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From Chang and Keisler's "Model Theory", section 7.2, we know that:

1) There is a sentence $\sigma$ in a suitable language $L$ such that for all infinite cardinals $\alpha$, $\sigma$ admits $(\alpha^+,\alpha)$ iff there exists a tree $T$ of height $\alpha^+$, with at most $\alpha$ elements at each level $\xi<\alpha^+$, and with no branch of length $\alpha^+$.

2) There is a sentence $\sigma$ in a suitable language $L$ such that for all infinite cardinals $\alpha$, $\sigma$ admits $(\alpha^{++},\alpha)$ iff there is a (Kurepa) family $F$ of subsets of $\alpha^+$ such that $|F|=\alpha^{++}$ and for every $\xi<\alpha^+$, $|\{X\cap\xi|X\in F\}|=\alpha.$

My question: Are there any "natural" statements that would be equivalent to a sentence $\sigma$ admitting $(\alpha^{+n},\alpha)$, $3\le n<\omega$? For Chang and Keisler, $\sigma$ has to be first-order. For our purposes, even $L_{\omega_1,\omega}$ is good enough.

Addition: Definition A sentence $σ$ in a language with a unary predicate $P$ admits $(κ,λ)$, if $σ$ has a model $M$ such that $|M|=κ$ and $|P^M|=λ$. Of course, λ≤κ. So, not only we need a model of a specific size, but we need the predicate P to have a specific size too.

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Does $2^\alpha\geq\alpha^{+n}$ qualify? –  Péter Komjáth May 15 '12 at 4:36
    
For the problem we are working on, we assume GCH. Nevertheless, I would be interested to hear what you have in mind. –  Ioannis Souldatos May 15 '12 at 12:14
    
Can you remind us what it means to "admit $(\kappa,\lambda)$"? Does it mean that the language has a predicate and you are asking for a model of size $\kappa$, where the predicate has size $\lambda$? –  Joel David Hamkins May 15 '12 at 13:13
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I wonder if using a morass instead of a Kurepa tree might be a source of examples, since here one uses a system of small structures to approximate a large structure. The gap-n morass case is similar to your question, with a gap-n difference in the cardinals. –  Joel David Hamkins May 15 '12 at 15:11
2  
See en.wikipedia.org/wiki/Morass_(set_theory) for some information and a list of references. –  Joel David Hamkins May 15 '12 at 21:08
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1 Answer

up vote 4 down vote accepted

Letting $\alpha$ be an infinite cardinal and $3\leq n\lt\omega$, I think Peter Komjath's proposed statement $2^\alpha\geq\alpha^{+n}$ is the simplest and most natural equivalent of a first-order sentence admitting $(\alpha^{+n},\alpha)$: just let $\sigma$ say a binary relation is extensional with domain given by a predicate and range equal to the universe. However, you asked for a GCH example in your comment, so I suggest the following statement, that a generalized kind of Kurepa family exists.

  • $(*_{\alpha,n})$ says there exists $\mathcal{X}\subset\mathcal{P}(\alpha^{+n-1})$ of size $\alpha^{+n}$ such that $|\{X\cap A: X\in\mathcal{X}\}|\leq\alpha$ for all $A\subset\alpha^{+n-1}$ of size $\leq\alpha$.

To get our first-order sentence $\sigma$, we use the fact that $[\alpha^{+n-1}]^{\leq\alpha}$ has nice cofinal subsets, and the equivalence of $(*_{\alpha,n})$ to its formal weakenings where $A$ is quantified over a cofinal set. Our first-order sentence $\sigma$ says that

  1. the universe is $L_n$,
  2. $(L_0,\lt_0),\ldots,(L_n,\lt_n)$ are linear orders,
  3. $f_i(x,\bullet)\colon L_i \rightarrow \{ y : y\leq_{i+1} x\}$ is always onto,
  4. $g(x,\bullet)\colon L_{n-1}\rightarrow \{0,1\}$ for all $x\in L_n$,
  5. $g(x,\bullet)\not=g(y,\bullet)$ for all $x\not= y$,
  6. $h(x_{n-1},\ldots,x_0,\bullet)\colon L_0\rightarrow 2$ for all $x\in\prod_{i=0}^{n-1} L_i$, and
  7. every $g(x_n,f_{n-2}(x_{n-1},f_{n-3}(x_{n-2},\cdots,f_0(x_1,\bullet)\cdots)))$ equals some $h(x_{n-1},\ldots,x_0,\bullet)$.

$(*_{\alpha,n})$ holds iff $\sigma$ has a model with size $\alpha^{+n}$ with $L_0$ of size $\alpha$.

Jensen proved that something stronger than $(*_{\alpha,n})$ holds if $V=L$.

  • KH($\kappa,\lambda$) says that there exists $\mathcal{F}\subset\mathcal{P}(\kappa)$ of size $\kappa^+$ such that $|\{X\cap A:X\in\mathcal{F}\}|\leq|A|+\aleph_0$ for all $A\subset\kappa$ of size $\lt\lambda$.

Clearly, KH($\alpha^{+n-1},\alpha^+$) implies $(*_{\alpha,n})$.

Jensen proved that if $V=L$, then KH($\kappa,\lambda$) holds for all regular uncountable $\kappa$ and all uncountable $\lambda<\kappa$. In particular, $(*_{\alpha,n})$ is always true in $L$. Jensen proved that $V=L$ also implies KH($\kappa,\kappa$) for all regular uncountable cardinals that are not ineffable. Jensen and Kunen proved that if $\kappa$ is ineffable, then KH($\kappa,\kappa$) fails. Devlin's exposition of the proofs is available here; see the chapter "Ineffable cardinals and the generalised Kurepa hypothesis."

On the other hand, $(*_{\alpha,n})$ is directly refuted by the Chang conjecture variant $(\alpha^{+n},\alpha^{+n-1})\twoheadrightarrow(\alpha^+,\alpha)$. For regular $\alpha$, we can force this Chang conjecture easily using the method of Levinski-Magidor-Shelah (MR1045371). Assume GCH and epsilon more than a huge embedding: $j\colon V\prec N\supset {}^{\lambda^+}N$ where $\kappa=cp(j)$ and $\lambda=j(\kappa)$. Since $N$ knows that $j''\mathfrak{A}\prec j(\mathfrak{A})$ for all structures $\mathfrak{A}$ of the form $(H(\lambda^+),\in,P)$, we have $(\lambda^+,\lambda)\twoheadrightarrow (\kappa^+,\kappa)$ in $V$ by elementarity of $j$. The two-step iteration $\mathbb{P}=\mathrm{Coll}(\alpha,\kappa)*\mathrm{Coll}(\kappa^{+n-2},\lt\lambda)$ preserves GCH and forces $(\alpha^{+n},\alpha^{+n-1})\twoheadrightarrow(\alpha^+,\alpha)$. The key point is that because there are only $\lambda$-many nice $\mathbb{P}$-names for elements of $\lambda$, this set of names Chang-transfers to a $\kappa$-sized set of names for elements of $\lambda$, implying our desired transfer from $\alpha^{+n-1}$ to $\alpha$ in the generic extension $V[G]$: if in $V$ we have

  • $\mathbb{P}\in M\prec(H(\lambda^+),\in,\dot{P})$,
  • $|M|=\kappa^+$, and
  • $|M\cap\lambda|=\kappa$,

then in $V[G]$ we have

  • $M[G]\prec(H(\lambda^+),\in,\dot{P}_G)=(H(\alpha^{+n}),\in,\dot{P}_G)$,
  • $|M[G]|=\kappa^+=\alpha^+$, and
  • $|M[G]\cap\alpha^{+n-1}|=|M[G]\cap\lambda|=|\kappa|=\alpha$.
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@David Milovich: Sorry I was gone for vacations and didn't see your answer until now. Give me a few days and I will be back to it. –  Ioannis Souldatos Jun 30 '12 at 13:34
    
Very nice answer. This is what I was looking for. Just to clarify one point: In clause (7), where you define $\sigma$, I am assuming that $x_{n-1},\ldots,x_1$ name the same elements in both functions ($g(x_n,f_{n-2}(x_{n-1},\ldots,f_0(x_1,\cdot)\ldots)))$ and $h(x_{n-1},\ldots,x_0,\cdot)$. So, the difference is that the first function contains $x_1,\ldots,x_n$, while the the second function contains the (same) $x_1,\ldots,x_{n-1}$ and $x_0$. Do we agree here? –  Ioannis Souldatos Jul 10 '12 at 17:31
    
Yes, we agree. A more formal version of (7) would start with "for all $x_n,\ldots, x_1$, there exists $x_0$ such that..." –  David Milovich Jul 10 '12 at 20:10
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