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 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 . Let $𝒞$ be the class of infinite cardinals that fail to be idempotent (with respect to $\cdot$). Suppose $𝒞\ne \mathrm{\varnothing }$. We shall derive a contradiction. Since $𝒞$ 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$\mathrm{\times }$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 well-order $⪯$ denotes the canonical ordering on $K$. Let $\lambda =\max ({\alpha }_1,{\alpha }_2)$. 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 $\max ({\beta }_1,{\beta }_2)\le \lambda$. 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 $|{\lambda }^{+}|\notin 𝒞$, or $|{\lambda }^{+}|\cdot |{\lambda }^{+}|=|{\lambda }^{+}|$. Therefore , again contradicting that $L$ is (order) isomorphic to $\kappa$.

Therefore, the assumption $𝒞\ne \mathrm{\varnothing }$ 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 Schroder-Bernstein’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 Schroder-Bernstein 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: .