How can one show that ${\rm Hom}\Bigl(\prod\limits_{i\geqslant 1} \Bbb Z,\Bbb Z\Bigr)$ has cardinality less than $2^{\mathfrak c}$?

Question

I have read here that it has cardinality $\aleph_0$, which follows from a theorem of Specker, which I couldn't find. I am looking for a less accurate bound to achieve the same conclusion, that $\prod\limits_{i\geqslant 1} \Bbb Z$ is not a free $\Bbb Z$-module. So for example $\leqslant \mathfrak c$ is good, or $<2^{\mathfrak c}$. However, I couldn't devise such a proof, and neither could I find a reference to Specker's proof. Any of those would suffice.

This is discussed in this MO question/answer. See also this one. — user160609, Jul 10 '14 at 15:13

score 2 · Accepted Answer · edited Apr 13 '17 at 12:58

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Here is the references for Specker's proof: Specker, Ernst (1950), "Additive Gruppen von Folgen ganzer Zahlen", Portugaliae Math. 9: 131–140, MR 0039719. However, there are many other proofs as well, see for example here and here. Don Zagier has given a short argument, too: http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Problems.html, second day, algebra, problem 3. Richard Stanley has this an exercise for some of his courses, see his solution here.

edited Apr 13 '17 at 12:58

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answered Jun 12 '14 at 15:21

Dietrich Burde

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Wikipedia calls $\prod\limits_{i\geqslant 1} \Bbb Z$ the Baer–Specker group and has references to the original papers too. – Jeppe Stig Nielsen Jun 12 '14 at 15:32
Certainly this will do! Thanks. – Pedro Jun 12 '14 at 15:45

score 1 · Answer 2 · answered Jun 15 '14 at 23:37

This is an alternative proof to the links provided I found in a problem sheet from a course I am taking, but the idea is slightly similar.

Suppose $F=\prod \Bbb Z$ is free, and let $M$ be the submodule of $x=(x_i)$ such that for each $n$, the set $\{i:2^n\nmid a_i\}$ is finite. Since $\Bbb Z$ is a PID, $M$ is free. Since $\# M=\mathfrak c$, $M\simeq \prod \Bbb Z$. But $M/2M\simeq \bigoplus \Bbb Z_2$ while $F/2F\simeq \prod \Bbb Z_2$, which yields a contradiction.

How can one show that ${\rm Hom}\Bigl(\prod\limits_{i\geqslant 1} \Bbb Z,\Bbb Z\Bigr)$ has cardinality less than $2^{\mathfrak c}$?

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