Show an arithmetic progression has infinitely many elements

Question

How would I Show this arithmetic progression has infinitely many elements.

$$A_{a,b}=\left\lbrace a+nb:n\in\mathbb{Z}\right\rbrace$$

Answer 1

Okay, first assume $b\neq 0$, because if $b=0$, then $A_{a,b}=\{a\}$ which is clearly finite, contradicting the statement. Then define a function $f:\mathbb{Z}\to A_{a,b}$ by $f(n)=a+bn$. By construction, this is surjective, meaning that each $x\in A_{a,b}$ has a corresponding $n\in\mathbb{Z}$ so that $f(n)=x$. Now, suppose $f(n)=f(m)$. Then $a+bn=a+bm\Rightarrow bn=bm\Rightarrow n=m$ since $b\neq 0$. This shows that $f$ is injective, meaning that no there are no two integer $n\neq m$ such that $f(n)= f(m)$, or that whenever $n\neq m$, $f(n)\neq f(m)$.

Now, by definition, $f$ is bijective (injective + surjective). This means that $A_{a,b}$ is infinite (the definition of a set $S$ being [countably] infinite is that there exists a bijection $\mathbb{Z}\to S$).

Intuitively, all this says is that $A_{a,b}$ has the same "size" as the integers, because each element of $A_{a,b}$ is uniquely determined by an integer $n$. Since the integers are infinite, this set must be infinite too.