5
$\begingroup$

I'm reading Todorcevic's "Localized Reflection and Fragmens of PFA" and have a rather specific question about the proof. To avoid being too specific, I'll ask a more general question first, and afterwards include my more specific question. Let me also say that I'm just as happy to get references to papers/books where I can find answers to my questions as I am happy to get direct answers here.

Let $C$ denote the canonical constructible square sequence -- that is, "$y = C(x)$" is equivalent to some formula which, when $x$ is restricted to ordinals which are singular in $L$, defines a class function $\langle C_{\alpha} : \alpha \in \mathrm{Ord}, \mathrm{cf}^L(\alpha) < \alpha \rangle$ satisfying:

  1. $C_{\alpha}$ is a club in $\alpha$ of order type $< \alpha$, and
  2. $C_{\alpha} = C_{\beta} \cap \alpha$ whenever $\alpha$ is a limit point of $C_{\beta}$

General Questions:

  1. What is the complexity of $C$? By "complexity," I mean in the same sense that the canonical constructible well-order is $\Sigma _1$.
  2. How absolute is the definition of $C$? Do $V$, $L$, $V _{\alpha}$, $H _{\alpha}$, $L _{\alpha}$ agree on what $C$ is up to $\alpha$, for arbitrary ordinals (or in the case of $H _{\alpha}$, cardinals) $\alpha$?

The reason I ask this question comes from a more specific question, but to ask that question I need to give some background -- all of this comes from Stevo's article mentioned above:

Let $\mathrm{PFA}(\omega _1)$ be the statement that if $P$ is a proper notion of forcing, and $\mathcal{D}$ an $\omega _1$-sequence of predense subsets of $P$, each of size at most $\omega _1$, then there exists a $(P, \mathcal{D})$-generic filter.

Lemma: $\mathrm{PFA}(\omega _1)$ holds iff for every $\Delta _0$-formula $\varphi$ and a parameter $a \in H _{\omega _2} $, if some proper poset forces that $\exists x \varphi (x, a)$ is true in some transitive model then $H _{\omega _2}$ already satisfies $\exists x \varphi (x, a)$.

In a later lemma, Stevo claims that the following formula is $\Delta _0$ in order to apply the lemma quoted above: $\psi (x,y,a)$ is the formula which says:

  • $x$ is a set of ordinals, all singular in $L$
  • $x$ has order type $\omega _1$
  • $x$ is closed in its supremum
  • $y$ is a function from $x$ into $\mathbb{N}$ such that $y(\alpha) \neq y(\beta)$ whenever $\alpha, \beta \in x$ and $\alpha \in \mathrm{lim}(C_{\beta})$
  • $L_{\mathrm{sup}(x)} \vDash \varphi (a)$ where $\varphi$ was some arbitrary formula from earlier in the proof (with no restrictions on its complexity)

I can't believe this formula is $\Delta _0$, but in order to apply the previous lemma we really only need it to be $\Sigma _1$, which I might believe is the case. We would also need to throw in $\omega _1$ as a parameter, but we're free to do that since we could take $\{ a, \omega _1 \}$ to be our parameter from $H _{\omega _2}$ in applying the above lemma. I can see why most of this is $\Sigma _1$ but there is a subformula for which I can't see this, and the problem I'm having is that I don't know the complexity of $C$.

Specific Questions:

Consider the subformula: $\forall \alpha, \beta \in x$, $(\alpha \in \mathrm{lim}(C_{\beta}) \rightarrow y(\alpha) \neq y(\beta))$

  1. Is it $\Sigma _1$, and if so why? It would suffice to know that $c = C_{\beta}$ is $\Sigma _1$ in $c$ and $\beta$.

  2. Later on in the proof it seems to be tacitly assumed that $H _{\omega _2}$ computes $C$ correctly. Can I be sure that $H _{\omega _2}$ computes $C$ correctly (up to $\omega _2$ of course)?

  3. Stevo proves the lemma in question using some pretty big guns - the Covering Lemma and the canonical constructible square sequence. He also uses a proper forcing which, unless you knew how it was going to be used beforehand, appears like it would be of no use in proving the lemma. Does anyone know of a more direct proof of the lemma I'm talking about? Or, since I often find in reading Stevo's work that he uses forcing posets which I would never think to use, can anyone put this proof in perspective, and motivate why it makes sense to do the proof this way?

$\endgroup$
4
  • $\begingroup$ Amit, I haven't looked at everything here carefully, but I can get you started: The global square sequences as here tend to be fairly absolute in their definitions, and particularly so in $L$. Essentially they are built by looking for each cardinal $\kappa^+$ at the $L_\gamma$ that are elementary in $L_{\kappa^+}$ (this is absolute), and using them to define fine structural embeddings (which are completely determined by a few parameters, which in turn are also determined from the situation). The actual clubs are then defined in terms of the critical points of the resulting embeddings. $\endgroup$ Dec 22, 2010 at 2:42
  • 1
    $\begingroup$ Actually there is an issue that concerns me: The global sequence has the property that if $\kappa$ is a cardinal, then the sequence defined in $L_\kappa$ or $V_\kappa$ or $H_\kappa$ is just the restriction to $\kappa$ of the global sequence. But this is far from meaning it is absolute (it is important that $\kappa$ is a cardinal, not just that it "looks" like a cardinal, so if $\kappa$ is not a cardinal but $L_\gamma$ thinks that $\kappa$ is a cardinal, for "many" $\gamma$ past $\kappa$, this still does not guarantee that the sequence as defined in $L_\kappa$ coincides with the global one.) $\endgroup$ Dec 24, 2010 at 2:59
  • $\begingroup$ Hi Andres, why does this concern you? It sounds like a pretty good answer to my General Question 2, and a positive answer to my Specific Question 2 (since we're talking about the true $\omega _2$, and not mereley $\omega _2 ^L$). Do you know a good reference where I can learn about the fact you're mentioning here? $\endgroup$ Dec 24, 2010 at 3:31
  • $\begingroup$ @Amit: Oh, sure, the proof of this fact answers your question 2. The issue is the $\Delta_0$ (or $\Sigma_1$) complexity of the formula. I think that Jensen's original paper deals with global square. I have been thinking of a good additional reference; my problem is that the ones I know are written for more general contexts, where you need to know more fine structure (since the constructions work for inner models with large cardinals). $\endgroup$ Dec 24, 2010 at 3:45

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.