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Let $\frak c$ be the cardinality of the reals. I know that in ZF the set of endomorphisms of $(\mathbb R,+)$ can have at least two different cardinalites:

  • If we allow the axiom of choice, you can consider the reals as a vector space over the rationals, and get $\frak{c^c}$ homomorphisms. Note that with the axiom of choice this is the same as $\frak c!$(cardinality of set of bijections) and $2^\frak c$.
  • On the other hand there is the Solovay model which is a model where every set of reals is Baire measurable. By theorem of Pettis all Baire measurable homomorphisms between Polish groups are continuous. There are $\frak c$ continuous homomorphisms $(\mathbb R,+)$ to itself.
  • Asaf Karagila brought up, in a version of this question on math.se, that it is consistent that $\aleph_1 \times \frak c$ can not support a group structure. This means the set of endomorphisms can not have that cardinality(in such models).

Questions: Can the cardinality of the set of endomorphisms,$\frak e$, be strictly between $\frak c$ and $\frak{c^c}$? Somewhat vague—is there a sort of classification of what cardinals $\frak e$ could be(for example you need to be able to put a group structure on it, are there any other restrictions)? I am also interested in the same questions except looking at isomorphisms and the cardinals are $\frak c$ and $\frak {c} !$.

(this is an edited version of a question I asked on math.se, I figured most of the active math.se set theory people already saw it there and I probably should have asked it here anyways)

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    $\begingroup$ Martin Sleziak pointed out, correctly, on MSE that choice is unnecessary for the cardinal arithmetic at the end of the first bullet point. $\endgroup$ – Asaf Karagila Sep 29 '19 at 18:00
  • $\begingroup$ I hate to interject with an uninformed comment, but what does cardinality mean without choice? Are we talking about sets modulo existence of a bijection? Or something about maps to or from cardinals? $\endgroup$ – Tim Campion Sep 29 '19 at 20:33
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    $\begingroup$ @TimCampion Tim, yes, I assume we are talking about the large poset that is the posetal collapse of the preorder defined by $X \leq Y$ iff there exists an injection from $X$ to $Y$ (by Cantor-Schroeder-Bernstein, $X \leq Y$ and $Y \leq X$ iff $X$ and $Y$ are isomorphic). The question is certainly sensible under this assumption. $\endgroup$ – Todd Trimble Sep 29 '19 at 21:55
  • $\begingroup$ @Tim: Cardinals, without choice, refer to (and should refer to) the general notion of cardinality. But since ordinals are just so damn useful, cardinals are defined as initial ordinals for well-ordered sets, and Scott equivalence classes otherwise (i.e., take the class-sized equivalence class modulo bijections, and cut it on the least rank it's not empty). $\endgroup$ – Asaf Karagila Sep 30 '19 at 9:14
  • $\begingroup$ @Todd: Actually it's the only sensible way to read this question, since if you talk about cardinals as well-ordered sets only, then either $\Bbb R$ can be well-ordered, in which case we fallback to the AC situation, or it's not a cardinal to begin with, in which case the question is meaningless. Since the question is not quite meaningless, reading it as "cardinals are initial ordinals" makes no sense. :-) $\endgroup$ – Asaf Karagila Sep 30 '19 at 9:15

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