# Schroeder-Bernstein theorem, proof of

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 $(C_{n})$ of subsets of $A$ 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

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

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 $h$ is well-defined; $h$ is injective by construction. Let $b\in B$. If $b\notin C$, then $h(b)=b$. Otherwise, $b\in C_{k}=f(C_{k-1})$ for some $k>0$, and so there is some $a\in C_{k-1}$ such that $h(a)=f(a)=b$. Thus $h$ is bijective; in particular, if $B=A$, then $h$ is simply the identity map on $A$.

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 $h^{\prime}:S\to g(T)$. The injectivity of $g$ implies that $g^{-1}:g(T)\to T$ exists and is bijective. Define $h:S\to T$ by $h(z)=g^{-1}h^{\prime}(z)$; this map is a bijection, and so $S$ and $T$ have the same cardinality.

Title Schroeder-Bernstein theorem, proof of SchroederBernsteinTheoremProofOf 2013-03-22 12:49:56 2013-03-22 12:49:56 mps (409) mps (409) 19 mps (409) Proof msc 03E10 proof of Cantor-Bernstein theorem proof of Cantor-Schroeder-Bernstein theorem AnInjectionBetweenTwoFiniteSetsOfTheSameCardinalityIsBijective