question about the axiom of choice.

i am trying to understand the axiom of choice.

def from book: suppose that $\displaystyle C$ is a collection of nonempty sets. Then there exist a function. $\displaystyle f: C \rightarrow \bigcup_{A\in C}{A}$ such that $\displaystyle f(A) \in A$ for each $\displaystyle A\in C$.

isnt this trivially obvious becouse ex.

if $\displaystyle C = \{\{1\},\{3,4\}\}$ then $\displaystyle \bigcup_{A\in C}{A} = \{1,3,4\}$

and if $\displaystyle f = x$ then $\displaystyle f(\{1\}) = \{f(1) \in \{1\} \}$ and $\displaystyle f(\{3,4\}) = \{f(3) \in \{3,4\},$$\displaystyle f(4) \in \{3,4\} \}$

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.

Re: question about the axiom of choice.

Quote:

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.

Re: question about the axiom of choice.

Re: question about the axiom of choice.

Quote:

Originally Posted by

**engpro** isnt this trivially obvious becouse ex.

if $\displaystyle C = \{\{1\},\{3,4\}\}$ then $\displaystyle \bigcup_{A\in C}{A} = \{1,3,4\}$

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

Quote:

Originally Posted by

**engpro** and if $\displaystyle f = x$

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.

Quote:

Originally Posted by

**engpro** then $\displaystyle f(\{1\}) = \{f(1) \in \{1\} \}$

The expression $\displaystyle \{f(1) \in \{1\} \}$ 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?

Quote:

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.

Quote:

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.

Re: question about the axiom of choice.

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}

Re: question about the axiom of choice.

Quote:

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 $\displaystyle f:C\to\bigcup_{A\in C}A$ and $\displaystyle f(A)\in A$ for each $\displaystyle A\in C$, as required.

Re: question about the axiom of choice.

thanks for the clarification :)

Re: question about the axiom of choice.

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

suppose $\displaystyle C = \{(-\infty,a): a \in \mathbb{R}\}$ 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 $\displaystyle \bigcup_{A \in C} A = \mathbb{R}$.

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 $\displaystyle C = \mathcal{P}(\mathbb{R})$?

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.