PlanetMath: Schroeder-Bernstein theorem, proof of

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proof of Schroeder-Bernstein theorem

(Proof)

We first prove as a lemma that for any $B\subset A$ , if there is an injection $f:A\to B$ , then there is also a bijection $h:A\to B$ .

Inductively define a sequence of subsets of by $C_0=A\setminus B$ and $C_{n+1}=f(C_n)$ . Now let $C=\bigcup_{k=0}^\infty C_k$ , and define $h:A\rightarrow B$ by

$\displaystyle h(z)=\begin{cases} f(z), & z\in C \ z, & z\notin C \end{cases}.$

If $z\in C$ , then $h(z)=f(z)\in B$ . But if $z\notin C$ , then $z\in B$ , and so $h(z)\in B$ . Hence

is well-defined;

is injective by construction. Let $b\in B$ . If $b\notin C$ , then

. Otherwise, $b\in C_k=f(C_{k-1})$ for some

, and so there is some $a\in C_{k-1}$ such that

. Thus

is bijective; in particular, if

, then

is simply the identity map on

To prove the theorem, suppose $f:S\to T$ and $g:T\to S$ are injective. Then the composition $gf:S\to g(T)$ is also injective. By the lemma, there is a bijection $h':S\to g(T)$ . The injectivity of implies that $g^{-1}:g(T)\to T$ exists and is bijective. Define $h:S\to T$ by $h(z)=g^{-1}h'(z)$ ; this map is a bijection, and so and have the same cardinality.