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I'm quite sure I'm missing something obvious, but I can't seem to work out the following problem (web search indicates that it has a solution, but I didn't manage to locate one -- hence the formulation):

Prove that there exists a surjection $2^{\aleph_0} \to \aleph_1$ without using the Axiom of Choice.

Of course, this surjection is very trivial using AC (well-order $2^{\aleph_0}$). I have been looking around a bit, but an obvious inroad like injecting $\aleph_1$ into $\Bbb R$ in an order-preserving way is impossible.

Hints and suggestions are appreciated.

Lord_Farin
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2 Answers2

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In what follows, by a "real", I mean a subset of $\omega\times\omega$, that is, a binary relation on $\omega$. (You can start with a bijection $\pi:\mathbb R\to\mathcal P(\omega\times\omega)$, which can be constructed without choice, so this is fine.)

If this relation happens to be a well-order of $\omega$ in order type $\omega+\alpha$, map the real to $\alpha$. Otherwise, map the real to $0$. This map is a surjection.

By the way, without choice, you cannot inject $\aleph_1$ into $\mathbb R$.

Andrés E. Caicedo
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    Why not map an order type of $\alpha$ to $\alpha$? Why write it as $\omega + \alpha$? (I mean, why ignore the subsets of $\omega\times \omega$ encoding finite well orders?) – Jason DeVito - on hiatus May 31 '13 at 16:41
  • Thank you. As I expected, it wasn't really hard, just an idea eluding me. – Lord_Farin May 31 '13 at 16:44
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    @JasonDeVito It is not important. I did it this way simply so I didn't have to say something to distinguish a well-order of $0$ from one of $1$ (both give empty relations). – Andrés E. Caicedo May 31 '13 at 16:44
  • @Andres: I see. Thanks for the clarification. – Jason DeVito - on hiatus May 31 '13 at 16:46
  • Andres, is there a nice analogy to this fact with $\omega_1^{CK}$, that is some surjection from $\omega$, or another "small" countable Polish space, onto $\omega_1^{CK}$ where the surjection is $\Delta^1_1$ but has no $\Delta^1_1$ inverse, or something like that? (Maybe replace the requirement by $\Delta^0_1$ instead?) – Asaf Karagila May 31 '13 at 17:54
  • @AsafKaragila Hmm... I don't know. I'll see if I can come up with something. – Andrés E. Caicedo May 31 '13 at 18:37
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    @AndresCaicedo Late to the game, but you can avoid the "order type 0/1" problem by using reflexive well-orders instead of irreflexive well-orders (for which order type zero is the empty set and order type 1 is a singleton of an ordered pair). – Mario Carneiro May 16 '15 at 03:57
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    @MarioCarneiro Sorry for the late reply. Yes, thank you, this simple change avoids the issue. (We map a reflexive well-ordering of a subset of $\omega$ to its order type, and all other binary relations to $0$.) – Andrés E. Caicedo Aug 10 '17 at 17:02
  • I didn't keep track of who posted the bounty, so I cannot personalize this message, my apologies. Many thanks! – Andrés E. Caicedo Aug 21 '17 at 20:01
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One of my favorite ways is to fix a bijection between $\Bbb N$ and $\Bbb Q$, say $q_n$ is the $n$th rational.

Now we map $A\subseteq\Bbb N$ to $\alpha$ if $\{q_n\mid n\in A\}$ has order type $\alpha$ (ordered with the usual order of the rationals), and $0$ otherwise.

Because every countable ordinal can be embedded into the rationals, for every $\alpha<\omega_1$ we can find a subset $\{q_i\mid i\in I\}$ which is isomorphic to $\alpha$, and therefore $I$ is mapped to $\alpha$. Thus we have a surjection onto $\omega_1$.

Asaf Karagila
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  • “Every countable ordinal can be embedded into the rationals” — is this true without choice? The standard proof of it I know uses (countable) choice, and I can’t see how to do without. – Peter LeFanu Lumsdaine May 31 '13 at 19:30
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    Peter, yes. It's true without the axiom of choice. You use transfinite induction over the ordinals. There is absolutely no use of countable choice. – Asaf Karagila May 31 '13 at 19:57
  • Hmmm… I think I see how to do it now, but unless I’m missing something, it’s subtle enough to be worth mentioning. Trying to simply prove, by induction, “every countable ordinal has some embedding into $\mathbb{Q}$”, surely requires countable choice in the induction step — having found some cofinal $\omega$-sequence in the new ordinal, one wants to choose embeddings for each ordinal in the sequence. (This is what I would think of as the standard proof.) [cont’d in next comment] – Peter LeFanu Lumsdaine May 31 '13 at 20:07
  • The simplest way I can see to avoid choice is to instead define by transfinite recursion a function giving, for every counted ordinal, a chosen embedding into $\mathbb{Q}$. Or am I missing a simpler way to do it? – Peter LeFanu Lumsdaine May 31 '13 at 20:09
  • @Peter: No, there's no use of the axiom of choice. The reason is that we don't require any sort of coherence between the functions. Just existence, and given $\beta$, the induction "unfolds" into a complete definition. It is true, however, that you cannot define an $\omega_1$-sequence of embeddings, but every countable ordinal can be mapped into the rationals, one at a time. – Asaf Karagila May 31 '13 at 20:28
  • @Peter: Also I'm not sure what a "counted ordinal" means. – Asaf Karagila May 31 '13 at 20:34
  • A counted ordinal (or counted set, etc.) is an ordinal together with a bijection with a subset of $\mathbb{N}$. (This is fairly standard terminology for working with countability without choice: e.g. without choice, one can’t prove “a countable union of countable sets is countable”, but one can prove “a countable union of counted sets is countable”.) – Peter LeFanu Lumsdaine May 31 '13 at 20:50
  • But on the main proof: I still don’t see how you’re avoiding choice in the induction step, if you’re taking just “every countable ordinal has some embedding into $\mathbb{Q}$” as the statement you’re proving inductively. Suppose we have some ordinal $\beta$, and an increasing $\omega$-sequence $\alpha_i$, cofinal in $\beta$, and we know that each $\alpha_i$ admits some embedding into $\mathbb{Q}$. How do you now give an embedding from $\beta$ into $\mathbb{Q}$, without using choice? – Peter LeFanu Lumsdaine May 31 '13 at 20:55
  • @Peter: First of all, every countable ordinal is counted. Now, suppose $\beta=\sup{\beta_n\mid n\in\omega}$ and for every $\alpha<\beta$ we have an embedding $f_\alpha\colon\alpha\to\Bbb Q$. Let $\alpha_n=\text{otp}[\beta_n,\beta_{n+1})$ (assume $\beta_0=0$ for simplicity). We know that for every $n$, $\Bbb Q\cap(n,n+1)\cong\Bbb Q$ (and in a "uniform" way, it's the composition of a scaling and a translation). Use this fact to map $\alpha_n$ into $(n+\varepsilon,n+1-\varepsilon)$ (say, $\varepsilon=\frac12$). It's easy to see the union of the ranges is indeed an embedding from $\beta$. – Asaf Karagila May 31 '13 at 21:01
  • “Every countable ordinal is counted”: for an individual ordinal, yes, the logical rules for $\exists$ let one choose a counting. But in the absence of choice, “a sequence of counted ordinals” is a very different thing from “a sequence of countable ordinals”; the difference is exactly an instance of countable choice, going from “for each $n$, there exists some counting of $\alpha_n$” to “there is a sequence $\langle f_n \rangle$, such that $f_n$ is a counting of $\alpha_n$”. [cont’d] – Peter LeFanu Lumsdaine May 31 '13 at 21:25
  • And similarly, when you say “use this fact to map $\alpha_n$ into $(n + \epsilon, n+1-\epsilon)$”, you are implicitly choosing a sequence of embeddings $f_n : \alpha_n \to \mathbb{Q}$, which requires countable choice if the inductive hypothesis just states “for each $n$ there exists some embedding from $\alpha_n$ to $\mathbb{Q}$”. – Peter LeFanu Lumsdaine May 31 '13 at 21:27
  • @Peter, that sequence of embedding exists. That's the induction hypothesis. – Asaf Karagila May 31 '13 at 21:30
  • Perhaps we’re interpreting the property “$\alpha$ has some embedding into ℚ” differently? I was interpreting it as “there exists some embedding from $\alpha$ to $\mathbb{Q}$”, and discussing what happens if one works with just this as the induction predicate. With that induction predicate, the induction hypothesis implies that (with $\beta$, $\alpha_n$, etc. as before) for each $n$, there exists some embedding from $\alpha_n$ to $\mathbb{Q}$. Without choice, this does not imply that there exists a sequence of embeddings $f_n : \alpha_n \to \mathbb{Q}$. Where do you differ from this? – Peter LeFanu Lumsdaine May 31 '13 at 21:39
  • @Peter: Kudos. You managed to get me slightly confused at the moment. Alas the result requires absolutely nothing of the axiom of choice. Note that we actually construct all the $f_\alpha$'s. Completely and explicitly. Yes, you are going to suggest the following conundrum, if $\omega_1$ is singular then pick some cofinal sequence, pick this embedding for each ordinal, embed each one into a disjoint interval, the union is an injection from $\omega_1$ into $\Bbb Q$. However the fact that we can construct this for every bounded sequence does not imply that we can for unbounded sequences. [...] – Asaf Karagila May 31 '13 at 21:49
  • [cont'd] This is very similar to the construction of an Aronszajn tree. We can always extend "a bit more", but we can't extend indefinitely. Given an arbitrary countable ordinal $\alpha$, in order to have $f_\alpha$ we go from $0$ all the way to $\alpha$, and we stop there. So this is fine. As much as it is confusing, compare this to the case of a Dedekind-finite set, where we can construct a tree that every point has a successor, but there is no infinite path. – Asaf Karagila May 31 '13 at 21:51
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    @Asaf The confusion IMO arose because you said "Induction" where you meant "Recursion". Peter's approach inducts over "there exists an embedding of $\alpha$ in $\Bbb Q$" and this argument, in its most natural form, requires choice. Whereas the recursive approach can avoid choice by using a fixed order-preserving bijection $\Bbb Q \to (0,1)$ (I think; there may be an issue with selecting a cofinal sequence without choice). So you were actually discussing different things. – Lord_Farin May 31 '13 at 23:06
  • @Lord_Farin: And had I said recursion someone would come and point out that there are non-recursive ordinals, big whoop. The distinction between "induction" and "recursion" exists, but it is more of a hindrance than it is a tool (this, for example, compared to the distinction between $\omega$ and $\Bbb N$, which is not useful for anything practical, but has deep philosophical roots, and is worth paying attention to). – Asaf Karagila May 31 '13 at 23:08
  • @Lord_Farin: Moreover, we indeed construct the embeddings by transfinite recursion, but we prove the correctness of this construction by a transfinite induction. Separating between the two would not have added clarity, in my opinion. – Asaf Karagila May 31 '13 at 23:10
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    I didn't say it was possible to avoid the confusion; I merely tried to put my finger on the cause. But it's bedtime; perhaps it could be insightful to have a QA pair address this, seeing as it causes extended confusion. But that's for tomorrow. – Lord_Farin May 31 '13 at 23:12
  • @Lord_Farin: I fairly believe that I already answered this question before on the site (including the details of the construction). Be sure to search before posting a new question. – Asaf Karagila May 31 '13 at 23:15
  • Ah — thankyou, @Lord_Farin; I think I finally see where I was misunderstanding @Asaf. Starting the proof as “Let $\beta$ be any countable ordinal; take some counting of $\beta$; then by recursion on $\alpha < \beta$, construct an embedding $f_\alpha$ from each such $\alpha$ into $\mathbb{Q}$; and then use this sequence to construct an embedding of $\beta$”, then I agree, it works fine without choice. – Peter LeFanu Lumsdaine May 31 '13 at 23:34
  • But, as I said back in my second comment, I don’t dispute that the result can be proven without choice. My point is that one very standard proof does rely on choice — that is, the proof starting “By induction on $\alpha < \omega_1$, there exists for each such $\alpha$ some embedding into $\mathbb{Q}$.”, which is what I thought @Asaf was sketching — and so it is worth noting that the result doesn’t really need the choice. – Peter LeFanu Lumsdaine May 31 '13 at 23:39
  • @Asaf Despite quite extensive search, I haven't been able to find your answer; hence I've posted my question, #408300. – Lord_Farin Jun 01 '13 at 08:37
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    I think the real point is that $\mathbb Q$ is universal in the sense that any countable linear order embeds into $\mathbb Q$. This has nothing to do with ordinals, and is proved by recursion (on an enumeration of the given linear order), with choice not playing a role anywhere ($\mathbb Q$ is countable, so there are canonical ways of choosing elements at each stage of the recursion); the argument resembles (but is simpler than) the well-known back-and-forth proof that any countable dense linear order without endpoints is isomorphic to $\mathbb Q$. – Andrés E. Caicedo Aug 02 '17 at 17:16