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Let $\mathfrak{c}$ be the cardinality of the continuum. It is well known that in ZFC that the reals under addition has $\frak{ c^c}$(cardinality of maps to itself, and in ZFC this is the same as $2^\frak c$) self homomorphisms which you can find by thinking of the reals as a vector space over the rationals and permuting basis elements(which requires some amount if choice)

It is also well known, at least in set theory circles, that it is consistent with ZF that there are only continuous homomorphisms and a little argument shows that there are $\mathfrak{c}$ of those.

Adding axiom of choice is a lot, so maybe you don't have to add so many new homomorphisms.

Question(s): Can the set of self homomorphisms of the reals have cardinality strictly between the two cardinalities discussed above?(see edit below but the more natural upper bound is $\frak {c^c}$) With cardinality gaps can you chose which cardinal in between arbitrarily? I would also be interested in the same question except isomorphisms instead of endomorphisms(where the more natural upper bound is $\frak{c}!$).

(This is inspired by a question I recently answered, and technically an answer to this could be an answer to that, although this is more focused)

Edit: There might be some subtlety in the upper bound cardinal, since without choice things get a weird where $\kappa^\kappa$, $\kappa !$(cardinality of bijections), and $2^\kappa$ might be different. Not sure if that is true for $\frak c$ but we at leat have the set of endomorphisms has cardinality less than or equal to $\frak {c^c}$. A paper by Dawson and Howard, Factorials of infinite cardinals, says that $\kappa! < 2^\kappa$, $2^\kappa < k!$, or the cardinals are incomparable are all possible.

I have asked this question on mathoverflow.

  • Assuming GCH, no, the cardinality would be $2^\mathfrak{c}$. Assuming the negation of GCH, maybe. – Brian Moehring Sep 24 '19 at 02:50
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    @BrianMoehring: I don't think this depends on whether GCH holds at the metalevel. Note that when we ask "can such-and-such be true" in set theory, this is always jargon for "is there a model of set theory that satisfies such-and-such", or equivalently, "is the theory ZF+such-and-such consistent". – hmakholm left over Monica Sep 24 '19 at 02:53
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    (In fact, I'm pretty sure that it doesn't depend on whether GCH holds at the metalevel, since consistency questions are arithmetical, and you can both make GCH hold and make GCH fail without modifying the naturals). – hmakholm left over Monica Sep 24 '19 at 02:56
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    @BrianMoehring Remember we're working without choice here - GCH is jumping the gun a bit. – Noah Schweber Sep 24 '19 at 03:01
  • @NoahSchweber Even on it's face, my comment was intended sort of tongue-in-cheek ("under this independent assumption, no; otherwise, shrugs"). However, I didn't know that ZF+GCH implies choice until after I posted initially and looked it up, so I wasn't aware of that level of absurdity in my comment. Either way, I have exclusively assumed GCH in every proof I've written in a related topic, so I'm bowing out. – Brian Moehring Sep 24 '19 at 03:45
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    Definitely not "every cardinal" is consistent. For example, if $\aleph_1\nleq\frak c$, then $\aleph_1\times\frak c<c^c$, but there is no group structure on that set, since a group structure implies comparability (as per the usual proof that group structure on every non-empty set implies the axiom of choice). – Asaf Karagila Sep 24 '19 at 07:22
  • @AsafKaragila Ah yes, I had wondered if comparability would be a problem but I was being to narrow and quick when initially considering it. Although, at the moment I am not seeing that inequality(I suspect it is easy but I haven't thought much about incomparable cardinals and the proof I have in mind uses comparabilitiy). –  Sep 24 '19 at 08:48
  • (I figured out why $\aleph_1 < \frak c^c$) –  Sep 24 '19 at 13:51
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    Can someone provide a reference, for those of us interested folks outside of set theory circles, of the statement "It is consistent with ZF that there are only continuous homomorphisms"? – Lee Mosher Sep 24 '19 at 16:35
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    @LeeMosher It follows from a couple things. First there is the Solovay model which provides a model of ZF where ever set of reals is Baire measurable(this is the hard set theory part). Then there is a theorem of Pettis, which "isn't hard" and can be found in Automatic continuity of group homomorphisms by Christian Rosendal Thm 2.2, which says that any Baire measurable homomorphism between Polish groups is actually continuous(nothing about ZF or ZFC). Putting this together you get that there are actually only continuous homomorphisms. –  Sep 24 '19 at 16:50
  • The proof of Pettis theorem is basically a Baire category argument for those interested, and only a couple of paragraphs. –  Sep 24 '19 at 16:55
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    @Lee:As suggested by Paul, this is a simple application of Pettis' theorem. You can find more information on the subject (with some proofs) in http://karagila.org/wp-content/uploads/2016/10/axiom-of-choice-in-analysis.pdf – Asaf Karagila Sep 25 '19 at 07:18
  • Thanks all. I had vaguely heard of the Solovay model, but its formulation and purpose were unknown to me. – Lee Mosher Sep 25 '19 at 17:29
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    @PaulPlummer I think that you have $\operatorname{Bij}(\mathbb R,\mathbb R)=2^{\mathfrak c}$ without choice. See some comments in chat and also answers to this question: $\Bbb {R^R}$ equinumerous to ${f\in\Bbb{R^R}\mid f\text{ surjective}}$. – Martin Sleziak Sep 29 '19 at 14:43
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    Also, referring to Howard's paper is kind of a red herring, since ordinals (and their power sets) tend to be incredibly well-behaved when it comes to these things. – Asaf Karagila Sep 29 '19 at 18:01
  • @MartinSleziak Thanks, I sort of figured, but I hadn't put that much though into it and was just being extra paranoid –  Sep 29 '19 at 18:05
  • @AsafKaragila Thanks, I just wasn't so sure as I didn't put much thought into finding explicit maps and was hedging my bets on the usual cardinal manipulations. –  Sep 29 '19 at 18:09

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