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

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 $k\in \mathbb{N}$ and a bijection
$f:\{1,\mathrm{\dots},k\}\to A\setminus B$. Since $B$ is finite, there
also exists a bijection $g:\{1,\mathrm{\dots},l\}\to B$. We can then define
a mapping $h:\{1,\mathrm{\dots},k+l\}\to A$ by

$h(i)$ | $=$ | $\{\begin{array}{cc}f(i)\hfill & \text{when}i\in \{1,\mathrm{\dots},k\},\hfill \\ g(i-k)\hfill & \text{when}i\in \{k+1,\mathrm{\dots},k+l\}.\hfill \end{array}$ |

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. $\mathrm{\square}$

Title | if $A$ is infinite and $B$ is a finite subset of $A\text{tmspace}+.1667em\text{tmspace}-.1667em,$ then $A\setminus B$ is infinite |
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Canonical name | IfAIsInfiniteAndBIsAFiniteSubsetOfAThenAsetminusBIsInfinite |

Date of creation | 2013-03-22 13:34:42 |

Last modified on | 2013-03-22 13:34:42 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 7 |

Author | mathcam (2727) |

Entry type | Theorem |

Classification | msc 03E10 |