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As you all know, cardinality of $\mathbb{R} = 2^{\aleph_0}$ can be proved in ZF, since cardinality of $\mathbb{N} \times \mathbb{N} = \aleph_0$ can be proved in ZF.

I know that the statement 'For any infinite set $A$, $|A\times A|=|A|$ is weaker than A.C.

I wonder if there is a way to prove in ZF that $|A\times A|=|A|$ when $|A|=2^{\aleph_0}$ specifically.

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This was essentially answered by Arturo here. –  Asaf Karagila Jul 17 '12 at 19:03
@Asaf I have learned that set of all infinite cardinals is isomorphic with set of von Neumann ordinals, thus mark infinite cardinals as $\aleph_0,...,\aleph_\omega,...$. –  Katlus Jul 18 '12 at 7:10
Here, I don't understand what's the difference between 'aleph's' and infinite cardinals. –  Katlus Jul 18 '12 at 7:10
Katlus, if the real numbers cannot be well-ordered then $|\mathbb R|$ is not $\aleph_\alpha$. Aleph cardinals are those cardinals which are $\aleph$ numbers, and the axiom of choice is equivalent to the assertion that every infinite set has a cardinality of some $\aleph$. Without the axiom of choice there are other cardinals too. –  Asaf Karagila Jul 18 '12 at 7:13
I wrote two answers about this issue which you may want to read: This and That. –  Asaf Karagila Jul 18 '12 at 7:16

2 Answers 2

up vote 7 down vote accepted

Yes, you can prove this in ZF. One way of seeing this is to note that $2^{\aleph_0}$ is the size of $A=\{0,1\}^{\mathbb N}$ (the set of functions from ${\mathbb N}$ to $\{0,1\}$), and $A\times A$ is easily seen to be in bijection with $\{0,1\}^{\mathbb N\sqcup\mathbb N}$, where $\sqcup$ denote disjoint union. But $\mathbb N\sqcup\mathbb N$ is in bijection with $\mathbb N$ (think even and odd numbers).

In fact, Cantor's classical proofs that $\mathbb R^2$ and $\mathbb R$ are in bijection do not use choice. For example, $\mathbb R$ and $(0,1)$ are easily seen to be in bijection (think $\arctan$ or somesuch), and one can find a bijection between $(0,1)^2$ and $(0,1)$ by looking at decimal expansions and intertwining. (Usually one needs to treat a small (countable) set a bit differently in these arguments.)

In general, the cardinal $2^\kappa$ is the size of the set of functions from a set of size $\kappa$ to $\{0,1\}$, so $2^\kappa\times 2^\kappa$ is $2^{\kappa+\kappa}$ (as with exponentiation of finite numbers), where the sum denotes the size of a disjoint union of two sets of size $\kappa$. So, it is enough to know that $\kappa+\kappa=\kappa$ to conclude that $2^\kappa\times 2^\kappa =2^\kappa$.

It is consistent with ZF to have infinite sets $A$ such that $A\times A$ is not in bijection with $A$. In fact, as already pointed out in Martin's answer, if there are no such exceptions, choice holds. It is also consistent to have sets $A$ such that $A\sqcup A$ and $A$ are in bijection, but $A\times A$ and $A$ are not. In that case, ${\mathcal P}(A)$ and ${\mathcal P}(A)\times{\mathcal P}(A)$ would be in bijection, as explained above.

Note that if $A$ is infinite and $A\times A$ and $A$ are in bijection, then so are $A\sqcup A$ and $A$ (by Schröder-Bernstein, which does not need choice). For the case you are asking, already $\mathbb N\times\mathbb N$ is in bijection with $\mathbb N$, so things work out nicely.

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All you need here is $$2^{\aleph_0}\cdot 2^{\aleph_0} = 2^{\aleph_0+\aleph_0}=2^{\aleph_0}.$$ There is no choice involved.

BTW the fact that $|A\times A|=|A|$ is equivalent to Axiom of Choice, this is a result due to Tarski (if I remember correctly). See this question: For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice.

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You remember correctly. It is a result by Tarski, see my answer to this question. –  Asaf Karagila Jul 17 '12 at 18:03
However, is there any "normal" set (i.e. any set which one would normally encounter) for which proving $\left|S\times S\right|=\left|S\right|$ would need the Axiom of Choice? –  celtschk Jul 17 '12 at 18:15
@celtschk: If you only deal with countable sets, and sets of size continuum, no. If you allow arbitrary subsets of the real line, or even just Borel, then yes. If you are a set theorist, in descriptive set theory and models of AD you may encounter them as well relatively often (e.g. $\aleph_1+\frak c$ has this property). –  Asaf Karagila Jul 17 '12 at 18:51
I'd consider Borel subsets of the real line as pretty "normal", so the answer is obviously "yes, it is needed". Thank you. –  celtschk Jul 17 '12 at 19:29
@celtschk: This is actually a fairly new result by A. Miller, it is consistent that there is a Dedekind-finite sets of real numbers which is Borel, and Dedekind-finite do not have the property that $|A|^2=|A|$. –  Asaf Karagila Jul 17 '12 at 20:47

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