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edited Sep 29, 2019 at 12:40

user35370

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, is that the only restrictionare 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 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)

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.

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, is that the only restriction)? 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)

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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asked Sep 29, 2019 at 0:33

user35370

Can the set of endomorphisms of $(\mathbb R,+)$ have cardinality strictly between $\frak c$ and $\frak{c^c}$?

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.

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, is that the only restriction)? 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)