Page:Cantortransfinite.djvu/125

Hence follows that $S_{1}\sim S$, and therefore ${\overline {\overline {S_{1}}}}=\aleph _{0}$.

From A and B the formula (4) results, if we have regard to §2.

From (2) we conclude, by adding $1$ to both sides,

$\aleph _{0}+2=\aleph _{0}+1=\aleph _{0}$,

and, by repeating this

We have also

[494] For, by (1) of §3, $\aleph _{0}+\aleph _{0}$ is the cardinal number ${\overline {\overline {(\{a_{\nu }\},\{b_{\nu }\})}}}$ because

${\overline {\overline {\{a_{\nu }\}}}}={\overline {\overline {\{b_{\nu }\}}}}=\aleph _{0}$.

Now, obviously

$\{\nu \}=(\{2\nu -1\},\{2\nu \})$,

$(\{2\nu -1\},\{2\nu \})\sim (\{a_{\nu }\},\{b_{\nu }\})$,

and therefore

${\overline {\overline {(\{a_{\nu }\},\{b_{\nu }\})}}}={\overline {\overline {\{\nu \}}}}=\aleph _{0}$.

The equation (6) can also be written

$\aleph _{0}\cdot 2=\aleph _{0}$;

and, by adding $\aleph _{0}$ repeatedly to both sides, we find that

We also have