Where does this proof of the Continuum Hypothesis use the axiom of choice?

Question

For simplicity, let $\mathbb N$ be the cardinality of the naturals, and $\mathbb R=2^{\mathbb N}$ be the cardinality of the reals.

Suppose we have a set $X$ with $\mathbb N < X < 2^{\mathbb N}.$ Multiplying by $2^{\mathbb N}$ gives us $\mathbb N \cdot 2^{\mathbb N} < X \cdot 2^{\mathbb N} < 2^{2\mathbb N} = 2^{\mathbb N}.$

The fact that $2\mathbb N=\mathbb N$ follows from a $1-1$ correspondence between integers and even integers, and does not require the axiom of choice.

However, every real number can be expressed in base $2$, and there are $\mathbb N$ possibilities for the integer part, and $2^{\mathbb N}$ possibilities for the fractional part, hence $\mathbb N \times 2^{\mathbb N} = \mathbb R = 2^{\mathbb N}.$

Then we get $2^{\mathbb N} < X \cdot 2^{\mathbb N} < 2^{\mathbb N},$ a contradiction, thus there does not exist such a set.

Also, why couldn't Cantor come up with such a proof? Afterall, since he didn't know about AC, he would have no trouble accepting such a proof.

Note: if $\mathbb R=2^{\mathbb N}$ itself requires AC, simply note that $\mathbb R=\mathbb N \times 2^{\mathbb N} = 2^{\mathbb N+\log_2 \mathbb N} = 2^{\mathbb N}.$

Answer 1

It is incorrect to deduce from $N<X<2^N$ that $N \cdot 2^N < X \cdot 2^N < 2^{2N}$. You can only deduce that $N \cdot 2^N \leq X \cdot 2^N \leq 2^{2N}$. In general, when dealing with potentially infinite cardinalities, arithmetic operations are very rarely guaranteed to preserve the strictness of inequalities.

(In any event, I don't know why you're asking about the axiom of choice. The axiom of choice is pretty much irrelevant here, and the continuum hypothesis cannot be proved from the axiom of choice as you seem to be assuming.)

Answer 2

There are numerous issues with your post.

The first and foremost, the Continuum Hypothesis is neither provable nor refutable from the standard axioms of set theory (read: $\sf ZFC$), and in particular whatever flavor of naive set theory you're seemingly using. As such, your proof fails, as pointed out by Eric in his answer.

The second is that the axiom of choice has absolutely no bearing on $|\Bbb R|=|2^\Bbb N|=2^{\aleph_0}$. This is easily deduced by using the Cantor–Bernstein theorem, which itself is provable without the axiom of choice, along with the standard construction of a chain in $\mathcal P(\Bbb N)$ which is order-isomorphic to the real numbers and the Cantor set.

The third, and perhaps the most terrible, is that you write $\log_2\Bbb N$. The logarithm is not defined for cardinals. Especially not for $\aleph_0$, which is what known as a strong limit cardinal. You might protest this, but note that $\log_2$ is not even defined for the natural numbers. Namely, $\log_23$ is not a natural number. Since the cardinal numbers extend the natural numbers and not the real numbers, there's no reason to believe expect $\log_2$ to be defined on the cardinal numbers. Specifically here, there is no such set $A$ such that $2^A$ is countably infinite, and again this requires no choice.

Finally, to the extent of the axiom of choice in the historical remark, this is also misleading. While the axiom itself was not formulated by Zermelo until a few decades after Cantor's seminal work on set theory, it is evident to us today that it was assumed implicitly, especially in Cantor's work. Although it is true that without the proper formalization some cardinal arithmetic were harder to prove in those early days of set theory.