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I need to prove the axiom of power set (existence of) using the axiom of replacement. X is the original set (say {0,1}) I tried to create a relation(Q) were each element of the set is assigned a true or a false value in all possible combinations. Then I created another function R which gives all the combination of values were the value of Q is true. By the axiom of replacement, this is a set. This is nothing but {{1},{0},{1,0},empty set}

Can this be made more elegant?

Please see this link: (exercise from Tao's analysis book) Proof of a lemma relating to power set of X

TheSilverBullet
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    Why do you think this is possible? (It isn't.) – spaceisdarkgreen Feb 21 '22 at 18:06
  • @spaceisdarkgreen Can you elaborate please? – TheSilverBullet Feb 22 '22 at 02:21
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    I mean the power set axiom is not provable from replacement (and any of the rest of the ZF axioms). For instance, the hereditarily countable sets satisfy ZF-Power set, but not Power set. So what you are asking is impossible. – spaceisdarkgreen Feb 22 '22 at 02:25
  • @spaceisdarkgreen You are correct. What I am trying to prove is this: Let X be a set. Then the set {:⊂} {Y:Y⊂X} is a set. – TheSilverBullet Feb 22 '22 at 02:52
  • That comment has some formatting issues, and a slight problem with wording, but isn't that just the power set axiom, the very thing I just said (and you agreed) one can't prove? – spaceisdarkgreen Feb 22 '22 at 02:56
  • @spaceisdarkgreen This is precisely what is required in an exercise. I stumbled on this question. Have a look: https://math.stackexchange.com/questions/3538905/power-set-axiom-terence-tao-analysis-3-4-6?rq=1 – TheSilverBullet Feb 22 '22 at 03:39
  • If you’d included the context of where you got the problem from, it would have been much easier to help. What you’re trying to prove is usually called the “power set axiom”. Tao calls something else (related) the “power set axiom”. The two are equivalent over ZF-powerset, since a subset is the same thing as a characteristic function, and $X^Y$ is a definable collection of subsets of $Y\times X$. – spaceisdarkgreen Feb 22 '22 at 04:09
  • Let us continue this discussion in chat. – TheSilverBullet Feb 22 '22 at 04:11

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