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For concreteness, let $A = \{\aleph_n : n < \omega\}$. We know $\max \mathrm{pcf}(A) \in [\aleph_{\omega+1},\Pi A]$. My question is, if $\Pi A$ is big (say, $\aleph_{\omega_1+1}$), then which cardinals in that interval can $\max \mathrm{pcf}(A)$ really be?

My second question, which motivates the first, is: Let $E = \{\aleph_{2n} : n < \omega\}$ and $O = \{\aleph_{2n+1} : n < \omega \}$; then is it possible for $\max \mathrm{pcf}(A) = \max \mathrm{pcf}(E) \gg \max \mathrm{pcf}(O)$ (where $\gg$ means "much larger than" in any way you care to make precise)?

If the answer to the first question is that $\max \mathrm{pcf}(A)$ can never really be that big, then the second question doesn't matter. But if it can be big, and the answer to the second question is that the even alephs and the odd alephs can have very different $\max\mathrm{pcf}$'s, I would find that somewhat disturbing.

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Is "pcf" an abbreviation that everyone doing set theory understands? –  Gerry Myerson Oct 11 '11 at 0:35
@Gerry, yes. For your interest: en.wikipedia.org/wiki/PCF_theory –  Amit Kumar Gupta Oct 11 '11 at 3:04
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1 Answer

Shelah proved that it is consistent that GCH holds below $\aleph_\omega$, while $2^{\aleph_\omega}=\aleph_{\omega+\alpha+1}$ for any countable ordinal $\alpha$ you care to choose. (See Theorem 36.5 of Jech's book, for example).

In such a model, ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$ as well. Now if you add $\aleph_{\omega_1+1}$ Cohen reals (which has no effect on the pcf structure) you end up with a model where

  • $|\prod A| = \aleph_{\omega_1+1}$, and

So ${\rm max pcf} (A)$ could potentially be any successor cardinal below $\aleph_{\omega_1}$.

(Of course, it's still unknown if $\aleph_{\omega_1}\leq{\rm max pcf }(A)$ is possible, so this is the best answer we can hope for given our current knowledge.)

I don't know the answer to your "evens and odds" question, but certainly you can split $A$ up into two disjoint pieces whose "gap" is as large as possible:

Let $\tau$ denote ${\rm max pcf}(A)$, and suppose $\aleph_{\omega+1}<\tau$.

We know there exists an unbounded $B\subseteq A$ such that $\prod B$ contains a scale (mod finite) of length $\aleph_{\omega+1}$. This implies ${\rm max pcf}(B)=\aleph_{\omega+1}$.

The set $A\setminus B$ cannot be in the ideal $J_{<\tau}[A]$ (otherwise, we contradict ${\rm max pcf}(A)=\tau$), and so we must conclude ${\rm max pcf}(A\setminus B)=\tau$ as well.

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I'll add my usual (for these days) warning that I've had no sleep because of young kids. I'll make sure I haven't said anything silly after I wake up a bit... –  Todd Eisworth Oct 11 '11 at 15:08
@Todd, thanks for the answer. Yes, certainly what you say in the last part of your answer is true, my question is: With the evens and odds, can the "gap" really be that big? –  Amit Kumar Gupta Oct 11 '11 at 20:29
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