Page:Cantortransfinite.djvu/126

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

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

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

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

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

(9)

${\displaystyle \lambda =\mu +{\frac {(\mu +\nu -1)(\mu +\nu -2)}{2}}}$.

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