How can I prove that $2^m \not\leq m^2$ for infinite cardinal $m$ without using Axiom of Choice? Andrés E. Caicedo's answer in Question about Generalized Continuum Hypothesis dealt with this problem, but I want to know the exact method of proving(specifically, the method called 'Specker's original argument' in the answer of linked article). Edited: I want to know if there is a simpler way than 'Specker's original argument'.

You have the HalbeisenShelah theorem which is a stronger result. – user10354138 Jul 31 '20 at 07:59

How can I prove it? I think it is quite complicated too. – Sphere Jul 31 '20 at 11:31

1*All* of this will be quite complicated. Proofs without AC tend to rely on complicated combinatorics. – Asaf Karagila Jul 31 '20 at 12:08

What would be a negative answer? It would require literally an infinite amount of work to go through all possible proofs and finding out which one is simpler. – Asaf Karagila Jul 31 '20 at 14:09
1 Answers
You can find the proof in Jech "The Axiom of Choice" as Lemma 11.10. The proof is due to Specker.
The idea is to show that if there was such an injection, then there is an injection from $\aleph(m)$ into $m$, where $\aleph(m)$ is the Hartogs' number of $m$, and that is by definition impossible.
Suppose that $f\colon\mathcal P(m)\to m^2$ is an injection, and fix $\{x_0,x_1,x_2,x_3,x_4\}$ to be some arbitrary four elements of $m$ ($m$ is infinite, but we really just need it to have $5$ or more elements). Now proceeding by recursion, first dealing with the finite index, $n\geq 5$:
Let $C_n$ denote the set $\{x_i\mid i<n\}$, then there is some $U\subseteq C_n$ such that $f(U)\notin C_n\times C_n$. This is because $n^2<2^n$ for finite integers where $n\geq 5$. Because we are given an order on $C_n$ (we are keeping track of the enumeration because this is a recursive process), there is a definable wellordering of $\mathcal P(C_n)$, so let $U$ be the first set as above. Now let $f(U)=(x,y)$, since $f(U)\notin C_n\times C_n$ either $x$ or $y$ are not in $C_n$. If $x\notin C_n$, let $x_n=x$, otherwise let $x_n=y$.
This leads us to $C_\omega$, which is a countably infinite set. For every $\alpha<\aleph(m)$, let $f_\alpha$ be the canonical bijection from $\alpha$ to $\alpha\times\alpha$. Suppose that $C_\alpha=\{x_\xi\mid\xi<\alpha\}$ was defined, we may assume that $f_\alpha\colon\alpha\to C_\alpha\times C_\alpha$ by applying the function to the indices (which again, due to the recursion process, are being tracked through the construction).
Define $g(\xi)=f^{1}(f_\alpha(\xi))$, namely decode an ordered pair from $\xi$, and take the preimage of that ordered pair under $f$. We can now diagonalise against $g$, to explicitly define $U\subseteq C_\alpha$ such that $f(U)=(x,y)\notin C_\alpha\times C_\alpha$, and define $x_\alpha$ to be $x$ or $y$ as above.
And this process may continue up to $\aleph(m)$ which defines an injection from the Hartogs' number of $m$ into $m$, which is impossible, of course.
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I understood from the beginning to the fourth paragraph(about $C_n$), but I am having difficulties going further since I have little knowledge about set theory. Especially I do not know(well) about transfinite recursion, canonical bijection, and 'diagonalise' in the sixth paragraph. Can you offer another way of proof or explain in more detail? – Sphere Jul 31 '20 at 11:29

No. To quote Shelah, if you ask a question, you must pay the terrible price and hear the answer. This is the proof, especially since you asked for Specker's argument. This is a technical result in a technical subject. The general idea was given in the first paragraph, but the details require you to know what's going on. – Asaf Karagila Jul 31 '20 at 11:30


