Re: Axiom of choice

On Mon, 18 Feb 2008 01:24:16 -0800 (PST), water
<waterloo2005@xxxxxxxxx> fed this fish to the penguins:

If Axiom of Choice is not assumed, non-Lebesgue measurable subsets of
the reals may cease to exist.

Why Axiom of Choice is deeply embedded in standard mathematics?

Because the axiom is needed to prove for many things.

Besides allowing you to prove that products of families of sets exist
(without AC, it may well be that prod_y X_y is empty) or that every
surjection has a section, here's a small list of results that need AC
in some form.

1. Tychonoff theorem, probably the single most important theorem in
general topology, since many spaces arise as closed subspaces of a
product of compact ones. Tychonoff theorem is equivalent to the axiom
of choice.

2. Baire category theorem. Most of the more spectacular theorems with
complete metric spaces use this one in one form or other. This
includes the open mapping theorem and the principle of uniform
boundedness. Countable choice is enough for it.

3. The Hahn-Banach theorem on extension of dominated linear
functionals. Strictly weaker than AC.

4. Stone's theorem on extension of ideals in Boolean algebras. This
creeps in analysis via the Stone space functor. Strictly weaker than
AC.

5. Krein-Millman theorem on extreme points of compact convex sets.
Strictly weaker than AC.

Two examples form algebra.

6. Every linear space has a basis.

7. Every field has an algebraic closure.

Regards,
G. Rodrigues
.

Relevant Pages

  • How much Choice is needed
    ... let me say that for me the Axiom of Choice is intuitively ... Let me start with the Tychonoff theorem. ... Stone's theorem on extension of ideals in Boolean algebras. ... I have read somewhere that this is also strictly weaker than AC but it ...
    (sci.math)
  • Re: Axiom of choice
    ... Because the axiom is needed to prove for many things. ... Strictly weaker than AC. ... algebras. ... Every linear space has a basis. ...
    (sci.math)
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