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It is obvious that there doesn't exist a maximal enumerable set in $\mathbb{R}$ (i.e a set $A$ that if $A \subset B \subset \mathbb{R}$, and $B$ is contable $\Rightarrow$ $A=B$ ).

I am searching for an example satisfying these conditions:

A family $\{A_i\}_{i\in I}$, such that each $A_i$ is a countable set contained in $\mathbb{R}$, and $\forall$ $i,j$ $\in$ $I$ $$A_i \subset A_j \quad\mbox{or} \quad A_j \subset A_i. $$But $\bigcup\limits_{i\in I} A_i$ is uncountable.

Using Zorn's lemma, it is easy to see that it must exist a family that satisfies the above conditions, otherwise would exist a maximal enumerable set in $\mathbb{R}$.

Is it possible to find an explicit example, or is it one of those cases in which the Axiom of Choice generates sets that exist but are impossible to construct?

Matheus Manzatto
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2 Answers2

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Assume there is an inaccessible cardinal in $V,$ and consider its Levy collapse extension $V[G].$ Famously, in the resulting universe, $M=HOD(\mathbb{R})$ is a model of ZF + DC + "all sets of reals are Lebesgue measurable" (LM) + "all sets of reals have the perfect set property" (PSP). I will show that in $M,$ for any chain $C$ of countable sets of reals, $\bigcup C$ is countable. This will show that ZFC does not imply the existence of a chain of countable sets of reals with uncountable union which is definable (even with real and ordinal parameters), and ZF + DC does not imply the existence of such a chain at all.

Working in $M,$ let $C$ be a chain of countable sets of reals. Suppose towards contradiction $\bigcup C$ is uncountable. By PSP, $|\bigcup C| = |\mathbb{R}|.$ Thus, we may assume without loss of generality $\bigcup C = [0, 1].$

Let $X = \{(x,y) \in [0,1]^2: \exists S \in C (x \in S, y \not \in S)\}.$ Then $X$ is measurable by LM, and from the Fubini theorem, we can compute that $X$ has measure 1 by integrating along $x,$ and measure 0 by integrating along $y,$ contradiction.

More generally, this argument shows that LM + PSP implies the ideal of countable sets of reals is closed under unions of chains. In fact, this can be proven just from LM or just from PSP, with significantly more effort.

Elliot Glazer
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Here's why I suspect this cannot be explicitly constructed in $\mathbf{R}$,

Suppose we had such an indexed family of sets. Let $A = \bigcup_{i \in I} A_i$. Then $A$ must have cardinality $\aleph_1$. To see this, biject $A$ to a cardinal $\kappa$, which we view also as an ordinal. Under this bijection, we will view each set $A_i$ as an ordinal contained in $\kappa$. So for $i \in I$, $A_i < \kappa$ and $A_i$ is countable. Since the union of all of the countable ordinals is $\aleph_1 = \omega_1$ (the smallest uncountable ordinal) we must have $\kappa \le \aleph_1$ and since $\aleph_0 < \kappa$ by assumption, we have $\kappa = \aleph_1$.

That means that this construction gives a subset of $\mathbf{R}$ with cardinality $\aleph_1$. Perhaps a set theorist can correct me, but I don't think there are choice-free constructions of subsets of $\mathbf{R}$ with cardinality $\aleph_1$ (assuming the continuum hypothesis is false).

On the other hand, one can construct such a family of sets outside of $\mathbf{R}$ without choice by simply considering the smallest uncountable ordinal, $\omega_1$ which is the union of all the countable ordinals. Ordinals are, by definition, linearly ordered.

Sera Gunn
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  • Why do you think that it has to be of size $\aleph_1$ without the axiom of choice? That is blatantly false. – Asaf Karagila Oct 21 '17 at 18:31
  • @Asaf Where did I say without the axiom of choice? I certainly didn't mean it if I did. – Sera Gunn Oct 21 '17 at 18:32
  • Yeah, but why are you so sure that you need to have a set of size $\aleph_1$ of reals in order to have such a chain? – Asaf Karagila Oct 21 '17 at 18:36
  • @AsafKaragila In ZFC we can prove that such a union has cardinality $\aleph_1$, right? (That's this question) So now the question is: can we construct a set in ZF which in ZFC we can prove has cardinality $\aleph_1$? I think that's reasonable. – Sera Gunn Oct 21 '17 at 18:40
  • Yes to the first question; the formulation of your second question is completely senseless. – Asaf Karagila Oct 21 '17 at 18:44
  • @AsafKaragila So do you suggest I delete my answer then? – Sera Gunn Oct 21 '17 at 18:46
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    I don't know about that. I'm just saying, that this is not entirely how I would interpret the question. – Asaf Karagila Oct 21 '17 at 18:46
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    Actually, your answer gives a wat of constructing such a family ! Let $f: \omega_1 \to \mathbb{R}$ be any injection (which you could find explicitly in terms of a choice function for $\mathbb{R}$ for instance), then the range of $f$ can be given as such a union – Maxime Ramzi Oct 25 '17 at 21:30
  • I agree with @Max, there are no more doubts – Matheus Manzatto Oct 26 '17 at 00:12
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    @AsafKaragila What can one say, in ZF, about unions of chains of countable subsets of the reals? Even when $\omega_1$ doesn't embed into $\mathbb R$, such a union might be uncountable; it might even be all of $\mathbb R$. For example, in Solovay's Lebesgue measure model, the sets in the chain could be the $\mathbb R$'s of intermediate extensions. Might something similar happen under AD? – Andreas Blass Oct 26 '17 at 17:14
  • @Andreas: I'm afraid that not a whole lot, and I don't quite know what is possible to say just in ZF. I did spend a few hours trying to at least some consistent interesting situation, but so far nothing. I suspect that under AD the situation with the reals of intermediate models can be replicated, but I am by no means an expert on that sort of thing. – Asaf Karagila Oct 26 '17 at 17:20
  • @AndreasBlass Actually as I show in my answer, there is no such chain in Solovay's model. In particular, the chain of intermediate $\mathbb{R}$'s does not exist inside the model. – Elliot Glazer Apr 09 '22 at 08:54