ProofOfPigeonholePrinciple 2003-03-14 00:25:37 2007-05-30 02:02:18 Proof pigeonhole principle 0 \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{psfrag} \usepackage{graphicx} \usepackage{amsthm} \usepackage{xypic} \begin{proof} It will first be proven that, if a bijection exists between two finite sets, then the two sets have the same number of elements. Let $S$ and $T$ be finite sets and $f \colon S \to T$ be a bijection. The claim will be proven by induction on $|S|$. If $|S|=0$, then $S=\emptyset$, and $f \colon \emptyset \to T$ can only be surjective if $T=\emptyset$. Assume the statement holds for any set $S$ with $|S|=n$. Let $|S|=n+1$. Let $x_1, \dots , x_{n+1} \in S$ with $S=\{x_1, \dots , x_{n+1}\}$. Let $R=S \setminus \{x_{n+1}\}$. Then $|R|=n$. Define $g \colon R \to T \setminus \{f(x_{n+1})\}$ by $g(x)=f(x)$. Since $R \subset S$, $f(x) \in T$ for all $x \in R$. Thus, to show that $g$ is well-defined, it only needs to be verified that $f(x) \neq f(x_{n+1})$ for all $x \in R$. This follows immediately from the facts that $x_{n+1} \notin R$ and $f$ is injective. Therefore, $g$ is well-defined. Now it need to be proven that $g$ is a bijection. The fact that $g$ is injective follows immediately from the fact that $f$ is injective. To verify that $g$ is surjective, let $y \in T \setminus \{f(x_{n+1})\}$. Since $f$ is surjective, there exists $x \in S$ with $f(x)=y$. Since $f(x)=y \neq f(x_{n+1})$ and $f$ is injective, $x \neq x_{n+1}$. Thus, $x \in R$. Hence, $g(x)=f(x)=y$. It follows that $g$ is a bijection. By the induction hypothesis, $|R|=|T \setminus \{f(x_{n+1})\}|$. Thus, $n=|R|=|T \setminus \{f(x_{n+1})\}|=|T|-1$. Therefore, $|T|=n+1=|S|$. \end{proof}