# The axiom of choice. Help

1. Sep 10, 2007

### sutupidmath

the axiom of choice. Help!!!

well i am having trouble understanding why we need the axiom of choice. So could anybody post here some problems, and their solutions, that include the axiom of choice, and explain how it works, so how it is used in solving problems, and explain why we actually need the axiom of choice at all.
I would also like to know how to show the equivalence between the axiom of choice and Zorn's lemma, and Cermelo's theorem( every non empty set X can be well ordered).

thnx in return.

2. Sep 10, 2007

### matt grime

You need the axiom of choice in order to guarantee that you can make a choice of infinitely many things in a consistent manner. It doesn't follow from the axioms of ZF (it is independent of them in fact). Choosing finitely many things you can do by induction, but infinitely many, well, that is tricky unless you happen to have something like the axiom of choice.

3. Sep 10, 2007

### Kummer

To add to matt-grime the axiom of choice is most nicely (for me) used in the syle of Zorn's Lemma. I have used it a few times before when I had to do something an infinitely number of times in an rigorous manner.

For example (I cannot find the thread) (I am going to skip much details because I do not know if you ever learned this, but you can still understand the structure of this argument. That is why I post it) every field has algebraic closure. Now here is how we prove this. Algebraic closure means (not exactly but for sake of simplicity) all elements are algebraic. Now given a field F if it is algebraically closed we are done. If not, i.e. there is an element "a" which is not algebraic we use Kroneckoer's theorem which says we can find a larger field having this "a" algebraic. Now if this new field, say E, is algebraically closed we are done. If not, i.e. there ia a "b" which is not algebraic we by simialr reasoning can extened it, to K say, where "b" is algebraic. The idea is that we do this infinitely number of times. And this is where Zorn's lemma enters.

But we have to be really really careful. It is not as simple as I say. In that thread (which I cannot find) math_wonk disfavors the approach I use above to prove the theorem because there are some technical set theory details that need to be met and I never address. But that is the general idea of how I seen it used and used it.

4. Sep 10, 2007

### mathwonk

you never need either the axiom of choice or the zorn lemma, if technical details do not bother you.

come on kummer, surely you learned more than that from my very detailed and thoroughly referenced comments on your erroneous argument.

it was in the post on algebraic closure in the linear and abstract algebra thread, and i moved it nearer the top for you, if you are now interested in reading it.

Last edited: Sep 11, 2007
5. Sep 11, 2007

### Kummer

When I made my post I forgot what you told me on that construction. I assumed you said that I need to be careful. But now you say that I am wrong. Maybe, I made a mistake, I do not know that much about axiomatic set theory so it very possible.

6. Sep 13, 2007

### LeoYard

Bertrand Russell stated :

"To choose one sock from each of infinitely many pairs of socks requires the Axiom of Choice, but for shoes the Axiom is not needed."

The idea is that the two socks in a pair are identical in appearance, and so we must make an arbitrary choice if we wish to choose one of them. For shoes, we can use an explicit algorithm -- e.g., "always choose the left shoe."

Let me ask you the following : What would happen if we didn't assume the Axiom of Choice?

7. Sep 13, 2007

### Tom Winter

The source of the quote is, I believe, Arthur C. Clark. I think you will find it in his __3001: The final Odyssey__. 'The tools of a mathematician are pencil, paper and waste paper basket. The waste paper basket distinguishes him from a philosopher' - ??? correct quote and attibution wanted." Clarke's version is even neater. Now you've got me rereading Arthur C. Clarke, for which I thank you.

Last edited: Sep 13, 2007
8. Sep 15, 2007

### fopc

Here's a plausable proposition.

Every infinite set includes a denumerbable subset.
Proof is straight forward with AC. What about without AC?

Here's a couple of definitions.

A set S is finite if there is a bijection f:{0,1,...,n-1} -> S. A set is infinite if it's not finite.

A set S is infinite if there is an injection f:S -> S such that f(S) is a proper subset of S. A set is finite if it's not infinite.

The definitions have been shown to be equivalent. Proof is based on AC. What about without AC?

9. Sep 17, 2007

### Tom Winter

Aha! The exact words -- I've got the book in front of me. Page 111 of Arthur C. Clarke's 3001, The Final Odyssey:
The Dean's complaining to his faculty. "Why do you scientists need such expensive equipment? Why can't you be like the Math Department, which only needs a blackboard and a wastebasket? Better still, like the Philosophy Department. That doesn'et even need a wastepaper basket."
In the novel, a character attributes this to the character Ted Khan.

10. Sep 18, 2007

### mathwonk

good work. there is also a famous terminology in algebraic geometry, theory of linear systems, where the general element is sometimes called the general elephant.

do you know where this comes from? ill give you a hint, the author is dickens.