Page:Cantortransfinite.djvu/120

with the cardinal number $\nu +1$. If the theorem is true for this aggregate, its truth for any other aggregate with the same cardinal number $\nu +1$ follows at once by § 1. Let $E'$ be any part of $E_{\nu }$; we distinguish the following cases:

(a) $E'$ does not contain $e_{\nu }$ as element, then $E$ is either $E_{\nu -1}$ [491] or a part of $E_{\nu -1}$, and so has as cardinal number either $\nu$ or one of the numbers $1,2,3,...,\nu -1$, because we supposed our theorem true for the aggregate $E_{\nu -1}$, with the cardinal number $\nu$,

(b) $E'$ consists of the single element $e_{\nu }$, then ${\overline {\overline {E'}}}=1$.

(c) $E'$ consists of $e_{\nu }$ and an aggregate $E''$, so that $E'=(E'',e_{\nu })$. $E''$ is a part of $E_{\nu -1}$ and has therefore by supposition as cardinal number one of the numbers $1,2,3,...,\nu -1$. But now ${\overline {\overline {E'}}}={\overline {\overline {E''}}}+1$, and thus the cardinal number of $E'$ is one of the numbers $2,3,...,\nu$.

Proof of A.—Every one of the aggregates which we have denoted by $E_{\nu }$ has the property of not being equivalent to any of its parts. For if we suppose that this is so as far as a certain $\nu$, it follows from the theorem D that it is so for the immediately following number $\nu +1$. For $\nu =1$, we recognize at once that the aggregate $E_{1}=(e_{0},e_{1})$ is not equivalent to any of its parts, which are here $(e_{0})$ and $(e_{1})$. Consider, now, any two numbers $\mu$ and $\nu$ of the series $1,2,3,...$; then, if $\mu$ is the earlier and $\nu$ the later, $E_{\mu -1}$ is a part of $E_{\nu -1}$. Thus $E_{\mu -1}$ and