Theorem 1

The relation $R\{\xi,\eta\}$: “$\xi$ and $\eta$ are cardinals and $\xi$ is equipotent to a subset of $\eta$” is a well-ordering relation (§ 2, no. 1).

Since $R\{\xi,\xi\}$ is true for every cardinal $\xi$, what must be proved is that, for every set $E$ of cardinals the relation “$\xi\in E$ and $\eta\in E$ and $R\{\xi,\eta\}$” is a well-ordering relation on $E$. Consider the set $A=\bigcup_{\xi\in E}\xi$.

Every cardinal $\xi\in E$ is then a subset of $A$. By § 2, Theorem 1 there exists a well-ordering relation on $A$, which we shall denote by $\xi\leq\eta$, and every subset of $A$ is equipotent to a segment of $A$ (§ 2, no. 5, Theorem 3, Corollary x). For every cardinal $\xi\in E$ consider the set of segments of $A$ which are equipotent to $\xi$; this set of segments is not empty and therefore has a least element (§ 2, no. 1, Proposition 2); let $\varphi(\xi)$ denote this least element. The relation

“$\xi\in E$ and $\eta\in E$ and $\xi$ is equipotent to a subset of $\eta$”

is then equivalent to

“$\xi\in E$ and $\eta\in E$ and $\varphi(\xi)\subset\varphi(\eta)$”.

For clearly the second relation implies the first. Conversely, if $\xi$ is equipotent to a subset of $\varphi(\eta)$, we cannot have $\varphi(\eta)\subset\varphi(\xi)$ and $\varphi(\eta)\neq\varphi(\xi)$; otherwise there would exist a segment of $\varphi(\eta)$ equipotent to $\xi$ (§ 2, no. 5, Theorem 3, Corollary 3), contrary to the definition of $\varphi(\xi)$. Since the set of segments of $A$ is well-ordered by inclusion (§ 2, no. 1, Proposition 2), the theorem follows.

¶ We shall denote the relation $R\{\xi,\eta\}$ by $\xi\leq\eta$. A set $X$ is equipotent to a subset of a set $Y$ if and only if $\operatorname{Card}(X)\leq\operatorname{Card}(Y)$.

Clearly we have $0\leq\xi$ for every cardinal $\xi$, and $1\leq\xi$ for every cardinal $\xi\neq0$.