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?
Asked
Active
Viewed 56 times
1
-
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
-
1https://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