Axiom of choice , Hartogs ordinals, well-ordering principle

Question

I'm trying to prove the following:

If it holds that if for any two sets $A$ and $B$, $A$ can be injected into $B$ or $B$ can be injected into $A$, then every set can be well-ordered (axiom of choice variant).

I intend to use Hartogs ordinal, but I am not very sure... my idea is that: Assuming that $A$ and $B$ have different cardinal number, and that $A$ can be injected into $B$, then $|A|<|B|$. Considering the Hartogs ordinals $H(A)$ and $H(B)$ of $A$ and $B$, I think that I can assume that $|H(A)|<|H(B)|$, then it should exist an injection from $H(A)$ to $B$, otherwise, $H(A)$ would be the Hartogs ordinal of $B$.

I'm not sure if this reasoning is correct how to continue it, I'm just beginning to study the axiom of choice...

Any ideas? Thanks!

By the way, Hartogs is the correct way to write that name, it's not Hartog. — Asaf Karagila, Mar 17 '13 at 13:53
You can find different answers here http://math.stackexchange.com/questions/268942/for-any-two-sets-a-b-a-leqb-or-b-leqa/268952#268952 — Camilo Arosemena-Serrato, Mar 17 '13 at 16:22

score 5 · Accepted Answer · answered Mar 17 '13 at 13:26

5

Hint: Compare $A$ with its Hartogs. Conclude that $A$ can be injected into an ordinal and can therefore be well-ordered.

answered Mar 17 '13 at 13:26

Asaf Karagila

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Axiom of choice , Hartogs ordinals, well-ordering principle

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