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About LinkBacksI've got a problem, I need to find the cardinality of set of every choice function forand i have no idea how to start. Can anyone help me,please?
I edited your LaTeX here is the code:Originally Posted by infinity003217
[tex]P(N)-\{\emptyset\} [/tex] gives
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?
Is it possible that you are translating characteristic function as choice function?Ok, i'll try to translate it one more time, I have the set of choise functionsand i need to find the cardinality of this 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(or
) non-empty subsets of N, so that would be the cardinality of any choice function on N. I'd suspect there are
many such choice functions on P(N) - {}, but that's just a wild guess.
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