The cardinality of set

• Dec 4th 2011, 03:39 AM
infinity003217
The cardinality of set
I've got a problem, I need to find the cardinality of set of every choice function for \$\displaystyle P(N)-\{\emptyset\}\$ and i have no idea how to start. Can anyone help me,please?(Crying)
• Dec 4th 2011, 03:43 AM
infinity003217
Re: The cardinality of set
it should be "P(N)-{empty set}" sorry i'm new here i don't know exactly how to write it.
• Dec 4th 2011, 04:45 AM
Plato
Re: The cardinality of set
Quote:

Originally Posted by infinity003217
it should be "P(N)-{empty set}" sorry i'm new here i don't know exactly how to write it.

I edited your LaTeX here is the code:
[tex]P(N)-\{\emptyset\} [/tex] gives \$\displaystyle P(N)-\{\emptyset\} \$

Here is a website that discusses the axiom of choice.

Now, I for one find the wording of this question a bit odd.
Have you posted the exact wording of the question as given to you?
• Dec 4th 2011, 07:02 AM
infinity003217
Re: The cardinality of set
Ok, i'll try to translate it one more time, I have the set of choise functions \$\displaystyle P(N)-\{\emptyset\} \$ and i need to find the cardinality of this set. Sorry but I'm not English so probably I make a lot of mistakes.
• Dec 4th 2011, 08:03 AM
Plato
Re: The cardinality of set
Quote:

Originally Posted by infinity003217
Ok, i'll try to translate it one more time, I have the set of choise functions \$\displaystyle P(N)-\{\emptyset\} \$ and i need to find the cardinality of this set.

Is it possible that you are translating characteristic function as choice function?
• Dec 4th 2011, 08:14 AM
infinity003217
Re: The cardinality of set
no, i don't think so. it is the choice function, related to the axiom of choice. I don't know if it helps.
• Dec 4th 2011, 09:46 AM
Annatala
Re: The cardinality of set
P(N) - {} will not give you a choice function. It will give you every possible non-empty subset of natural numbers.

A function is generally defined as a set of ordered pairs where the first element is unique (more precisely, a set of pairs (x,z) where if you have (x,z) and (y,z) it must be true that x = y). Also, in the absence of qualification, functions are left-total: every object in the domain must be defined. A choice function is a function defined with domain: a collection of non-empty sets, and target: elements of those sets (the union), with the special restriction that each set must map to one of its own elements.

Now P(N) - {} definitely has a choice function: you don't even need the Axiom of Choice for it! For each subset of N, simply take the smallest element. This works because the naturals are well-ordered. There are \$\displaystyle {2^\aleph^_0}\$ (or \$\displaystyle \beth_1\$) non-empty subsets of N, so that would be the cardinality of any choice function on N. I'd suspect there are \$\displaystyle \beth_2\$ many such choice functions on P(N) - {}, but that's just a wild guess.
• Dec 4th 2011, 11:30 AM
infinity003217
Re: The cardinality of set
ok, now I understand. Thank you.