14
$\begingroup$

Recall that the covering number $cov(B)$ is the least cardinal $\kappa$ such that $\kappa$ meagre sets cover the real line. Andreas Blass and John Irwin http://www.math.lsa.umich.edu/~ablass/bb.pdf prove, among many other results, that $\mathbb{Z}^{\omega}$ (the infinite abelian group of integer-valued sequences under addition) is never the union of a chain of proper subgroups each isomorphic to $\mathbb{Z}^{\omega}$, if the chain has length less than $cov(B)$. They note it might be possible to remove the hypothesis on the length of the chain. However, they raise the question whether e.g. under $CH$, $\mathbb{Z}^{\omega}$ could be the union of a chain of proper subgroups each isomorphic to $\mathbb{Z}^{\omega}$. If the chain $\lbrace A_{\alpha} : \alpha < \delta \rbrace$ has the additional property that $\mathbb{Z}^{\omega}/A_{\alpha}$ is cotorsion-free for all $\alpha < \delta$, the answer is negative.

Has there been any further progress on this question?

One could replace $\omega$ by any countable infinite index set $A$, for example $\mathbb{Q}$, if this might help. Section 6 of George Bergman's paper exploits this possibility to define unusual subgroups of $\mathbb{Z}^{\omega}$: http://math.berkeley.edu/~gbergman/papers/free_dual.pdf

My initial misguided thought had been to seek a forcing adding a "pawn", i.e. a new real $r \in {}^{\omega}{\omega}$ (or similar) such that the old and new Baer-Specker groups would nevertheless be isomorphic in the generic extension; one might then try to build a chain by adding enough pawns. Pawns do not exist, since every epimorphic image of $\mathbb{Z}^{\omega}$ is a direct sum of $\mathbb{Z}^{I}$ and a cotorsion group for some $I$, but $\mathbb{Z}^{\omega} \cap V$ is cotorsion-free. Forcings that add epimorphisms of $\mathbb{Z}^{\omega}$ onto $\mathbb{Z}^{\omega} \cap V$ appear not to add reals.

$\endgroup$
  • 1
    $\begingroup$ I assume that $\ \mathbb Z^\omega =\mathbb Z^{\aleph_0}\ $ is the additive group of all integer sequences (?). $\endgroup$ – Włodzimierz Holsztyński Oct 2 '14 at 2:57
  • $\begingroup$ A paper on group $\ \mathbb Z^{\aleph_0}\ $ was published a time ago by Stanisław Mrówka. $\endgroup$ – Włodzimierz Holsztyński Oct 2 '14 at 3:11
  • 1
    $\begingroup$ @Włodzimierz: I was just pointing out that $\Bbb Z^\omega$ is in fact the appropriate notation. Not $\Bbb Z^{\aleph_0}$. $\endgroup$ – Asaf Karagila Oct 2 '14 at 5:26
  • 1
    $\begingroup$ (2) While perhaps here $\ \mathbb Z^\omega\ $ is to me a preferable notation (specific), nevertheless this group is isomorphic to all such groups $\ \mathbb Z^A\ $ with $\ |A|=\aleph_0.\ $ Furthermore, selecting some other set $\ A\ $ may have advantages for a particular construction or some other applications. It's similar to square matrices, where index set doesn't have to be $\ \{1\ldots n\}^2\ $ but some $\ S^2\ $ related to the respective applications (in combinatorics or StatisticalmMechanics or other). $\endgroup$ – Włodzimierz Holsztyński Oct 2 '14 at 5:41
  • 1
    $\begingroup$ This paper arxiv.org/pdf/math/0508146v6.pdf might be of relevance. $\endgroup$ – Tomek Kania Oct 2 '14 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.