# if $A$ is infinite and $B$ is a finite subset of $A\tmspace+{.1667em}\tmspace-{.1667em},$ then $A\setminus B$ is infinite

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 $k\in\mathbb{N}$ 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

 $\displaystyle h(i)$ $\displaystyle=$ $\displaystyle\left\{\begin{array}[]{ll}f(i)&\mbox{when}\,i\in\{1,\ldots,k\},\\ g(i-k)&\mbox{when}\,i\in\{k+1,\ldots,k+l\}.\\ \end{array}\right.$

Since $f$ and $g$ are bijections, $h$ is a bijection between a finite subset of $\mathbb{N}$ and $A$. This is a contradiction since $A$ is infinite. $\Box$

Title if $A$ is infinite and $B$ is a finite subset of $A\tmspace+{.1667em}\tmspace-{.1667em},$ then $A\setminus B$ is infinite IfAIsInfiniteAndBIsAFiniteSubsetOfAThenAsetminusBIsInfinite 2013-03-22 13:34:42 2013-03-22 13:34:42 mathcam (2727) mathcam (2727) 7 mathcam (2727) Theorem msc 03E10