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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?

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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.

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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.

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  • $\begingroup$ Why do you think that it has to be of size $\aleph_1$ without the axiom of choice? That is blatantly false. $\endgroup$
    – Asaf Karagila
    Commented Oct 21, 2017 at 18:31
  • $\begingroup$ @Asaf Where did I say without the axiom of choice? I certainly didn't mean it if I did. $\endgroup$
    – Sera Gunn
    Commented Oct 21, 2017 at 18:32
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    $\begingroup$ I don't know about that. I'm just saying, that this is not entirely how I would interpret the question. $\endgroup$
    – Asaf Karagila
    Commented Oct 21, 2017 at 18:46
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    $\begingroup$ 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 $\endgroup$ Commented Oct 25, 2017 at 21:30
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    $\begingroup$ @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? $\endgroup$ Commented Oct 26, 2017 at 17:14

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