idempotency of infinite cardinalsIdempotencyOfInfiniteCardinals2009-04-14 03:09:512009-04-15 02:51:20Definitioncardinal arithmetic\usepackage{amssymb,amscd}
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\newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}}In this entry, we show that every infinite cardinal is idempotent with respect to cardinal addition and cardinal multiplication.
\begin{thm} $\kappa\cdot \kappa = \kappa$ for any infinite cardinal $\kappa$. \end{thm}
\begin{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 $\mathscr{C}$ be the class of infinite cardinals that fail to be idempotent (with respect to $\cdot$). Suppose $\mathscr{C}\ne \varnothing$. We shall derive a contradiction. Since $\mathscr{C}$ 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 \textbf{On}$\times$\textbf{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:
\begin{enumerate}
\item If $\lambda$ is finite, so is $\lambda^+ \times \lambda^+$, contradicting that $L$ is (order) isomorphic to $\kappa$, an infinite set.
\item If $\lambda$ is infinite, so is $|\lambda^+|$. Since $\lambda <\kappa$, and $\kappa$ is a limit ordinal, $|\lambda^+|<k$ as well, which means $|\lambda^+|\notin \mathscr{C}$, or $|\lambda^+|\cdot |\lambda^+| = |\lambda^+|$. Therefore $|L|\le |\lambda^+|\cdot |\lambda^+| = |\lambda^+|\le \lambda^+ < \kappa$, again contradicting that $L$ is (order) isomorphic to $\kappa$.
\end{enumerate}
Therefore, the assumption $\mathscr{C} \ne \varnothing$ is false, and the proof is complete.
\end{proof}
\begin{cor} If $0< \lambda \le \kappa$ and $\kappa$ is infinite, then $\lambda \cdot \kappa = \kappa$. \end{cor}
\begin{proof} $\kappa = 1 \cdot \kappa \le \lambda \cdot \kappa \le \kappa \cdot \kappa = \kappa$. By Schroder-Bernstein's Theorem, $\lambda \cdot \kappa = \kappa$.
\end{proof}
\begin{cor} If $\lambda \le \kappa$ and $\kappa$ is infinite, then $\lambda + \kappa = \kappa$. \end{cor}
\begin{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 $\kappa = \lambda+\kappa$.
\end{proof}
Since $\kappa\le \kappa$, we get the following:
\begin{cor} $\kappa + \kappa = \kappa$ for any infinite cardinal. \end{cor}
\textbf{Remark}. No cardinal greater than $1$ is idempotent with respect to cardinal exponentiation. This is a direct consequence of Cantor's theorem: $\kappa < 2^ \kappa \le \kappa ^ \kappa$.