Page:Cantortransfinite.djvu/108

ence. If $M$, $N$, $P$, $...$ are aggregates which have no common elements, $M'$, $N'$, $P'$, $...$ are also aggregates with the same property, and if

$M\sim M'$, $N\sim N'$, $P\sim P'$, $...$,

then we always have

$(M,N,P,...)\sim (M',N',P',...)$.

§2
"Greater" and "Less" with Powers

If for two aggregates $M$ and $N$ with the cardinal numbers ${\mathfrak {a}}={\overline {\overline {M}}}$ and ${\mathfrak {b}}={\overline {\overline {N}}}$, both the conditions:

(a) There is no part of $M$ which is equivalent to $N$,
(b) There is a part $N_{1}$ of $N$, such that $N_{1}\sim M$,

are fulfilled, it is obvious that these conditions still hold if in them $M$ and $N$ are replaced by two equivalent aggregates $M'$ and $N'$. Thus they express a definite relation of the cardinal numbers ${\mathfrak {a}}$ and ${\mathfrak {b}}$ to one another.

[484] Further, the equivalence of $M$ and $N$, and thus the equality of ${\mathfrak {a}}$ and ${\mathfrak {b}}$, is excluded; for if we had $M\sim N$, we would have, because $N_{1}\sim M$, the equivalence $N_{1}\sim N$, and then, because $M\sim N$, there would exist a part $M_{1}$ of $M$ such that $M_{1}\sim M,$ and therefore we should have $M_{1}\sim N$; and this contradicts the condition (a).

Thirdly, the relation of ${\mathfrak {a}}$ and ${\mathfrak {b}}$ is such that it makes impossible the same relation of ${\mathfrak {b}}$ and ${\mathfrak {a}}$; for if