Page:Cantortransfinite.djvu/108

Rh. 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