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Why here (Almost Free Modules: Set-theoretic Methods by P.C. Eklof, A.H. Mekler), on the page 181, in the 3rd line there is $$2^{\aleph_\omega}<\aleph_{\omega_2}$$? How the index $2$ in the r.h.s. was created?

The notation "pcf" stands for possible cofinality. Shelahintroduced a powerful tool into the study of cardinal arithmetic when he considered (for an infinite set $A$ of regular cardinals) the set $\operatorname{pcf}(A)$, defined to be the set of all cardinals $\lambda$ such that for some ultrafilter $U$ on $A$, $\lambda$ is the cofinality of the ultraproduct of ordered sets $\prod_{\kappa\in A} \kappa/U$. Among the striking results he proved (in ZFC) are: if $2^{\aleph_n}<\aleph_\omega$ for all $n\in\omega$ then $2^{\aleph_\omega}<\min\{\aleph_{(2^{\aleph_0})^+},\aleph_{\omega_4}\}$; in particular, if GCH holds below $\aleph_\omega$, then $2^{\aleph_\omega}<\aleph_{\omega_2}$. (See Shelah 1992, Shelah 1994 or Jech 1995 or Burke-Magidor 1990.)

page 180

page 181

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  • $\begingroup$ could someone tell me what $\omega_2$ is supposed to mean? I've never seen this notation before. $\endgroup$ Dec 18, 2018 at 21:41
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    $\begingroup$ @PinkPanther: It is the second uncountable cardinal, or the third infinite cardinal. $\endgroup$
    – Asaf Karagila
    Dec 18, 2018 at 22:07
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    $\begingroup$ I don't understand the question; the text you quoted gives the reason for the bound and provides references for the proof. It is a combination of two results of Shelah. One first appeared in the first edition of his "Proper forcing" book (it is not in the revised edition), and the other is in his "Cardinal arithmetic" book. $\endgroup$ Dec 19, 2018 at 12:26
  • $\begingroup$ OK then. I just wondered if $2$ should be read as $4$, but you say that "no". $\endgroup$
    – user122424
    Dec 19, 2018 at 15:03
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    $\begingroup$ Yes, in fact you do not need the results from "Cardinal arithmetic", just the earlier result in the Proper forcing book. $\endgroup$ Dec 19, 2018 at 16:35

1 Answer 1

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The $\aleph_{\omega_2}$ arises because under the assumption of GCH below $\aleph_\omega$, we have:

$$(2^{\aleph_0})^+ = \aleph_1^+ = \omega_2$$

so that the minimum $\min\{\aleph_{(2^{\aleph_0})^+},\aleph_{\omega_4}\}$ is actually not $\aleph_{\omega_4}$, but the other argument, which evaluates to $\aleph_{\omega_2}$.

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