proof the union of an infinetly countable set and an element is aslo countable If a set $A$ is a countably infinite and b $\notin$ $A$, then $A$ $\cup$ {b} is also countable.
I think that by assuming that the union is countable then $A$ is a subset of a countable set. thus it's true. however I feel this is wrong. Is there a better way to prove it. or how do I do my proof step by step.
 A: One way to show that $A\cup \{b\}$ is countably infinite is by defining a bijection $f:\Bbb{N}\to A\cup \{b\}$.
Now, since $A$ is countably infinite, there exist $g:\Bbb{N}\to A$ which is bijective. Let's call $g(1)=a_1, g(2)=a_2,...,g(n)=a_{n},..$ So lets define $f$ in the following way: 
$$f(1)=b, f(2)=a_1,...,f(n)=g(n+1)=a_{n+1},...$$ and let's see that $f$ is in fact bijective. 
Surjective:
We already know that $f(1)=b$, so let's take $a\in A$. Since $g$ is bijective, there exist $n\in \Bbb{N}$ such that $f(n)=a_n=a$. But by definition, $g(n+1)=f(n)=a$ and of course $n+1 \in \Bbb{N}$.
Injective:
Take $n,m \in \Bbb{N}$ such that $m \neq n$ then if both $m,n \neq 1$ then:
$g(n)=f(n-1)\neq f(m-1)=g(m)$ since $f$ is bijective. 
Now if $m=1$ and $n \neq 1$ then $f(1)=b \neq f(n)=a_{n-1}$ for every $n\geq 2$ since $b \notin A$. (The case $n=1$ and $m \neq 1$ is of course analogous).
Therefore $g$ is bijective and therefore $A \cup \{b\}$ is countably infinite
A: You made a logical mistake. Assuming something true and getting a true result just tell you that your assumption is consistent in that way. Actually you need to check every connection to other truths to precisely say that your assumption is consistent. 
 With consistency you can define new things onto others. If you don't define it, then everything is still fine.
 However proving means showing that it must be true because of other known truths. One of the famous way of doing this is assuming something true and getting a false result, implies that your assumption is inconsistent, that is it can't be true with other known truths and it must be false.
