Hacker News new | ask | show | jobs
by tomp 2379 days ago
I'm tempted to exclude the Axiom of Choice (AC) from any math I do, and instead include the Axiom of Determinacy (AD) [1] (which contradicts AC), so that all subsets of R^n are measurable [2] (thus precluding the Banach–Tarski paradox), and the Axiom of Dependent Choice (DC), which is weaker than AC but sufficient to develop most of real analysis. Like, I don't really care if not all vector spaces have a basis; it's enough for me that all interesting vector spaces do (I think).

But then we have this (from [3]):

> For each of the following statements, there is some model of ZF¬C where it is true:

> - In some model, there is a set that can be partitioned into strictly more equivalence classes than the original set has elements, and a function whose domain is strictly smaller than its range. In fact, this is the case in all known models.

So it's really, pick your own poison - either one of these:

- you can take apart a ball and put the pieces back together into two balls

- there exists a function whose range is larger than its domain

Math is weird.

[1] https://en.wikipedia.org/wiki/Axiom_of_determinacy

[2] https://en.wikipedia.org/wiki/Solovay_model

[3] https://en.wikipedia.org/wiki/Axiom_of_choice#Statements_con...

2 comments

In fact it's an open problem (very likely the oldest open problem in set theory) whether in every model of not AC there is a set with such a paradoxical partition! Here's a nice introduction to the problem by Asaf Karagila http://karagila.org/2014/on-the-partition-principle/
"The Axiom of Choice is obviously true, the Well–ordering theorem is obviously false; and who can tell about Zorn’s Lemma?"