Question

Show that the axiom of choice is equivalent to the following statement.

Answer 1

If \[\space \{A_i\}_{i\in I}\] is any indexed collection of non empty sets, then
\[\large X_{i\in I}\small \space A_i=\emptyset\]

Answer 2

\[\large X_{i\in I}\small \space A_i=\large\{\large \cup_{i\in I}\small A_i\small: x(i)\in A_i,i\in I\large\}\]

Answer 3

@zzr0ck3r whats ur problem

Answer 4

Give me a moment while I work on your question!

Answer 5

that should say
\[\large X_{i\in I}\small A_i\ne \emptyset\]

Answer 6

I understand the argument I think. If I have a indexed set of non empty sets then for sure the cross product is not empty, and thus I can choose an element from each set. Conversely if I can choose an element from each set in a collection of non empty sets, then surely I can have at least one element in my cross product.
How do I make this a formal proof of equivalency...

Answer 7

@zzr0ck3r I was reading about axiom of choice and zorn's theorem yesterday and I didn't know what a partially ordered set was. Will you, sir mr.last year math major, tell me please? =]

Answer 8

I guess I need to know if I need to show reflexivity, transitivity, and symmetry?

Answer 9

if you have some set A
then the relation ~ on A is called a partial ordering if for all x,y,z in A
you have
x~x
x~y and y~x implies x =y
x~y and y~z imples x~z
(A,~) is the partially ordered set

Answer 10

if we have (A,~)
then
a chain is a subset of A we can call it C such that for each a,b in C
a~b or b~a

Answer 11

\[(R,\le)\]is a partially ordered set

Answer 12

if P is the powerset of some set A, then
\[(P(A),\large\subset)\]
is a partially ordered set

Answer 13

the power set of A, P(A) is the set of all subsets of A

Answer 14

Yes I totally understood.

Answer 15

are you making fun of me?

Answer 16

wat u think son

Answer 17

good now prove the Zorn thing:)

Answer 18

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