# proof of pigeonhole principle

###### 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, $|T|=n+1=|S|$. ∎

Title proof of pigeonhole principle ProofOfPigeonholePrinciple 2013-03-22 13:31:09 2013-03-22 13:31:09 Wkbj79 (1863) Wkbj79 (1863) 8 Wkbj79 (1863) Proof msc 03E05