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Homeidempotency of infinite cardinals

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# 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 $\lambda=1\cdot\lambda\leq\lambda\cdot\lambda$. 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 $\mathscr{C}\neq\varnothing$. 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 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 $K$, $L=\{(\beta_{1},\beta_{2})\mid(\beta_{1},\beta_{2})\prec(\alpha_{1},\alpha_{2})\}$ for some $(\alpha_{1},\alpha_{2})\in K$. The well-order $\preceq$ 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 $\max(\beta_{1},\beta_{2})\leq\lambda$. Therefore $L\subseteq\lambda^{+}\times\lambda^{+}$, or $|L|\leq|\lambda^{+}\times\lambda^{+}|\leq|\lambda^{+}|\cdot|\lambda^{+}|$. 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 $|\lambda^{+}|\notin\mathscr{C}$, or $|\lambda^{+}|\cdot|\lambda^{+}|=|\lambda^{+}|$. Therefore $|L|\leq|\lambda^{+}|\cdot|\lambda^{+}|=|\lambda^{+}|\leq\lambda^{+}<\kappa$, again contradicting that $L$ is (order) isomorphic to $\kappa$.

Therefore, the assumption $\mathscr{C}\neq\varnothing$ is false, and the proof is complete. ∎

###### Corollary 1.

If $0<\lambda\leq\kappa$ and $\kappa$ is infinite, then $\lambda\cdot\kappa=\kappa$.

###### Proof.

$\kappa=1\cdot\kappa\leq\lambda\cdot\kappa\leq\kappa\cdot\kappa=\kappa$. By Schroder-Bernstein’s Theorem, $\lambda\cdot\kappa=\kappa$. ∎

###### Corollary 2.

If $\lambda\leq\kappa$ and $\kappa$ is infinite, then $\lambda+\kappa=\kappa$.

###### Proof.

$\kappa=0+\kappa\leq\lambda+\kappa\leq\kappa+\kappa=2\cdot\kappa\leq\kappa\cdot% \kappa=\kappa$ by the corollary above (since $2\leq\kappa$). Another application of Schroder-Bernstein gives $\kappa=\lambda+\kappa$. ∎

Since $\kappa\leq\kappa$, 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: $\kappa<2^{\kappa}\leq\kappa^{\kappa}$.

## Mathematics Subject Classification

03E10*no label found*

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