If $\mathfrak{p}\ge5$, then $2^\mathfrak{p}\not\le\mathfrak{p}^2$?

Question

I don't understand some passages of the following demonstration which is present in "The Axiom of Choice" by Thomas Jech.

I don't understand

• why $|\mathscr{P}(C_n)|=2^n>n^2$ if we originally assumed $2^\mathfrak{p}\le\mathfrak{p}^2$,
• why $U=\{x_\xi\in C_\alpha:x_\xi\not\in g(\xi)\}\not=\varnothing$ and $f(U)\not\in C_\alpha\times C_\alpha$,
• why the constructed sequence $\{x_0,...,x_4,...,x_{(n-1)},...\}$ of length $\aleph$ is contrary to the definition of $\aleph$.

Could someone help me?

Answer 1

• why $|\mathscr{P}(C_n)|=2^n>n^2$ if we originally assumed $2^\mathfrak{p}\le\mathfrak{p}^2$,

The assumption that $2^\mathfrak{p}\le\mathfrak{p}^2$ is irrelevant. Here $n$ is some finite ordinal greater than $4$, and so you can show by elementary methods (say, induction on $n$ and some simple algebra) that $2^n>n^2$.

• why $U=\{x_\xi\in C_\alpha:x_\xi\not\in g(\xi)\}\not=\varnothing$ and $f(U)\not\in C_\alpha\times C_\alpha$,

The proof never claims that $U$ is nonempty and nothing goes wrong if $U$ is empty. The idea here is just that you are implementing Cantor's diagonal argument: by Cantor's theorem, there does not exist any surjection $\alpha\to\mathscr{P}(\alpha)$, and thus there is also no surjection $C_\alpha\times C_\alpha\to \mathscr{P}(C_\alpha)$ since $C_\alpha$ is in bijection with $\alpha$ and $\alpha\times\alpha$ is in bijection with $\alpha$. If $f$ mapped every element of $\mathscr{P}(C_\alpha)$ to an element of $C_\alpha\times C_\alpha$ then we could invert $f$ to get a surjection $C_\alpha\times C_\alpha\to \mathscr{P}(C_\alpha)$ and thus a contradiction by Cantor's theorem. The definition of the set $U$ is just the explicit "diagonal element" of $\mathscr{P}(C_\alpha)$ that is a counterexample to surjectivity which is produced by the proof of Cantor's theorem.

To go through the details, let me first correct a couple errors in Jech's definition of $g$. The definition should be $g(\xi)=f^{-1}(x_\eta,x_\zeta)$ where $(\eta,\zeta)=f_\alpha(\color{red}\xi)$ (here I correct a typo of Jech), or $\emptyset$ if $(x_\eta,x_\zeta)$ is not in the image of $f$ (here Jech forgets that $f$ may not be surjective). Note that by this definition $g$ may not be injective as Jech claims but that is irrelevant to the argument.

Now suppose $f(U)\in C_\alpha\times C_\alpha$. Then $f(U)=(x_\eta,x_\zeta)$ for some $\eta,\zeta<\alpha$ and there exists $\xi<\alpha$ such that $(\eta,\zeta)=f_\alpha(\xi)$. By definition, then, $g(\xi)=U$. But now by definition of $U$, $x_\xi\in U$ iff $x_\xi\not\in g(\xi)=U$, which is a contradiction.

• why the constructed sequence $\{x_0,...,x_4,...,x_{(n-1)},...\}$ of length $\aleph$ is contrary to the definition of $\aleph$.

This is literally the definition of $\aleph$. By definition, $\aleph$ is the least ordinal such that there is no injection $\aleph\to X$.