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mathguy1

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proof of same cardinality

Hey guys, I am stuck on this proof so if anyone can show me how to do it that would be fantastic.

The question:

is a surjective map between non-empty sets where B is infinite. Suppose that

 

Prove

assuming the axiom of choice.

My working:

So from

we know that

.

If we can show

then

.

So we need an injection

.

But I am unsure how to go from here. And also how to set up the proof in the correct manner.

Can someone run me through this proof?

Thanks heaps

Alg Num Theory

Member

Registered: 2017-11-24

Posts: 693

Website

Re: proof of same cardinality

Note that

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is a partition of A since f is surjective and (as a bona fide function) is well defined. So, use the axiom of choice to pick for each b ∈ B exactly one aᵇ ∈ f⁻¹(b). This will give you an injection

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Me, or the ugly man, whatever (3,3,6)