Proposition 8

If $\mathfrak{a}$ and $\mathfrak{b}$ are cardinals such that $\mathfrak{a} + 1 = \mathfrak{b} + 1$, then $\mathfrak{a} = \mathfrak{b}$.

Let $X = \mathfrak{a} + 1 = \mathfrak{b} + 1$. Then there exist subsets A, B of X with cardinals $\mathfrak{a}$, $\mathfrak{b}$, respectively, such that the complements $X - A, X - B$ each consist of a single element. Let $u$, $v$ be these elements. The intersection $C = A \cap B$ has as complement in X the set $\{u, v\}$. If $u = v$, then $A = B = C$, so that $\mathfrak{a} = \mathfrak{b}$. If $u \neq v$, then $A = C \cup \{v\}$, $B = C \cup \{u\}$, and therefore $\mathfrak{a} = 1 + \mathrm{Card}\ (C) = \mathfrak{b}$.

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The reader should beware of assuming that $\mathfrak{a} + \mathfrak{m} = \mathfrak{b} + \mathfrak{m}$ implies $\mathfrak{a} = \mathfrak{b}$ for all cardinals $\mathfrak{m}$ (cf. § 6); \* we shall see later, however, that this implication is true if $\mathfrak{m}$ is finite (§ 5, no. 2, Proposition 3, Corollary 4 and § 6, no. 3, Theorem 3, Corollary 4). \*