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The Axiom of Choice

by Bachelier
Tags: axiom, choice
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Bachelier
#1
Feb9-13, 09:34 PM
P: 376
When I read the AC, "that the ∏ of a coll. of non-∅ sets is itself non-∅" I understand its meaning, yet I come short from understanding its cardinal importance in Axiomatic set theory.

I have no exposure "yet" in ZFC but I was hoping if someone could clarify to me why is it that AC is such an important axiom especially that Zermelo used it to formulate the well-ordering theorem. Being also that Set Theory is regarded as the foundation of Mathematics. (Disregarding Godel's work of course)

Thank you
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pwsnafu
#2
Feb10-13, 03:51 AM
Sci Advisor
P: 839
AC has a large number of equivalent statements. You touched on well-ordering but there is also
"every surjective function has a right inverse",
"every non-trivial unital ring has a maximal ideal",
"every vector space has a basis", and
"two set either have the same cardinality or one is greater than the other."
micromass
#3
Feb10-13, 07:09 AM
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Quote Quote by pwsnafu View Post
AC has a large number of equivalent statements. You touched on well-ordering but there is also
"every surjective function has a right inverse",
"every non-trivial unital ring has a maximal ideal",
"every vector space has a basis", and
"two set either have the same cardinality or one is greater than the other."
And then there are a myriad of statements which require AC but are not equivalent to it. Some of those statement look pretty innocent. For example:
- A function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] is continuous at a fixed point x (from the [itex]\varepsilon-\delta[/itex] condition) if and only if for each sequence [itex]x_n\rightarrow x[/itex] holds that [itex]f(x_n)\rightarrow f(x)[/itex]. (fun fact: if we change the "at fixed x" by "at every point x", then we don't require AC anymore!)
- For every set X holds that X is either finite or there exists an injection [itex]\mathbb{N}\rightarrow X[/itex]
- [itex]\mathbb{N}[/itex] is Lindelof: every open cover of [itex]\mathbb{N}[/itex] has a countable subcover
- [itex]\mathbb{R}[/itex] is not a countable union of countable sets
- Any two bases in a vector space must have the same cardinality
- The Hahn-Banach theorem
- The Ascoli-Arzela theorem
- The existence of the Cech-Stone compactification
- Lebesgue measure is [itex]\sigma[/itex]-additive
- Every unbounded subset of [itex]\mathbb{R}[/itex] contains an unbounded sequence

Bachelier
#4
Feb11-13, 09:06 PM
P: 376
The Axiom of Choice

Quote Quote by micromass View Post
- For every set X holds that X is either finite or there exists an injection [itex]\mathbb{N}\rightarrow X[/itex]
- [itex]\mathbb{N}[/itex] is Lindelof: every open cover of [itex]\mathbb{N}[/itex] has a countable subcover
- [itex]\mathbb{R}[/itex] is not a countable union of countable sets
- Any two bases in a vector space must have the same cardinality
- The Hahn-Banach theorem
- The Ascoli-Arzela theorem
- The existence of the Cech-Stone compactification
- Lebesgue measure is [itex]\sigma[/itex]-additive
- Every unbounded subset of [itex]\mathbb{R}[/itex] contains an unbounded sequence
These will keep me busy reading for awhile.
Bachelier
#5
Feb11-13, 09:11 PM
P: 376
Quote Quote by pwsnafu View Post
AC has a large number of equivalent statements. You touched on well-ordering but there is also
"every surjective function has a right inverse",
"every non-trivial unital ring has a maximal ideal",
"every vector space has a basis", and
"two set either have the same cardinality or one is greater than the other."
So where does AC play a role here? How did Mathematicians deduce these corollaries from it?
For instance with respect to the right inverse? Where is its role?

Thanks
Bachelier
#6
Mar5-13, 10:38 AM
P: 376
Quote Quote by micromass View Post
And then there are a myriad of statements which require AC but are not equivalent to it. Some of those statement look pretty innocent. For example:
- A function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] is continuous at a fixed point x (from the [itex]\varepsilon-\delta[/itex] condition) if and only if for each sequence [itex]x_n\rightarrow x[/itex] holds that [itex]f(x_n)\rightarrow f(x)[/itex]. (fun fact: if we change the "at fixed x" by "at every point x", then we don't require AC anymore!)
- For every set X holds that X is either finite or there exists an injection [itex]\mathbb{N}\rightarrow X[/itex]
- [itex]\mathbb{N}[/itex] is Lindelof: every open cover of [itex]\mathbb{N}[/itex] has a countable subcover
- [itex]\mathbb{R}[/itex] is not a countable union of countable sets
- Any two bases in a vector space must have the same cardinality
- The Hahn-Banach theorem
- The Ascoli-Arzela theorem
- The existence of the Cech-Stone compactification
- Lebesgue measure is [itex]\sigma[/itex]-additive
- Every unbounded subset of [itex]\mathbb{R}[/itex] contains an unbounded sequence
Just finished reading your great blog entries w.r.t. the subject. Very clear and informative read indeed. Thank you.
Bacle2
#7
Mar5-13, 12:44 PM
Sci Advisor
P: 1,172
Check too, the construction of a nonmeasurable set using AC. Still, I believe non-measurable subsets can be constructed in theories that do not use AC.

There is also Tikunov's theorem, Lowenheim-Skolem, which allows you to construct models of the reals of any infinite cardinality (have you heard of

the non-standard reals?). For more, see, e.g: http://plato.stanford.edu/entries/ax.../#MatAppAxiCho
micromass
#8
Mar5-13, 12:56 PM
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Quote Quote by Bacle2 View Post
Check too, the construction of a nonmeasurable set using AC. Still, I believe non-measurable subsets can be constructed in theories that do not use AC.
But it is consistent with ZF that all sets are measurable. People have actually constructed models of set theory in which all sets are measurable. So you need some additional axiom which allows for nonmeasurable sets. I agree that the full AC is not necessary, but we do need some other form.
Bacle2
#9
Mar5-13, 01:04 PM
Sci Advisor
P: 1,172
Actually, AFAIK, using forcing, you can come up with models that satisfy ZF+ ~AC, and these models contain nonmeasurable subsets. But I have not seen this in a while, and it would take me a while to produce more arguments.


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