Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2011-03-13 12:26:43

mathguy1
Guest

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

#2 2019-05-17 02:33:16

Alg Num Theory
Member
Registered: 2017-11-24
Posts: 693
Website

Re: proof of same cardinality

Note that

[list=*]
[*]

[/*]
[/list]

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 bB exactly one aᵇf⁻¹(b). This will give you an injection

[list=*]
[*]

.[/*]
[/list]


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

Offline

Board footer

Powered by FluxBB