Page:Cantortransfinite.djvu/126

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