I've been working through Robert Stoll "Set Theory and Logic" (Dover, 1979). Chapter 2, Section 10 "Some Theorems Equivalent to the Axiom of Choice" has an exercise (exercise 2, specifically) that reads in full as follows:
Another of Tarski's results concerning the axiom of choice asserts that it is equivalent to the proposition "if $2u < u+v$, then $u<v$," while the proposition "if $2u > u + v$, then $u > v$" can be proved without the axiom of choice. Prove this result.
It isn't stated in the problem, but I think it's safe to assume that $u$ and $v$ are infinite cardinals.
The forward direction of the first statement is easy since choice makes cardinal arithmetic trivial, but the reverse direction is lost on me. My initial guess is related to the fact that, to each infinite cardinal $c$, there is an aleph (transfinite cardinal that is well-ordered) $\aleph(c)$ which is not less than or equal to $c$ (this is Thm 1 of the same section), do some clever arithmetic, and show that $c$ must be an aleph (from which choice follows as a result of other theorems), much in the manner of Tarski's original proofs in his 1923 paper "On some theorems which are equivalent to the axiom of choice" (though I'm not sure what his exact definition of aleph is, and his central lemma regarding alephs is slightly different to the one in Stoll. Tarski states it as "to each cardinal $m$ there is an aleph which is neither greater than nor less than $m$").
In any case, Stoll says in the notes for Chapter 2 that
The various propositions which are equivalent to the axiom of choice and which have been credited to A. Tarski appear in Tarski (1923).
The paper "Tarski (1923)" is the one mentioned above, "On some theorems which are equivalent to the axiom of choice." However, the result the exercise asks for does not appear in that paper, unless I'm misreading something horribly. My question then is, if someone could point me toward a paper in which this result is proved, or assist in a proof, I would be greatly appreciative.
Best Answer
There's no clever arithmetic here.
Take a set $a$, and let $u=a\times\omega$. Note that $u+u=a\times\omega\times 2$, so $2u=u$. Now let $v=\aleph(u)$, the Hartogs number of $u$, then $2u=u<u+v$, therefore $u<v$, but that means that $u$ injects into an ordinal, and since $a$ injects into $u$ we get that $a$ injects into an ordinal and can be well-ordered.