Page:Cantortransfinite.djvu/121

102 $$E_{\nu-1}$$ are not equivalent, and accordingly their cardinal numbers $$\mu = \overline\overline{E}_{\mu-1}$$ and $$\nu = \overline\overline{E}_{\nu-1}$$ are not equal.

Proof of B.—If of the two finite cardinal numbers $$\mu$$ and $$\nu$$ the first is the earlier and the second the later, then $$\mu < \nu$$. For consider the two aggregates $$M = E_{\mu-1}$$ and $$N=E_{\nu-1}$$; for them each of the two conditions in § 2 for $$\overline\overline{M} < \overline\overline{N}$$ is fulfilled. The condition (a) is fulfilled because, by theorem E, a part of $$M = E_{\mu-1}$$ can only have one of the cardinal numbers $$1, 2, 3, ..., \mu-1$$, and therefore, by theorem A, cannot be equivalent to the aggregate $$N = E_{\mu-1}$$. The condition (b) is fulfilled because $$M$$ itself is a part of $$N$$.

Proof of C.— Let $$\mathfrak{a}$$ be a cardinal number which is less than $$\nu+1$$. Because of the condition (b) of §2, there is a part of $$E_\nu$$ with the cardinal number $$\mathfrak{a}$$. By theorem E, a part of $$E_\nu$$ can only have one of the cardinal numbers $$1, 2, 3, ..., \nu$$. Thus $$\mathfrak{a}$$ is equal to one of the cardinal numbers $$1, 2, 3, ..., \nu$$. By theorem B, none of these is greater than $$\nu$$. Consequently there is no cardinal number $$\mathfrak{a}$$ which is less than $$\nu+1$$ and greater than $$\nu$$. Of importance for what follows is the following theorem:

F. If $$K$$ is any aggregate of different finite cardinal numbers, there is one, $$\kappa_1$$, amongst them which is smaller than the rest, and therefore the smallest of all.

[492] Proof—The aggregate $$K$$ either contains