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)

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- Dec 4th 2011, 03:39 AMinfinity003217The 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 AMinfinity003217Re: 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 AMPlatoRe: The cardinality of set
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 AMinfinity003217Re: 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 AMPlatoRe: The cardinality of set
- Dec 4th 2011, 08:14 AMinfinity003217Re: 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 AMAnnatalaRe: 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 AMinfinity003217Re: The cardinality of set
ok, now I understand. Thank you.