It is true that the criterio of comparation of sets is equivalent to Axiom choice?

Question

We say that two sets $A$ and $B$ are comparables if and only if $|A|\leq |B|$ or $|B|\leq |A|$.

I want prove that this criterio is equivalent to axiom of choice. If I use Zorn lemma it is simple prove that A-C implies that criterio but I don’t know how to prove that that criterio implies A-C. Some ideas?

score 3 · Accepted Answer · answered Jul 27 '19 at 18:38

3

HINT:

First, show that if $A$ is a set, $\alpha$ is an ordinal, and there is an injection from $A$ to $\alpha$, then $A$ can be well-ordered.
Now fix a set $A$. Note that comparability means that for every ordinal $\alpha$, either $\alpha$ injects into $A$ or $A$ injects into $\alpha$. Is it possible that every ordinal injects into $A$?
Do you see how to get a principle you already know is equivalent to the axiom of choice from the two bulletpoints above? (What does well-orderability have to do with choice?)

answered Jul 27 '19 at 18:38

Noah Schweber

245,398

Why it is not possibile that any ordinal can be injected to A? – Federico Fallucca Jul 27 '19 at 19:42
@FedericoFallucca Intuitively, if every ordinal injected into $A$ then we could run the Burali-Forti paradox. This takes a bit of argument but does work (think about the set of well-orderings of subsets of $A$, and use Powerset and Separation). – Noah Schweber Jul 27 '19 at 19:50

score 2 · Answer 2 · answered Jul 27 '19 at 18:42

Let $\aleph(X)$ denote the Hartogs number of $X$, then we have $\aleph(X)\not\leq |X|$, but since by assumption we have that every two cardinals are comparable we obtain $|X|<\aleph(X)$.

But this now implies that $|X|$ can be well-ordered, since $\aleph(X)$ is always well-orderable, so $X$ can be well-ordered, and since $X$ was arbitrary we conclude that the well ordering principle, which is equivalent to choice, holds.

It is true that the criterio of comparation of sets is equivalent to Axiom choice?

2 Answers2