Theorem 1   $\kappa\cdot \kappa = \kappa$ for any infinite cardinal $\kappa$ .

Proof. For any non-zero cardinal $\lambda$ , we have $\lambda = 1\cdot \lambda \le \lambda \cdot \lambda$ . So given an infinite cardinal $\kappa$ , either $\kappa = \kappa\cdot\kappa$ or $\kappa < \kappa \cdot \kappa$ . Let be the class of infinite cardinals that fail to be idempotent (with respect to $\cdot$ ). Suppose . We shall derive a contradiction. Since consists entirely of ordinals, it is therefore well-ordered, 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 well-ordering inherited from the canonical ordering on On$\times$ On. Let $\alpha$ be the ordinal isomorphic to $K$ . Since $\kappa < \kappa \cdot \kappa = |K|$ , there is an initial segment $L$ of $K$ that is order isomorphic to $\kappa$ .

Since $L$ is an initial segment of $K$ , $L=\lbrace (\beta_1,\beta_2) \mid (\beta_1,\beta_2) \prec (\alpha_1,\alpha_2) \rbrace$ for some $(\alpha_1,\alpha_2)\in K$ . The well-order $\preceq$ denotes the canonical ordering on $K$ . Let $\lambda = \max(\alpha_1,\alpha_2)$ . Since $L \subset K = \kappa\times \kappa$ , $\alpha_1<\kappa$ and $\alpha_2<\kappa$ , and therefore $\lambda <\kappa$ .

For any $(\beta_1,\beta_2)\in L$ , we have $(\beta_1,\beta_2) \prec (\alpha_1,\alpha_2)$ , which implies that $\max(\beta_1,\beta_2) \le \lambda$ . 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 $\lambda <\kappa$ , and $\kappa$ is a limit ordinal, $|\lambda^+|<k$ as well, which means , or $|\lambda^+|\cdot |\lambda^+| = |\lambda^+|$ . Therefore $|L|\le |\lambda^+|\cdot |\lambda^+| = |\lambda^+|\le \lambda^+ < \kappa$ , again contradicting that $L$ is (order) isomorphic to $\kappa$ .

Therefore, the assumption is false, and the proof is complete.

Corollary 1   If $0< \lambda \le \kappa$ and $\kappa$ is infinite, then $\lambda \cdot \kappa = \kappa$ .

Proof. $\kappa = 1 \cdot \kappa \le \lambda \cdot \kappa \le \kappa \cdot \kappa = \kappa$ . By Schroder-Bernstein's Theorem, $\lambda \cdot \kappa = \kappa$ .

Corollary 2   If $\lambda \le \kappa$ and $\kappa$ is infinite, then $\lambda + \kappa = \kappa$ .

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 Schroder-Bernstein gives $\kappa = \lambda+\kappa$ .

Corollary 3   $\kappa + \kappa = \kappa$ for any infinite cardinal.