Clarification prepended below. Essentially the original post follows it.

Whether or not CH is true in V, the well ordering theorem lets us prove there exists a set of reals equinumerous with the countable ordinals, i.e. of cardinality $\aleph_1$; well-order the reals, and then take the reals with only countably-many predecessors in that order. I am saying "reals" because it is more concise than "subsets of $P(\mathbb{N})$", but it is the latter that I care about. As far as I know, this doesn't affect anything significant below.

I have heard the above construction of a set of $\aleph_1$ reals is about as "explicit" as we know how to do, by any means. But of course "explicit" doesn't have a unique precise interpretation. So, I have tried to ask here a precise question that sheds light on that. The question should be a faithful formalization of: Without AC, can you prove there exists a set of reals that is equinumerous with the countable ordinals? Unfortunately, AC is needed to prove some things about the coincidence of equinumerosity and cardinal numbers (see the linked-to stack exchange questions), so we have to be careful about how we formalize the question. I wasn't careful enough in my first go. Here is attempt 2.

Out of an abundance of caution, I will switch to asking about $P(\mathbb{N})$. Let $X<Y$ be the relation: "There is an injection from X to Y and no injection from Y to X." Let A be a sentence in the language of set theory that expresses: There exists $X \subseteq P(\mathbb{N})$ such that $\mathbb{N} < X$ and there is no $Y \subseteq P(\mathbb{N})$ such that $\mathbb{N} < Y < X$.

Then we have a 2-part question:

  • Q1: Does ZF prove A?
  • Q2: If ZF does not prove A, adding what axiom (as weak as possible) suffices? Ideally, this would be a well-known sentence B such that ZF $\vdash A \iff B$, but any $B$ weaker than $AC$ is helpful.

Original post

I'm attempting to ask roughly if ZF proves the existence of a subset of $\mathbb{R}$ of size $\aleph_1$, while avoiding the language of cardinality in light of complications discussed here: There's non-Aleph transfinite cardinals without the axiom of choice? and Well-orderings of $\mathbb R$ without Choice and Defining cardinality in the absence of choice.

My understanding is that ZF does prove such an X exists if X is not required to be a subset of $\mathbb{R}$ (there exists an uncountable set X all of whose uncountable subsets have bijections with X). I'm assuming that follows in ZF from https://proofwiki.org/wiki/Existence_of_Minimal_Uncountable_Well-Ordered_Set#Proof_Without_Using_Choice, but correct me if I'm wrong.

I've made a couple easy observations. Let A be a sentence in the language of set theory that's clearly equivalent to the question in the title, minus the informal part: there exists an uncountable set of reals X all of whose uncountable subsets have bijections with X.

  1. ZF $\vdash$ A follows from ZF + $\neg$AC + $\neg$CH $\vdash$ A, since AC and CH both separately suffice to prove A. ZF + AC $\vdash$ A using the well-ordering theorem; the reals having only countably-many predecessors in a well-ordering of $\mathbb{R}$ works for X. ZF + CH $\vdash$ A since CH implies $\mathbb{R}$ works for X.

  2. A seems to be equivalent in ZF to "$P(\mathbb{R})$ with the partial order ($X < Y$ iff $X \subset Y$ and there's no injection from Y to X) has no infinite descending chain." I'm not entirely confident in this since ZF not proving the Partition Principle gives me doubts about my ability to sketch a proof from ZF without accidentally using an independent axiom.

If the answer to ZF $\vdash$ A is no, is there a well-known axiom, weaker than AC, that together with ZF suffices to prove A? I had a look at https://en.wikipedia.org/wiki/Axiom_of_choice#Weaker_forms but didn't find one that obviously suffices.

  • 1
    $\begingroup$ Please, use MathJax to type the math in your questions. Some symbols of your question are not rendering properly. $\endgroup$
    – jjagmath
    Commented Aug 12, 2023 at 17:36

1 Answer 1


There are a lot of small misconceptions here, so let me address them one by one, while explaining why the answer is very much "no, it does not prove that".

  1. $\sf ZF$ is consistent with "There exists a Dedekind-finite set of reals", e.g. in Cohen's first model, in which case we have a subset of the reals which is not countable, not comparable with $\aleph_1$, and when we remove a point from it, it shrinks in cardinality.

  2. $\sf ZF$ is also consistent with "There exists an $\aleph_1$-amorphous set of reals", where $\aleph_1$-amorphous means that every subset is countable or co-countable, and we also implicitly mean that every infinite subset of our set is Dedekind-infinite, to avoid trivial constructions as above. The interesting thing about $\aleph_1$-amorphous sets is that they are uncountable, but every subset is just a countable set away from the whole thing, and since adding a countable set to a Dedekind-infinite set will not change its cardinality, an $\aleph_1$-amorphous set is in fact minimal, nevertheless, it is incomparable with $\omega_1$, so we get that it is consistent to have multiple minimal uncountable cardinals below the continuum (in fact, we can easily arrange to have a large number of these simultaneously).

  3. $\sf ZF$ is also consistent with "There is no set of reals of size $\aleph_1$", which we usually prove through the Perfect Set Property, which means "every uncountable set of reals contains a perfect set", which also implies that every uncountable set of reals has size continuum, which is what you're asking about. So in fact you can get a sense in which the Continuum Hypothesis holds, and indeed the continuum itself is the minimum uncountable cardinals below the continuum, but there is no set of reals of size $\aleph_1$ to begin with, therefore equating "minimum" or even "minimal" size with "size $\aleph_1$" is not provable in $\sf ZF$.

  4. $\sf ZF$ is also consistent with "There is a descending sequence of cardinals below the continuum, and every uncountable set contains a subset of size $\aleph_1$". This means that equating "there is a minimum uncountable cardinal" with the cardinals below the continuum being well-founded is not provable in $\sf ZF$ either.

  • $\begingroup$ +1. Do you know whether ZF proves "there is a minimal uncountable cardinal $\leq\mathfrak{c}$"? $\endgroup$ Commented Aug 13, 2023 at 19:58
  • $\begingroup$ Isn't that answered by (1)? $\endgroup$
    – Asaf Karagila
    Commented Aug 14, 2023 at 1:02
  • $\begingroup$ No, I wonder whether there is a model of ZF in which every uncountable caridnal $\leq\mathfrak{c}$ has another uncountable cardinal strictly below it. So for example there is no set of reals of size $\aleph_1$, not every set of reals has the PSP, ... $\endgroup$ Commented Aug 14, 2023 at 7:05
  • $\begingroup$ Ohhh, I see what you mean. If you take a Cohen model extension of the Solovay model, I believe you'll get exactly that. $\endgroup$
    – Asaf Karagila
    Commented Aug 14, 2023 at 7:36
  • 1
    $\begingroup$ No. I don't think this was ever discussed or done, or perhaps even thought about before this morning. If you want to get into the details, drop me an email. $\endgroup$
    – Asaf Karagila
    Commented Aug 14, 2023 at 14:30

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