Originally Posted by mathcnc
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
$S$ is finite with $n$ elements if and only if there is a bijection $f$ between $S$ and $\{1,2,..,n\}$

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

Define:

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

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

CB