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Is obvious that doesn't exists 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'm searching for a 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 must exist a family that satisfies the above conditions, otherwise would exists a maximal enumerable set in $\mathbb{R}$.

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

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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 Oct 21 '17 at 18:31
  • $\begingroup$ @Asaf Where did I say without the axiom of choice? I certainly didn't mean it if I did. $\endgroup$ – Trevor Gunn Oct 21 '17 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 Oct 21 '17 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 $\endgroup$ – Maxime Ramzi Oct 25 '17 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? $\endgroup$ – Andreas Blass Oct 26 '17 at 17:14

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