Page:Cantortransfinite.djvu/126

Proof.—By (6) of §3, $\aleph _{0}\cdot \aleph _{0}$ is the cardinal number of the aggregate of bindings

$\{(\mu ,\nu )\}$,

where $\mu$ and $\nu$ are any finite cardinal numbers which are independent of one another. If also $\lambda$ represents any finite cardinal number, so that $\{\lambda \}$, $\{\mu \}$, and $\{\nu \}$ are only different notations for the same aggregate of all finite numbers, we have to show that

$\{(\mu ,\nu )\}\sim \{\lambda \}$.

Let us denote $\mu +\nu$ by $\rho$; then $\rho$ takes all the numerical values $2,3,4,...$, and there are in all $\rho -1$ elements $(\mu ,\nu )$ for which $\mu +\nu =\rho$, namely:

$(1,\rho -1),(2,\rho -2),...,(\rho -1,1)$.

In this sequence imagine first the element $(1,1)$, for which $\rho =2$, put, then the two elements for which $\rho =3$, then the three elements for which $\rho =4$, and so on. Thus we get all the elements $(\mu ,\nu )$ in a simple series:

$(1,1);(1,2),(2,1);(1,3),(2,2),(3,1);(1,4),(2,3),...$,

and here, as we easily see, the element $(\mu ,\nu )$ comes at the $\lambda$th place, where

The variable $\lambda$ takes every numerical value $1,2,3,...$, once. Consequently, by means of (9), a