Page:Cantortransfinite.djvu/109

90 in (a) and (b) the parts played by $$M$$ and $$N$$ are interchanged, two conditions arise which are contradictory to the former ones.

We express the relation of $$\mathfrak{a}$$ to $$\mathfrak{b}$$ characterized by (a) and (b) by saying: $$\mathfrak{a}$$ is "less" than $$\mathfrak{b}$$ or $$\mathfrak{b}$$ is "greater" than $$\mathfrak{a}$$; in signs (1) We can easily prove that, (2) Similarly, from the definition, it follows at once that, if $$P_1$$ is part of an aggregate $$P$$, from $$\mathfrak{a}<\overline\overline{P_1}$$ follows $$\mathfrak{a}<\overline\overline{P}$$ and from $$\overline\overline{P}<\mathfrak{b}$$ follows $$\overline\overline{P_1}<\mathfrak{b}$$.

We have seen that, of the three relations

$\mathfrak{a}=\mathfrak{b}$, $\mathfrak{a}<\mathfrak{b}$, $\mathfrak{b}<\mathfrak{a}$, each one excludes the two others. On the other hand, the theorem that, with any two cardinal numbers $$\mathfrak{a}$$ and $$\mathfrak{b}$$, one of those three relations must necessarily be realized, is by no means self-evident and can hardly be proved at this stage.

Not until later, when we shall have gained a survey over the ascending sequence of the transfinite cardinal numbers and an insight into their connexion, will result the truth of the theorem:

A. If $$\mathfrak{a}$$ and $$\mathfrak{b}$$ are any two cardinal numbers, then either $$\mathfrak{a}=\mathfrak{b}$$ or $$\mathfrak{a}<\mathfrak{b}$$ or $$\mathfrak{a}>\mathfrak{b}$$.

From this theorem the following theorems, of which, however, we will here make no use, can be very simply derived: