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This question already has an answer here:

$\{E_\alpha\}_{\alpha\in A}\ ,\ E_\alpha\subset E\subset\mathbb{R}$ where $A$ is an index set (may be uncountable)

is a condition for the axiom of choice in my lecture notes, along with $E_\alpha\cap E_\beta=\emptyset$ if $\alpha\ne\beta$

Why can't we simply write $\{E_\alpha\}_{\alpha\in A}\ ,\ E_\alpha\subset\mathbb{R}$ instead? Cause for any collection of $E_\alpha$'s their union contains them all and is a subset of $\mathbb{R}$

Just for context, our version of the axiom of choice concludes by saying that $\exists V$ that contains one and only one element from each $E_\alpha$

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marked as duplicate by Asaf Karagila axiom-of-choice Feb 19 '18 at 18:15

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  • $\begingroup$ You need to add more information, this question is unanswerable in its current form. $\endgroup$ – Tony S.F. Feb 19 '18 at 15:18
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    $\begingroup$ @TonyS.F. I cannot add anything new, I mean it is a very simple axiom $\endgroup$ – John Cataldo Feb 19 '18 at 15:21
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If you ditch pairwise disjointness, there are counterexamples: Consider $E_1 = \{1\}, E_2 = \{2\}, E_3=\{1,2\}$. Then any choice set $V$ for $\{E_1,E_2,E_3 \}$ must contain $1$ and $2$ but then $V \cap E_3$ contains two elements.

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  • $\begingroup$ Yes I understand the disjointness condition, but why do $E_1, E_2$ and $E_3$ have to belong to 1 set $E$? $\endgroup$ – John Cataldo Feb 19 '18 at 15:23
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    $\begingroup$ @StanislasHildebrandt Oh, now I understand your question: Indeed, $E$ is unnecessary here. You can always take $E = \mathbb R$. $\endgroup$ – Stefan Mesken Feb 19 '18 at 15:26

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