if $A$ is infinite and $B$ is a finite subset of $A+.1667em-.1667em,$ 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

$h(i)$ $=$

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 $A+.1667em-.1667em,$ then $A\setminus B$ is infinite
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