# What is the largest cardinal which can inject into $\mathbb{R}$ in ZF?

This question takes place in ZF. Assume some mild large cardinals; then it is consistent (in fact, it follows from AD, the consistency of which follows from mild large cardinals) that there are very large well-ordered cardinals onto which $\mathbb{R}$ surjects (cf. http://mathoverflow.net/questions/47028/value-of-theta-in-zfad). I am interested in the converse: what sorts of sets can inject into $\mathbb{R}$?

My main question is, Is it consistent with ZF that there is some measurable well-ordered cardinal $\mu$ and an injection $i: \mu\rightarrow\mathbb{R}$? I suspect that this question has an easy negative answer which I'm just not seeing right now, hence my asking here as opposed to at the Overflow.

A secondary, and vaguer, question, is the following: in ZF (or ZF+AD), what sorts of sets can inject into $\mathbb{R}$?

Thank you all very much in advance!

-
This was as much [real-analysis] as a question about data sets is set theory :-) – Asaf Karagila Dec 17 '11 at 6:33
I am also not sure whether [large-cardinals] or [cardinals] or both fit here. – Asaf Karagila Dec 17 '11 at 11:47

Although it is true under ZF+AD that there is no $\omega_1$-sequence of distinct reals, and also that $\omega_1$ is a measurable cardinal, in fact you do not need any AD hypothesis to show that a measurable cardinal can never inject into $\mathbb{R}$.

Theorem.(Assume ZF only) A measurable cardinal $\kappa$ cannot inject into $\mathbb{R}$, or indeed, into $P(\delta)$ for any $\delta\lt\kappa$.

Proof. Suppose that $\kappa$ injects into $P(\delta)$ for some $\delta\lt\kappa$. (Since $\mathbb{R}$ is bijective with $P(\omega)$, this includes the case of the question.) By giving the rest of $P(\delta)$ outside the range of the injection measure $0$, it follows that there is a $\kappa$-complete nonprincipal ultrafilter $\mu$ on $2^\delta$, the binary sequences of length $\delta$. Note that for any particular $\alpha\lt\delta$, we have $2^\delta=\{S\in 2^\delta\mid s(\alpha)=0\}\sqcup\{s\in 2^\delta\mid s(\alpha)=1\}$, and so exactly one of these sets is in $\mu$. Thus, we define a digit $d_\alpha$ such that $X_\alpha=\{ s\in 2^\delta\mid s(\alpha)=d_\alpha\}\in \mu$ for each $\alpha\lt\delta$. But since $\mu$ is $\kappa$-complete, it follows that $\bigcap_{\alpha\lt\delta}X_\alpha\in \mu$. But this intersection is precisely $\{d\}$, where $d(\alpha)=d_\alpha$, which contradicts the assumption that $\mu$ is nonprincipal. QED

This argument is essentially the same as the argument used in ZFC to prove that measurable cardinals are strong limits.

-
This is very nice! – Asaf Karagila Dec 17 '11 at 15:44

To your second question, there is no provable bound in ZF on the cardinal.

Consider the Feferman-Levy model, in which $\mathbb R$ is a countable union of countable sets, Cohen proved that in that model there are no $\aleph_1$ many reals; that is only countable subsets of $\mathbb R$ can be well ordered.

On the other hand, if we start with a model in which $2^{\aleph_0}=\kappa$ and go through the usual Cohen method of adding of a Dedekind-finite subset of reals then we ensure that:

1. The real numbers cannot be well ordered;
2. There is an injection from $\kappa$ into $\mathbb R$, which is true since we first add countably many Cohen reals (so no change in cardinality) and we then reduce to an inner model (of ZF, of course) which has the same $\aleph$ numbers and contains the ground model.

Lastly, if I recall correctly under AD we have that every set of reals is measurable and has the perfect set property. This implies that there are no cardinalities between $\aleph_0$ and the continuum, but also that the continuum cannot be well ordered, ergo $\aleph_1$ cannot be injected into the real numbers.

One final remark is that assuming ZFC we only have one limitation on the continuum, namely $\operatorname{cf}(2^{\aleph_0})>\omega$.

-
Asaf, under AD, the cardinal $\omega_1$ is measurable, since the club filter is a measure. So while your assertion that measurable cardinals are inaccessible is fine in ZFC, it is not correct in the context of ZF. – JDH Dec 17 '11 at 12:38
@JDH: You are correct, of course. I will try to think of a different reason why a measurable cardinal cannot be injected into the continuum (or try to see why it is possible, although I'd be surprised to hear that). – Asaf Karagila Dec 17 '11 at 13:04
I cannot find an immediate answer (nor I found any reference in the literature at this point). I believe that the answer can be deduced from the fact that if $\aleph_1$ is measurable then for every real number $a$, $\omega_1$ is inaccessible in $L[a]$. I just can't see how exactly right now. – Asaf Karagila Dec 17 '11 at 13:26
I posted an answer showing that one can use essentially the same argument as in the ZFC case showing that measurables are strong limits. – JDH Dec 17 '11 at 15:39