0

Im looking for a detailed proof for $\kappa \cdot \kappa = \kappa $ with $\kappa $ beeing a infinite cardinal number. The problem is described in a book (Frank R. Drake, Set Theory: An Introduction to Large Cardinals) as a exercise but without any solution.

As far as i know i can proof that $\kappa \cdot \kappa = \kappa $ holds for the cardinal $\aleph_0$($\mathbb{N}$) wich can be proven with "Cantor's Diagonal Proof". But i need it for any infinite cardinal so it should also be valid for $\aleph_1,\aleph_2,...$

So im missing the induction part.

I know that it should be possible cause very roughly explained to show something like $| \mathbb{N} \cdot \mathbb{R}| = |\mathbb{R}|$. We can make a bijection if we put all the $\mathbb{N}$ numbers in between 0,1 of the $\mathbb{R}$ and still have enough space to adress all $\mathbb{R}$.

TL;DR im looking for the proof of $\kappa \cdot \kappa = \kappa $ and $\kappa \cdot \lambda = max(\kappa, \lambda) $

GEnGEr
  • 103
  • 3
  • I don't think Cantor's diagonal argument is how it is proved for $\mathbb N.$ – Thomas Andrews Jun 26 '18 at 19:15
  • @ThomasAndrews: I've seen Cantor's zig-zag enumeration of $\mathbb N^2$ described as a "diagonal method" before, but I agree that is not what is usually meant. – hmakholm left over Monica Jun 26 '18 at 19:17
  • Zig-zag is not a diagonal argument. @HenningMakholm, and is certainly never referred to as "Cantor's Diagonal Proof" in quotes, and capitalized. That can really only refer to one thing. – Thomas Andrews Jun 26 '18 at 19:17
  • If you want to emulate the zig-zag argument, you can take the smallest ordinal of the cardianal $\kappa$ and then show that the cardinal of $\kappa\times\kappa$ can be written as a union of $\kappa$ sets, each of cardinality strictly less than $\kappa.$ – Thomas Andrews Jun 26 '18 at 19:21
  • Ahh okey sorry i thought it counts as a proof. – GEnGEr Jun 26 '18 at 22:37

1 Answers1

1

$\kappa\lambda = \max\{\kappa, \lambda\}$ is a consequence of $\kappa\kappa = \kappa$.

Indeed, say for instance $\kappa \geq \lambda$. Then for $\lambda\neq 0$, $\kappa \leq \kappa\lambda\leq \kappa\kappa = \kappa$, so $\kappa\lambda= \kappa$.

Then for $\kappa\kappa= \kappa$, you may want to look at my answer here for instance. The idea is a transfinite induction where you well-order $\kappa\times\kappa$ nicely and show that its order type must be $\kappa$ (because each proper initial segment is $<\kappa$)

Maxime Ramzi
  • 43,598
  • 3
  • 29
  • 104