Quote Originally Posted by mathcnc View Post
ok im confused on how to start with this
we had previously proved that A is a denumerable set and x is an element not in A, then A union {x} is also deumerable.
Prove that if S = {x1, x2, ...,xn} is a set of n elements, none of which are in A, then A union S is denumerable
$\displaystyle S$ is finite with $\displaystyle n$ elements if and only if there is a bijection $\displaystyle f$ between $\displaystyle S$ and $\displaystyle \{1,2,..,n\}$

A set $\displaystyle A$ is denumerable (countably infinite) if and only if there exists a bijection $\displaystyle g$ between $\displaystyle A$ and $\displaystyle \mathbb{N}$ (for convienience I will take $\displaystyle \mathbb{N}$ to denote the positive integers rather than the alternative non-negative integers).


h(x) = \begin{cases} f(x), & x \in S\\ g(x)+n, & x \in A \end{cases}

Because there are no common elements in $\displaystyle A$ and $\displaystyle S$ this is well defined and $\displaystyle h$ is a bijection between $\displaystyle A\cup S$ and $\displaystyle \mathbb{N}$