Axiom of choice condition. [duplicate]

$\{E_\alpha\}_{\alpha\in A}\ ,\ E_\alpha\subset E\subset\mathbb{R}$ 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 $E_\alpha$

marked as duplicate by Asaf Karagila♦ axiom-of-choice StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 19 '18 at 18:15

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