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Paradox with elementary submodels of the constructible tower

  1. May 20, 2012 #1
    This is an argument I thought up after a class on combinatrical properties of the model [itex]\textbf{L}[/itex]. Our course is about set theory, not logic, so this paradox desn't seem relevant in its context. Can you help me figure out where I got it wrong?

    The constructible heirarchy of sets is a series [itex]L_{\alpha}[/itex] that is defined for all ordinal numbers [itex]\alpha[/itex]. The important properties for my argument are:
    1. [itex]L_{\alpha}[/itex] is transitive for every [itex]\alpha[/itex]
    2. If [itex]\alpha < \beta[/itex], then [itex]L_{\alpha}\subset L_{\beta}[/itex]
    3. The transitive collapse (aka Montowski collapse) of every elementary submodel [itex]M \prec L_{\alpha}[/itex] is [itex]L_{\beta}[/itex] for some [itex]\beta[/itex]
    4. [itex]L_{\omega_{1}}[/itex] satisfies "every set is countable" and [itex]L_{\omega_{2}}[/itex] does not
    5. [itex]L_{\alpha}[/itex] is coutable iff [itex]\alpha[/itex] is countable

    So, we take an countable elementary submodel (CESM) [itex]M_{1} \prec L_{\omega_{1}}[/itex], and look at its transitive collapse, [itex]L_{\alpha_{1}}[/itex] for some countable [itex]\alpha_{1}[/itex]. We then take an CESM [itex]M_{2} \prec L_{\omega_{2}}[/itex] that contains [itex]L_{\alpha_{2}}[/itex], and collapse it to get [itex]L_{\alpha_{2}}[/itex] with countable [itex]\alpha_{2}[/itex]. Then the same procedure yields a model [itex]L_{\alpha_{3}} \supset L_{\alpha_{2}}[/itex] that has an elementary embedding into [itex]L_{\omega_{1}}[/itex]. We generate an infinite series, switching between modelling [itex]L_{\omega_{1}}[/itex] and [itex]L_{\omega_{2}}[/itex].

    The limit [tex]L_{\alpha}=L_{\lim_{n<\omega}\alpha_{n}}=\bigcup_{n<\omega}L_{\alpha_{n}}[/tex] is then the union of both subseries [itex]\{L_{\alpha_{n}}\}_{n=1,3,\ldots}[/itex] and [itex]\{L_{\alpha_{n}}\}_{n=2,4,\ldots}[/itex]. But a union of a series of elementary submodels is itself an elementary submodel, since it is a direct limit. In particular [itex]L_{\alpha}[/itex] should be elementary equivalent to both [itex]L_{\omega_{1}}[/itex] and [itex]L_{\omega_{2}}[/itex]. This is impossible because of property (4), namely there is a statement true in one and not in another.

    Where did I go wrong in my reasoning? All kinds of tips are appreciated...
     
  2. jcsd
  3. May 21, 2012 #2

    AKG

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    "The union of elementary submodels is itself an elementary submodels" is only true when those submodels are elementary substructures of one another. This means more than just the fact that [itex]L_{\alpha_n} \subset L_{\alpha_{n+2}}[/itex] and [itex]L_{\alpha_n} \equiv L_{\alpha_{n+2}}[/itex], it requires that the inclusion map is the elementary embedding. This would be the case if the [itex]L_{\alpha_n}[/itex] were elementary substructures of their respective [itex]L_{\omega_i}[/itex], but they're generally not. You do get that they're elementarily equivalent subsets of their respective [itex]L_{\omega_i}[/itex], but the inclusion maps are not typically the elementary embeddings. The embeddings here are the inverses of the Mostowski collapses.

    Nothing I said above proves that the inclusion maps aren't elementary embeddings, although nothing you said proves that they are. But here's why they're definitely not. Let's take for example [itex]L_{\omega_2}[/itex]. For the same that GCH holds in L, we know that [itex]L_{\omega_2}[/itex] thinks there is exactly one cardinal after [itex]\omega[/itex], and that this cardinal is [itex]\omega_1[/itex]. If [itex]L_{\alpha} \equiv L_{\omega_2}[/itex] with [itex]\alpha[/itex] countable, then this structure will also think there's a unique cardinal after [itex]\omega[/itex], but it won't think [itex]\omega_1[/itex] is it.
     
  4. May 21, 2012 #3
    I see what you mean...

    So, it seems that for every ordinal [itex]\alpha[/itex], the set [itex]\{\delta < \omega_{1} \mid L_{\delta} \prec L_{\alpha}\}[/itex] is closed w.r.t taking limits. I thought about it some more and it's not hard to see this set is unbounded for [itex]\alpha = \omega_{1}[/itex], since for each [itex]\beta < \omega_{1}[/itex] you can find find some [itex]L_{\beta'}[/itex] with [itex]\beta<\beta'<\omega_{1}[/itex] that is closed w.r.t Skolem functions of [itex]L_{\omega_{1}}[/itex]. This natually means that the set of δs can't be unbounded for [itex]|\alpha|>\aleph_{1}[/itex].
     
    Last edited: May 21, 2012
  5. May 21, 2012 #4

    AKG

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    Correct. Another simple argument is to do the following construction:

    • [itex]X_0 = Hull(\{\beta\},L_{\alpha})[/itex]
    • [itex]\beta_0 = \min\{\gamma : X_0 \subset L_\gamma\}[/itex]
    • [itex]X_{n+1} = Hull(L_{\beta_n},L_\alpha)[/itex]
    • [itex]\beta_{n+1} = \min\{\gamma : X_n \subset L_\gamma\}[/itex]
    • [itex]L_{\beta'} = \bigcup X_n[/itex]
    Here [itex]Hull(X,M)[/itex] denotes some Skolem Hull of the set X in the structure M. Since we're working with L, there's a canonical choice for such hull.
    Indeed, the set of such [itex]\delta[/itex]'s is empty!
     
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