# Do you believe in the Axiom of Choice?

## Do you believe in the Axiom of Choice?

9 vote(s)
81.8%

1 vote(s)
9.1%
3. ### Undecided

1 vote(s)
9.1%
1. Mar 16, 2008

### mathboy

Feel free to give your reasons.

I voted yes, because too many useful theorems are thrown out the window if Axiom of Choice is rejected. I believe that these useful theorems outweigh the surprising (strange?) results that also arise from AC (e.g. every set can be well-ordered). Also, if AC is truly a failure (based on what I have no idea), then shouldn't it have failed by now, over 100 years later? I'm assuming that there has been no physical experiment available to disprove the Axiom of Choice, am I right? Will there ever be such a physical experiment?

Last edited: Mar 16, 2008
2. Mar 16, 2008

### Hurkyl

Staff Emeritus
It's not really something to "believe in". ZFC is one theory, ZF-C is another theory.

3. Mar 16, 2008

### mathboy

Then I guess I'm asking which theory you prefer: ZFC or ZF without AC?

4. Mar 16, 2008

### Hurkyl

Staff Emeritus
Models of ZF that satisfy choice are generally simpler objects than models of ZF that do not; you can usually say more interesting things about models of ZFC.

5. Mar 18, 2008

### QuantumGenie

Can someone explain to me what Zorn's lemma is?

6. Mar 18, 2008

### Hurkyl

Staff Emeritus
7. Mar 19, 2008

### ice109

another axiom used to prove the linear ordering of the cardinals i.e. for any two sets A and B there exists f: A->B or f:B->A , f an injection. this of course non-exclusive or. if you want i can type out the proof in my book for you.

8. Mar 19, 2008

### QuantumGenie

Well I would like to take a look if its not too long!

9. Mar 19, 2008

### andytoh

Proving the Law of Dichotomy is easier using the axiom of choice. The proof using Zorn's lemma is also very elegant too:
Let K be the set of all bijections from a subset of A to a subset of B. Then every totally ordered subset L of K contains an upper bound in K (the union of the bijections in L) (showing that the upper bound in K is the most crucial part). So there is a maximal function F, which you then show has either domain A or image B.

Here's a Zorn's Lemma problem that I posed earlier:
(Show that if r partially orders X, then there exists a total order relation m such that m contains r and m totally orders X.)