This is from a proof in "Introduction to Cardinal Arithmetic," pages 48-49 using induction and transfinite recursion. I am trying to see how things work in this instance.
It is a proof that:
If for any set $a$ and $n\in \omega\setminus\{0\}$, $a^n$ is defined as $a^1:=a$ and $a^{k+1}:=a^k\times a$, then we have for any infinite set $a$:
$|a|=|a\times a|=|a^n|$ and $|^{n}a|=|a|$
The proof proceeds asserting that $a^n$ and $^{n}a$ are sets of all $n$-tuples of members of $a$. And assumes
as inductive hypothesis that $g$ is a bijection from $^{n}a$ onto $a$.
Here is the transfinite recursion to which my question pertains:
then $h(f)=(g(f\upharpoonright n),f(n))$ defines a bijection from $^{n+1}a$ onto $a\times a$.
I was trying to see how things work in a specific case, say $n=4$, to get $^{3+1}a=a\times a$.
Using the induction hypothesis for $n=3$, $g(3)$ is a bijection from $a\times a\times a=^{3}a$ onto $a$.
From here on I am lost. I can pose several questions:
-- is $h(f)$ a function from $g(f\upharpoonright n)$ to $f(n)$ or something that becomes clear based on the following question
-- what would $f\upharpoonright n$ and $f(n)$ and $g(f\upharpoonright n)$ be.
Then perhaps I can see the bijection from $^{3+1}a$ onto $a\times a$. I presume the $a$ in the first position corresponds to the induction hypothesis of the bijection from $^{3}a$ onto $a$.
Thanks
Note that here $f$ is a function from $n+1$ to $a$. So it's not $g(3)$, but rather $g(a_0,a_1,a_2)$.
This is really the "obvious" way of doing that. Composing the bijection between $a\times a$ and $a$ over itself inductively.