5
votes
8answers
690 views

Result that follows from ZFC and not ZF but are strictly weaker than choice

A number of results that people use that require the axiom of choice (i.e. do not follow from ZF alone) are known to actually imply the axiom of choice. Therefore, one might naturally wonder whether ...
6
votes
3answers
2k views

Is it possible to show that an infinite set has a countable (infinite) subset, without using the Axiom of Choice?

Let X be an infinite set. Is it possible to show the existence of a countably infinite subset of X without using the Axiom of Choice?
14
votes
11answers
3k views

Does the Axiom of Choice (or any other “optional” set theory axiom) have real-world consequences? [closed]

Or another way to put it: Could the axiom of choice, or any other set-theoretic axiom/formulation which we normally think of as undecidable, be somehow empirically testable? If you have a particular ...
3
votes
2answers
1k views

Axiom of Computable Choice versus Axiom of Choice

What would be the consequence of requiring that any choice function be computable; i.e. using as the foundational basis ZF + ACC? Does it make a difference if we admit definable functions? I guess I ...