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Do we need axiom of choice to prove $A \subset B \subset A \Rightarrow A = B$ ?

The proof is trivial, define $C = B \backslash A$. If $A \neq B$ then $C$ is nonempty, so pick any element $x$ of $C$, then $x \in C \Rightarrow x \in B, x \not \in A$, but as $B \subset A$, thus $x \in A$, a contradiction.

Do we need actually need AoC to justify the bolded part ?

Andrés E. Caicedo
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Lelouch
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    The definition of $C$ non-empty says that there exists some $x \in C$. So no, you don't need the AoC... The AoC is needed when you know that you can pick an element from each set, but you need to make infinitely many such choices/picks at the same time. – N. S. Aug 25 '19 at 20:24
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    Can I just point out that there is nothing to prove because this statement is an axiom of set theory ? – Maxime Ramzi Aug 25 '19 at 20:30
  • It’s essentially the axiom of extensionality, but perhaps you’re being asked to prove the statement using the axiom of extensionality and the definition of subset. – Joe Aug 25 '19 at 20:44

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Intuitively, you don't need the Axiom of Choice to make a single choice. You need Choice when you have to make infinitely many choices at once (at least in general).

So no, you don't need Choice here. If a set is non-empty, then you can pick an arbitrary element in it.

Arthur
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