# Is the axiom of choice needed to show that $a^2=a$?

A comment on this answer states that choice is needed for the statement that $a^2=a$ for all infinite cardinals $a$. In Thomas Jech's Set Theory (3rd edition), his theorem 3.5 proves this statement when $a = \aleph_b$ for some $b$.

It's not clear to me in the proof where the axiom of choice is used - can someone clarify that?

I would also like to know if this statement is true without choice when we are working with a set that is already well-ordered. Specifically: suppose we have a well-ordered set $A$, then is it true that for any infinite $B \subset A$, we have $$\text{card}(B)^2 = \text{card}(B) ?$$

I believe that Jech's proof of 3.5 would hold in that case without the need for AC, but I would like to check this.

• Of course the statement that $a^2=a$ for all infinite cardinals is stronger than the statement that it is true for $a=\aleph_b$ if there are cardinals which are not of the form $a=\aleph_b$ (and as I recently learned without choice such cardinals exist). Therefore it would not be surprising if the proof of the general statement needed more conditions than the proof of the more special one. (I don't know if choice is used by Jech or needed at all, though). – celtschk Jul 24 '12 at 18:32
• @celtschk: Jech is proving the theorem for $\aleph$ numbers, and no choice is needed there. These sets are already well-ordered. – Asaf Karagila Jul 24 '12 at 18:38

## 1 Answer

For [infinite] every well-ordered set $A$ it holds that $A^2\sim A$. Furthermore if $B\subseteq A$ and $A$ can be well-ordered then $B$ can be well-ordered as well, and therefore this property is true for $B$ as well (if it is infinite).

However it is possible that for a non well-ordered set it might not be true anymore, in fact we can easily generate such set from every non well-orderable set. If $A$ is a non well-ordered set and $\aleph(A)$ is an ordinal such that $\aleph(A)\nleq A$ (e.g. the Hartogs number of $A$) then $A+\aleph(A)$ has the property: $$\big(A+\aleph(A)\big)^2>A+\aleph(A)$$

One should remark that it is not true for all non well-ordered sets. Even if $\mathbb R$ cannot be well-ordered it still holds that $\mathbb R^2\sim\mathbb R$.

Some proofs:

Let me add a bit of history, and refer to the Jech citation. Zermelo formulated the axiom of choice and proved (without it) that every [infinite] well-orderable set has this property in $1904$. The proof appears in the second link above. When one proves that the axiom of choice is equivalent to the fact that every set can be well-ordered, one immediately sees that the axiom of choice implies that every infinite set is bijectible with its square.

On the other hand, Tarski proved in $1923$ that the opposite holds, in particular the proof relies on an interesting lemma which abuses the example I gave for a set without this squaring property. The details appear in the first answer I have linked.

Interestingly enough, in the early $1970$'s two proofs were announced that $A+A\sim A$ does not imply the axiom of choice. I only know that one was published, it was in the Ph.D. dissertation of Sageev, and my advisor told me that he (an undergrad at the time) remembers Sageev sitting on the bench and working on this proof, and that it took him a long time to finish it. (Every footnote referring to this model suggests that the proof is very hard.)

• The Hartogs number is the least such ordinal, not just any. – tomasz Jul 24 '12 at 19:28
• @tomasz: Indeed. However any ordinal which is at least the Hartogs number will do. The Hartogs is just an example. – Asaf Karagila Jul 24 '12 at 20:20
• Sageev's result appears here: Sageev, Gershon. "An independence result concerning the axiom of choice." Ann. Math. Logic 8 (1975), 1–184. – Andrés E. Caicedo Jul 24 '12 at 22:40
• @Will: Despite my flawless Engrish, I is not live America. Video blocked me to, yes? I agree. Not good noeghu service. – Asaf Karagila Jul 24 '12 at 23:22
• I think that the first too letters would stand for "Really? This Four-Beers Mode?" – Asaf Karagila Jul 24 '12 at 23:39