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Let $\mathcal E$ be a set having the same cardinality as $\mathbb{R}$, and let $\mathcal E _\sigma:=\{\cup_{n=1}^{\infty}E_n:(E_n)\subset\mathcal E\}$. The goal is to show that $\mathcal E _\sigma$ has the same cardinality as $\mathbb{R}$.

The notes am reading give the following proof:

Since $\mathcal E\subset\mathcal E _\sigma$, $\mathcal E$ has cardinality less than $ \mathcal E _\sigma$. Conversely, there is an obvious map from $\prod_{n=1}^{\infty} \mathcal E$ onto $ \mathcal E _\sigma$, so $ \mathcal E _\sigma$ has cardinality less than $\prod_{n=1}^{\infty} \mathcal E$. But from a previous exercise $\text{card}\big( \prod_{n=1}^{\infty} \mathcal E\big)=\text{card}\big( \prod_{n=1}^{\infty} \mathbb R\big)=\text{card}(\mathbb R)$, so we are done.

The converse part of the proof shows that there exist a surjection from $\mathbb R$ onto $\mathcal E _\sigma$, but to apply Schroeder-Bernstein I need an injection from $\mathcal E _\sigma$ into $\mathbb R$, and to obtain such an injection from the previous surjection usually requires the axiom of choice (see here).

Is there a way to avoid using the axiom of choice for this proof? The problem I see is that given $E\in\mathcal E _\sigma$ there is no definitive rule for choosing a sequence $(E_n)\subset\mathcal E$ with $E=\cup_{n=1}^{\infty}E_n$.

Alphie
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  • What exactly is your $\mathcal{E}_\sigma$? Is it just the set of all countable unions of subsets of $\mathcal{E}$? In that case, it is just the power set of $\mathcal{E}$, right? In that case, the statement is not true then? – Aryaman Maithani Apr 30 '21 at 14:26
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    @AryamanMaithani $\mathcal{E}_\sigma$ is the set of all countable unions of elements of $\mathcal E$. – Alphie Apr 30 '21 at 14:31
  • Oh, I see. The "$(E_n) \subset \mathcal{E}$" confused me but now I see that you mean that you have a sequence $(E_n)_n$ such that $E_n \in \mathcal{E}$ for all $n$. – Aryaman Maithani Apr 30 '21 at 14:33

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No. There's no way to avoid using AC.

Consider $\cal E$ to be simple $\{\{x\}\mid x\in\Bbb R\}$. Clearly, the same size as $\Bbb R$. Now $\cal E_\sigma$ is the set of all countable sets of reals.

It is consistent with $\sf ZF$ (and $\sf DC$) that it has a strictly larger cardinality than $\Bbb R$. Even though there is always an injection and a surjection from $\Bbb R$ into this set.

Asaf Karagila
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  • Thank you for your answer. I also thought about the case where $\mathcal E$ is the set of open intervals of $\mathbb R$. In this case no AC would be needed right? (since I can write $(a,b)$ as the countable union its subintervals with rational endpoints). – Alphie Apr 30 '21 at 14:50
  • Yes. In the case of open intervals, this is fine. We can canonically replace each union by a pairwise disjoint union, and we can enumerate those canonically by choosing a rational from each interval (in a canonical way too). So we can assign each open set a unique sequence of rationals. So we can an injection, as wanted. – Asaf Karagila Apr 30 '21 at 14:54