If [TEX]f:A\to B[/TEX] is onto [TEX]B[/TEX], then [TEX]P(B)\leq P(A)[/TEX]

Set theory question.

I need to prove that if is onto , then . And under axiom of choice.

Thank You!

Re: If [TEX]f:A\to B[/TEX] is onto [TEX]B[/TEX], then [TEX]P(B)\leq P(A)[/TEX]

You mean |P(B)| <= |P(A)| ?

Re: If [TEX]f:A\to B[/TEX] is onto [TEX]B[/TEX], then [TEX]P(B)\leq P(A)[/TEX]

Quote:

Originally Posted by

**Also sprach Zarathustra** Set theory question.

I need to prove that if

is onto

, then

. And

under axiom of choice.

The way one proves this depends on the way is defined.

This might help. There must be an injection

Re: If [TEX]f:A\to B[/TEX] is onto [TEX]B[/TEX], then [TEX]P(B)\leq P(A)[/TEX]

We have a surjection f from A to B. We need an injection g from P(b) to P(A).

Let b in P(B), we define g like this : g(b)=f^-1(b). g is an injection, proof by contradiction :

let b1 and b2 two subsets of B such that b1=/=b2 and g(b1)=g(b2)=S (S is some subset of A).

Let c in b1\b2 (c is in b1 and not in b2) let x=f^-1(c). x exists because f is surjective. we know that x is in S and f(x)=c.

But c=f(x) belongs to b2 also (because the image of an element of S is in b1 and in b2). This is a contradiction because c is in b1/b2.

so we have an injection from P(b) to P(A) this means by definition |P(b)<=|P(A)|

Re: If [TEX]f:A\to B[/TEX] is onto [TEX]B[/TEX], then [TEX]P(B)\leq P(A)[/TEX]