Thread: 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!

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

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

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)|

Originally Posted by Also sprach Zarathustra

Set theory question.
I need to prove that if is onto , then under axiom of choice.

is an surjection from

Let . Now is a partition of the set .

There is a choice function, (why?) on so that .

Define by .

Show that is an injection.