Theorem. If $A$ is an infinite set and $B$ is a finite subset of $A$ , then $A\setminus B$ is infinite.

Proof. The proof is by contradiction. If $A\setminus B$ would be finite, there would exist a $k\in \sN$ and a bijection $f:\{1,\ldots, k\}\to A\setminus B$ . Since $B$ is finite, there also exists a bijection $g:\{1,\ldots, l\}\to B$ . We can then define a mapping $h:\{1,\ldots, k+l\} \to A$ by \begin{eqnarray*} h(i)&=& \left\{ \begin {array}{ll} f(i) & \mbox{when} \, i\in \{1,\ldots, k\}, \\ g(i-k) & \mbox{when}\, i\in\{k+1,\ldots, k+l\}. \\ \end{array}. Since $f$ and $g$ are bijections, $h$ is a bijection between a finite subset of $\sN$ and $A$ . This is a contradiction since $A$ is infinite.