Power of set of choice functions

[ben21]

Homework Statement

Find the power of set of all choice functions for P($N$) - {0}.

The Attempt at a Solution

I really don't know how to start with that. Hope you will give some clues.

 

[mtayab1994]

Try to look at P(N)=2^n

How many numbers are in N-{0}?

 

[ben21]

There are $\aleph$ (aleph zero) numbers in N-{0}...
So 2^$\aleph$ equals $\Im$ (continuum)... So it proves, that the power of P(N) equals $\Im$.. However, how to combine it with a choice function?

 

[mtayab1994]

Ok so you said P(N) has N amount of number except zero. Now what do you say for this?f(N)=??

 

[ben21]

by "P(N) has N amount of number except zero" are you meaning, that P(N) has $\aleph$ one-element subsets excpet zero?

f(N)? Do you mean one-element subsets by N as argument?

 

[mtayab1994]

by "P(N) has N amount of number except zero" are you meaning, that P(N) has $\aleph$ one-element subsets excpet zero?

f(N)? Do you mean one-element subsets by N as argument?

No i mean that P(N) has an uncountable amount of subsets besides zero.

You have to construct a choice function f and whose range is N-{0}, such that f(N) is an element of N.

 

[ben21]

OK, but.. amount of subset of P(N) is more than natural numbers.. it is continuum, not aleph zero...
edit::: wrong... I'm a little bit confused.. We need to find a amount of functions f: P(B) -> n, where B is a proper subset of N and n in B...

 

[ben21]

So I think we are able to construct aleph zero these functions.. P(N) contains as subset the set of N so..
We can build this function like this:
F: P(N) -> N;
F(N) = {case 1 - the power of N equals aleph zero: G(N);
{case 2 - in other case (the power of N is less than aleph zer): max(N)

where G: P(N) -> N.
G(B) -> n, where P(N) containn B, and b in B. We can build aleph zero g functions, because we use it only if the argument of f functions has aleph zero power, so B als has alpeh zero functions - so the set of values of function G has also alpeh zero elements.

It is correct?