Page:Cantortransfinite.djvu/124

Rh This rests on the following theorems:

A. Every transfinite aggregate $$T$$ has parts with the cardinal number $$\aleph_0$$.

Proof.—If, by any rule, we have taken away a finite number of elements $$t_1, t_2, t_3, ..., t_{\nu-1}$$, there always remains the possibility of taking away a further element $$t_{\nu}$$. The aggregate $$\{t_\nu\}$$, where $$\nu$$ denotes any finite cardinal number, is a part of $$T$$ with the cardinal number $$\aleph_0$$, because $$\{t_\nu\}\sim \{\nu\}$$ (§1).

B. If $$S$$ is a transfinite aggregate with the cardinal number $$\aleph_0$$, and $$S_1$$ is any transfinite part of $$S$$, then $$\overline\overline{S_1}=\aleph_0$$.

Proof.—We have supposed that $$S\sim \{\nu\}$$. Choose a definite law of correspondence between these two aggregates, and, with this law, denote by $$s_\nu$$ that element of $$S$$ which corresponds to the element $$\nu$$ of $$\{\nu\}$$, so that $S = \{s_\nu\}$ The part $$S_1$$ of $$S$$ consists of certain elements $$s_\kappa$$ of $$S$$, and the totality of numbers $$\kappa$$ forms a transfinite part $$K$$ of the aggregate $$\{\nu\}$$. By theorem G of §5 the aggregate $$K$$ can be brought into the form of a series $K = \{\kappa_\nu\}$ where $\kappa_\nu < \kappa_{\nu+1}$; consequently we have $S_1 = \{s_{\kappa_\nu}\}$.