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I'm pretty sure that, with the axiom of choice, there always exists a surjection from $\Bbb R$ to $\omega_1$ (well-order $\Bbb R$, send the first $\omega_1$ elements to the corresponding ordinals, if there are any reals left over send them to whatever). What happens in the absence of choice?

Akiva Weinberger
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2 Answers2

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There is still such a surjection: send $r$ to the countable ordinal it codes if it codes a countable ordinal, and to (say) $17$ otherwise.

Specifically, we can assign (exercise) a relation $R_r$ on $\mathbb{N}\times\mathbb{N}$ to each real $r$, in such a way that every binary relation on $\mathbb{N}$ winds up being of the form $R_r$ for at least one $r$. Now say $r$ codes a countable ordinal $\alpha$ if $(\mathbb{N},R_r)$ is a well-ordering of type $\alpha$.

A more computability-theoretic approach: to each real $r$, assign its relative Church-Kleene ordinal $\omega_1^{CK}(r)$ (this is the least ordinal with no copy computable from $r$). The set $\{\omega_1^{CK}(r): r\in\mathbb{R}\}$ is cofinal in $\omega_1$ (exercise - this is basically the previous paragraph!), and so we can "collapse" it to get a surjection $\mathbb{R}\rightarrow\omega_1$. By "collapse," I mean the following: for a set $S$ of ordinals, map $x\in S$ to the ordertype of $(\{y\in S: y\in x\}, \in)$. The image of $S$ under this map is an ordinal (exercise), and if $S$ is a cofinal subset of $\omega_1$ the image is in fact $\omega_1$ (exercise).

Interestingly, we don't need the regularity of $\omega_1$ to do this second exercise - which is good, since $\omega_1$ need not be regular in ZF alone!

Noah Schweber
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  • In the first construction, some additional grittywork is necessary to ensure all the finite ordinals are hit. (But this is of course not difficult). – hmakholm left over Monica Aug 02 '17 at 14:24
  • @HenningMakholm Eh, there's barely any gritty work at all. You can just apply the collapse at the end, as in the second paragraph. Or look instead at a bijection between the non-natural reals and the binary relations on $\mathbb{N}$, and send each natural to itself. Or for those reals which don't code ordinals, send them to the absolute value of their integer part (which technically only works with "nice" bijections between reals and binary relations on $\mathbb{N}$, but it's easy to cook up such a bijection). – Noah Schweber Aug 02 '17 at 15:14
  • I suppose grit is in the eye of the beholder. :-) – hmakholm left over Monica Aug 02 '17 at 15:16
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Yes, such a surjection always exists. Without choice, there is a bijection from $\mathbb{R}$ to $\mathcal{P}(\mathbb{N}\times\mathbb{N})$, the set of relations on $\mathbb{N}$. There is also a surjection $\mathcal{P}(\mathbb{N}\times\mathbb{N})\to\omega_1$ which sends each well-ordering of a subset of $\mathbb{N}$ to its order type and every relation that is not a well-ordering of a subset of $\mathbb{N}$ to $0$. Composing these gives a surjection $\mathbb{R}\to\omega_1$.

Eric Wofsey
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