Timeline for Is $\mathbb{Z}^{\omega}$ ever the union of a chain of proper subgroups each isomorphic to $\mathbb{Z}^{\omega}$?
Current License: CC BY-SA 3.0
19 events
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| Jan 26, 2016 at 15:19 | comment | added | YCor | And without the cardinality argument: every endomorphism of the discrete group $Z^N$ is continuous for the natural Polish (product) topology. So their images are far from arbitrary. | |
| Jan 26, 2016 at 15:13 | comment | added | YCor | In a comment to an erased message, I asked whether it is true that every subgroup of $Z^N$ containing $2Z^N$ (I write $Z=\mathbb{Z}$, $N=\mathbb{N}$) is isomorphic to $Z^N$. (This would have implied a positive answer to the question.) This is not true, because, writing $c$ for continuum, the number of subgroups of $Z^N$ containing $2Z^N$ is $2^c$ while the number of subgroups of $Z^N$ isomorphic to $Z^N$ is $c$ (because the cardinal of Hom$(Z^N,Z^N)$ is $c$). Still I'm wondering if there's a reasonable description of which subgroups of $Z^N$ containing $2Z^N$ are isomorphic to $Z^N$. | |
| Jan 26, 2016 at 1:45 | history | edited | YCor |
edited tags
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| Jan 26, 2016 at 1:03 | comment | added | Boaz Tsaban | @TomekKania: You have made my day. :) | |
| Oct 2, 2014 at 18:41 | history | edited | Avshalom | CC BY-SA 3.0 |
Some information added about a wrong try at a solution
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| Oct 2, 2014 at 12:38 | history | edited | Avshalom | CC BY-SA 3.0 |
Note on an intial idea for a solution
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| Oct 2, 2014 at 12:34 | comment | added | Tomasz Kania | This paper arxiv.org/pdf/math/0508146v6.pdf might be of relevance. | |
| Oct 2, 2014 at 12:02 | comment | added | Avshalom | Thanks; I have clarified the notation to avoid any doubt. (The subgroup of all bounded sequences is free; $\mathbb{Z}^{\omega}$ is only $\aleph_{1}$-free, not $\aleph_{2}$-free.) | |
| Oct 2, 2014 at 11:56 | history | edited | Avshalom | CC BY-SA 3.0 |
Add reference and clarified notation
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| Oct 2, 2014 at 5:49 | comment | added | Włodzimierz Holsztyński | About Stanisław Mrówka's paper: the paper was not about the group of all sequences but about all bounded sequences. (This time I remember better, I believe). I hope that you will find his paper still interesting in the given context. | |
| Oct 2, 2014 at 5:41 | comment | added | Włodzimierz Holsztyński | (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). | |
| Oct 2, 2014 at 5:32 | comment | added | Włodzimierz Holsztyński | (1) It so happens, @Asaf, that Andreas Blass and John Irwin in their paper used notation $\ \mathbb Z^{\aleph_0}.\ $ Just in case, I wanted to make sure (without studying the issue too much ahead of time) that there is no other group $\ \mathbb Z^{\omega}\ $ hidden somewhere deep further in the paper. | |
| Oct 2, 2014 at 5:26 | comment | added | Asaf Karagila♦ | @Włodzimierz: I was just pointing out that $\Bbb Z^\omega$ is in fact the appropriate notation. Not $\Bbb Z^{\aleph_0}$. | |
| Oct 2, 2014 at 5:24 | comment | added | Włodzimierz Holsztyński | Oh, @Asaf, really?! | |
| Oct 2, 2014 at 5:15 | comment | added | Asaf Karagila♦ | @Włodzimierz: $\aleph_0$ denotes a cardinal, $\omega$ denotes an ordinal. Sequences are naturally indexed by ordinals. | |
| Oct 2, 2014 at 3:11 | comment | added | Włodzimierz Holsztyński | A paper on group $\ \mathbb Z^{\aleph_0}\ $ was published a time ago by Stanisław Mrówka. | |
| Oct 2, 2014 at 2:57 | comment | added | Włodzimierz Holsztyński | I assume that $\ \mathbb Z^\omega =\mathbb Z^{\aleph_0}\ $ is the additive group of all integer sequences (?). | |
| Oct 2, 2014 at 2:15 | history | edited | Avshalom | CC BY-SA 3.0 |
Punctuation
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| Oct 2, 2014 at 0:29 | history | asked | Avshalom | CC BY-SA 3.0 |