{\alpha} : \alpha < \omega_{1} \rangle$ is in $L(\mathbb{R})$. If $a$ forces each $\tau_{\alpha}$ to be a mod-$U$ equivalence class in the Baire space, then the members of each $B_{\alpha}$ must agree mod-finite on $a$. So, restricting to $a$, we get an $\omega_{1}$-sequence of disjount countable subsets of the Baire space, which contradicts the conjunction of the Perfect Set Property and the Baire Property. So there are no $\omega_{1}$-sequences of distinct mod-$U$ equivalence classes in $L(\mathbb{R})[U]$, regardless of the relationship between the classes.

A similar argument gives the Perfect Set Property in $L(\mathbb{R})[U]$, which was originally proved by Di Prisco and Todorcevic. It seems that the same approach should give a negative answer to the second question, but I don't see it.

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Welcome to MathOverflow, Paul! –  Joel David Hamkins Feb 27 '13 at 19:54
    
Thanks for the great answer! –  Philipp Schlicht Feb 28 '13 at 6:52
    
Paul, I am curious to know whether your line of reasoning can be used to answer the following somewhat related MO question that has remained unsolved: mathoverflow.net/questions/24047/… –  Ali Enayat Mar 1 '13 at 0:42
    
Joel, Philipp : thanks. Ali, I have thought about that question, and about the question of discontinuous automorphisms of $(\mathbb{R}, +)$, but I haven't had any success. –  Paul Larson Mar 7 '13 at 21:22

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