idempotency of infinite cardinals
In this entry, we show that every infinite^{} cardinal is idempotent^{} with respect to cardinal addition^{} and cardinal multiplication.
Theorem 1.
$\kappa \cdot \kappa =\kappa $ for any infinite cardinal $\kappa $.
Proof.
For any nonzero cardinal $\lambda $, we have $\lambda =1\cdot \lambda \le \lambda \cdot \lambda $. So given an infinite cardinal $\kappa $, either $\kappa =\kappa \cdot \kappa $ or $$. Let $\mathcal{C}$ be the class of infinite cardinals that fail to be idempotent (with respect to $\cdot $). Suppose $\mathcal{C}\ne \mathrm{\varnothing}$. We shall derive a contradiction^{}. Since $\mathcal{C}$ consists entirely of ordinals^{}, it is therefore wellordered, and has a least member $\kappa $.
Let $K=\kappa \times \kappa $. As $K$ is a collection^{} of ordered pairs of ordinals, it has the canonical wellordering inherited from the canonical ordering on On$\mathrm{\times}$On. Let $\alpha $ be the ordinal isomorphic^{} to $K$. Since $$, there is an initial segment $L$ of $K$ that is order isomorphic to $\kappa $.
Since $L$ is an initial segment of $K$, $L=\{({\beta}_{1},{\beta}_{2})\mid ({\beta}_{1},{\beta}_{2})\prec ({\alpha}_{1},{\alpha}_{2})\}$ for some $({\alpha}_{1},{\alpha}_{2})\in K$. The wellorder $\u2aaf$ denotes the canonical ordering on $K$. Let $\lambda =\mathrm{max}({\alpha}_{1},{\alpha}_{2})$. Since $L\subset K=\kappa \times \kappa $, $$ and $$, and therefore $$.
For any $({\beta}_{1},{\beta}_{2})\in L$, we have $({\beta}_{1},{\beta}_{2})\prec ({\alpha}_{1},{\alpha}_{2})$, which implies that $\mathrm{max}({\beta}_{1},{\beta}_{2})\le \lambda $. Therefore $L\subseteq {\lambda}^{+}\times {\lambda}^{+}$, or $L\le {\lambda}^{+}\times {\lambda}^{+}\le {\lambda}^{+}\cdot {\lambda}^{+}$. There are two cases to discuss:

1.
If $\lambda $ is finite, so is ${\lambda}^{+}\times {\lambda}^{+}$, contradicting that $L$ is (order) isomorphic to $\kappa $, an infinite set.

2.
If $\lambda $ is infinite, so is ${\lambda}^{+}$. Since $$, and $\kappa $ is a limit ordinal^{}, $$ as well, which means ${\lambda}^{+}\notin \mathcal{C}$, or ${\lambda}^{+}\cdot {\lambda}^{+}={\lambda}^{+}$. Therefore $$, again contradicting that $L$ is (order) isomorphic to $\kappa $.
Therefore, the assumption^{} $\mathcal{C}\ne \mathrm{\varnothing}$ is false, and the proof is complete^{}. ∎
Corollary 1.
If $$ and $\kappa $ is infinite, then $\lambda \mathrm{\cdot}\kappa \mathrm{=}\kappa $.
Proof.
$\kappa =1\cdot \kappa \le \lambda \cdot \kappa \le \kappa \cdot \kappa =\kappa $. By SchroderBernstein’s Theorem, $\lambda \cdot \kappa =\kappa $. ∎
Corollary 2.
If $\lambda \mathrm{\le}\kappa $ and $\kappa $ is infinite, then $\lambda \mathrm{+}\kappa \mathrm{=}\kappa $.
Proof.
$\kappa =0+\kappa \le \lambda +\kappa \le \kappa +\kappa =2\cdot \kappa \le \kappa \cdot \kappa =\kappa $ by the corollary above (since $2\le \kappa $). Another application of SchroderBernstein gives $\kappa =\lambda +\kappa $. ∎
Since $\kappa \le \kappa $, we get the following:
Corollary 3.
$\kappa +\kappa =\kappa $ for any infinite cardinal.
Remark. No cardinal greater than $1$ is idempotent with respect to cardinal exponentiation. This is a direct consequence of Cantor’s theorem: $$.
Title  idempotency of infinite cardinals 

Canonical name  IdempotencyOfInfiniteCardinals 
Date of creation  20130322 18:53:30 
Last modified on  20130322 18:53:30 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  8 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 03E10 
Related topic  CanonicalWellOrdering 