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
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:

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

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