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I read somewhere that the statement "$x \cup y$ exists for every $x$ and $y$" can be proven in $ZFC$ set theory without using the full strength of the union axiom, which says that arbitrary unions exist, not just binary unions. Can that statement also be proven from $ZF$ set theory? Is this an open problem?

user107952
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  • Can you sketch, or give a reference for, the ZFC proof? How does it use the axiom of choice? – Nate Eldredge Aug 28 '17 at 17:10
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    https://math.stackexchange.com/questions/498256/can-we-prove-the-existence-of-a-cup-b-without-the-union-axiom?rq=1 – user107952 Aug 28 '17 at 17:12
  • @user107952 The post you linked contains a complete answer to your question posted in great detail by Andrés E. Caicedo. Hence I vote to close this question as a duplicate. – Stefan Mesken Aug 28 '17 at 18:29

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