Mathematical questions whose answer depends on the Axiom of Choice

Question

Question inspired by the following surprising claim:

The chromatic number of the $R^n$ hyperplane may depend on whether the Axiom of Choice is available or not.

https://shelah.logic.at/papers/E33/

See more on the chromatic number (Hadwiger-Nelson) problem:

https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem

Are there any interesting non-artificial claims out there (like well known theorems) whose veracity critically depends on the Axiom of Choice - they comletely fall apart (or the answer changes) if the AC is removed?

Answer 1

Of course:

1. There exists a free ultrafilter.
2. There exists a Lebesgue non-measurable set.
3. There exists a non-Borel set.
4. Countable unions of countable sets are countable.
5. The real numbers are not a countable union of countable sets.
6. Every complete metric space is a Baire space.
7. Every vector space has a Hamel basis.
8. Every field has an algebraic closure, and it is unique up to isomorphism.
9. Urysohn's lemma.
10. Every infinite set has a countably infinite subset.
11. Product of compact spaces are compact.
12. Every tree of height $\omega$ in which every level is finite has a branch.
13. Continuity of a real function at $x$ is equivalent to sequential continuity.
14. Banach-Tarski theorem.

And so on and so forth. There's about infinity of them.

Also relevant:

• Advantage of accepting the axiom of choice
• Advantage of accepting non-measurable sets

And books:

1. Herrlich, H. Axiom of Choice. Lecture Notes in Mathematics, Springer, 2006.

2. Jech, T. The Axiom of Choice. North-Holland (1973).

3. Howard, P. and Rubin, J.E. Consequences of the Axiom of Choice. American Mathematical Soc. (1998). Also see the online database for the book.

4. Moore, G. H. Zermelo's Axiom of Choice. Springer-Verlag (1982).

Answer 2

I'd say the Banach-Alaoglu theorem is another example (BA : the unit ball of the dual of a Banach space is weakly-* compact).

In the very common case where the space is also separable, the Axiom of choice is not needed, but we there are spaces which are not separable (for example, $L^\infty(\mu)$). So (even though it is a bit far fetched), if you have a probability space $(\Omega, \mu)$ with no natural locally compact topology, and complicated enough so that $L^\infty(\mu)$ is not separable, then you'd need the axiom of choice to extract weak-* convergent subsequences of a sequence of bounded linear forms on $L^\infty(\mu)$.