Thread: question about the axiom of choice.

i am trying to understand the axiom of choice.

def from book: suppose that is a collection of nonempty sets. Then there exist a function. such that for each .

isnt this trivially obvious becouse ex.
if then

and if then and
have i understod this correct?.
now what i dont understand is that certain mathematician refuses to use this theorem, they think that this theorem cant be trusted.
why?

also about latex, when i tried to see what the code would look like when i posted it, stuff like f(< fontsize = \{1.... appeard.
so i thought something was off with the size, so i marked the text and clicked on 2. that removed the problem.
maby this is a bug, in this sites latex interpreter.
also when i clicked size, the [size] parameters did not appear.

Originally Posted by engpro

i am trying to understand the axiom of choice.
now what i dont understand is that certain mathematician refuses to use this theorem, they think that this theorem cant be trusted.
why?

This webpage is one of the best I have seen.

thanks

Originally Posted by engpro

isnt this trivially obvious becouse ex.
if then

For finite C, the axiom of choice is indeed provable.

Originally Posted by engpro

and if

Do you mean f(x) = x? The f maps sets to sets, while for this C (which is a collection of sets of numbers), it should maps sets to numbers.

Originally Posted by engpro

then

The expression does not makes much sense. First, f is defined on sets of numbers, not numbers themselves, so f(1) is not defined. Second, f(1) ∈ {1} is a proposition, i.e., something that is either true or false. Are you considering a set that contains one proposition?

Originally Posted by engpro

now what i dont understand is that certain mathematician refuses to use this theorem, they think that this theorem cant be trusted.
why?

Plato has already provided the link.

Originally Posted by engpro

also about latex, when i tried to see what the code would look like when i posted it, stuff like f(< fontsize = \{1.... appeard.

LaTeX on this site does not like when a formula has newline characters in it, i.e., the whole formula must be on a single line. Also, sometimes it helps to insert spaces every 50-60 characters.

i meant f(1) = 1 and we know that 1 is in {1,2}, also i thought that you have to write : (such that) in order for it to be a proposition.
{x: proposition for x}

Originally Posted by engpro

i meant f(1) = 1 and we know that 1 is in {1,2}

For C = {{1}, {2, 3}}, one possible f is defined as follows: f({1}) = 1 and f({3, 4}) = 3. Then and for each , as required.

thanks for the clarification

here is a better example illustrating why the axiom of choice is viewed as "suspect".

suppose that is, C is a collection of open intervals that include every real number less than a particular real number.

it's not hard to see that .

the question is now: how do you define f? given the interval (-∞,a) we have to choose some real number in it. how do we do that? think about this for a while.

the axiom of choice in this setting essentially means: "well, we just pick one. who cares which one it is?"

of course, if you're clever, you might say:

ok, how about we define f(A) = floor(a). so, in a sense, this C is "still too simple". what if the elements of C are ANY subset of the real numbers? that is ?

put another way, when we define a function, do we have to "know something about it"? (like, how to compute f(x) given x). arbitrary functions f:X-->Y can be quite strange, and even when we DO know how to calculate f(x) = y, there may not be any clear way to choose a particular x that f maps to y as "distinguished". (example: pick some very large uncountable set, S. define the function f:S-->{1} by f(s) = 1, for all s in S. how do we pick an element in f^-1(1)?).

personally, i don't like to think about "arbitrary sets" because they ARE so arbitrary. but some people do, and are very good at it.