Page:Cantortransfinite.djvu/125

106 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 (5) We have also (6) [494] For, by (1) of §3, $$\aleph_0+\aleph_0$$ is the cardinal number $$\overline\overline{(\{a_\nu\}, \{b_\nu\})}$$ because $\overline\overline{\{a_\nu\}} = \overline\overline{\{b_\nu\}} = \aleph_0$.|undefined Now, obviously $\{\nu\}=(\{2\nu-1\}, \{2\nu\})$, $(\{2\nu-1\}, \{2\nu\})\sim (\{a_\nu\}, \{b_\nu\})$, and therefore $\overline\overline{(\{a_\nu\}, \{b_\nu\})} = \overline\overline{\{\nu\}} = \aleph_0$.|undefined

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 (7) We also have (8)