Page:Cantortransfinite.djvu/119

100 number of $$N_1$$ is equal to one of the preceding numbers $$1, 2, 3,. . ., \nu-1$$.

Proof of D. Suppose that the aggregate $$(M, e)$$ is equivalent to one of its parts which we will call $$N$$. Then two cases, both of which lead to a contradiction, are to be distinguished:

(a) The aggregate $$N$$ contains $$e$$ as element; let $$N = (M_1, e)$$; then $$M_1$$ is a part of $$M$$ because $$N$$ is a part of $$(M, e)$$. As we saw in § 1, the law of correspondence of the two equivalent aggregates $$(M, e)$$ and $$(M_1, e)$$ can be so modified that the element $$e$$ of the one corresponds to the same element $$e$$ of the other; by that, then, $$M$$ and $$M_1$$ are referred reciprocally and univocally to one another. But this contradicts the supposition that $$M$$ is not equivalent to its part $$M_1$$.

(b) The part $$N$$ of $$(M, e)$$ does not contain $$e$$ as element, so that $$N$$ is either $$M$$ or a part of $$M$$. In the law of correspondence between $$(M, e)$$ and $$N$$, which lies at the basis of our supposition, to the element $$e$$ of the former let the element $$f$$ of the latter correspond. Let $$N = (M_1, f)$$; then the aggregate $$M$$ is put in a reciprocally univocal relation with $$M_1$$. But $$M_1$$ is a part of $$N$$ and hence of $$M$$. So here too $$M$$ would be equivalent to one of its parts, and this is contrary to the supposition.

Proof of E.—We will suppose the correctness of the theorem up to a certain $$\nu$$ and then conclude its validity for the number $$\nu+1$$ which immediately follows, in the following manner:—We start from the aggregate $$E_\nu = (e_0, e_1, ..., e_\nu)$$ as an aggregate