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One of cantors theorem watch

1. Lol trying to prove that
|X|<|P(X)|

So yeah, to show its not a surjection I'm having a problem.

Define

But, lol let A=f(a)

But, I don't see understand the argument that if , then contradiction and then if .

But, lol its confusing. Anyone got an easier proof.

P.S. Also, what if A was empty?
2. (Original post by Simplicity)

But, I don't see understand the argument that if , then contradiction and then if .
hello brother ..here it is http://en.wikipedia.org/wiki/Axiom_of_power_set
3. (Original post by simplexity)
hello brother ..here it is http://en.wikipedia.org/wiki/Axiom_of_power_set
Wikipedia article on power set was a help.

P.S. But, what if T or A is empty?
4. Even if A is empty, since f is a surjection there exists x such that f(x) = A. So either x is in A or x is not in A. In the former case we have a contradiction since A is empty it can't contain A, in the later case, we then have found an element x such that x is not in f(x) but A (the set of such elements) is empty. Contradiction!
5. (Original post by SimonM)
Even if A is empty, since f is a surjection there exists x such that f(x) = A. So either x is in A or x is not in A. In the former case we have a contradiction since A is empty it can't contain A, in the later case, we then have found an element x such that x is not in f(x) but A (the set of such elements) is empty. Contradiction!

Hmm, but if A is empty, how can you validly sat let f(x)=A, as yeah, since that would be mapping an element to the empty set?

arrghhhhhhhhhhh. That didn't help.

Got some picture?
6. OK, so the empty set belongs to the power set (since it is by definition a subset). We've assumed f to exist, and that it is surjective, so we know there should exist an element x which maps to it
7. (Original post by SimonM)
OK, so the empty set belongs to the power set (since it is by definition a subset). We've assumed f to exist, and that it is surjective, so we know there should exist an element x which maps to it
Yes, but if A was empty you would have found that set that isn't mapped onto, hence not a bijection?

P.S. Hmm, I'm still annoyed. Stupid proof.

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