TITLE: $f: \mathbb N \to \mathbb N $ is injective $ \implies $ $\lim\Big(f(n)\Big)=+ \infty $?
QUESTION [6 upvotes]: If $f: \mathbb N \to \mathbb N $ is an injective function , then for the sequence $\Big(f(n)\Big)$ do we have $\lim\Big(f(n)\Big)=+ \infty  $ ?

REPLY [11 votes]: Yes. You want to show that for any $m\in \mathbb N$, there is some $N$ such that 
$$n> N\implies f(n)>m.$$
Since $f$ is injective, the set $f^{-1}(\{0,\ldots,m\})$ has at most $m+1$ points. Let $N$ be the largest of these. Then if $n> N$, $n\notin f^{-1}(\{0,\ldots,m\})$ so $f(n)>m$.

REPLY [3 votes]: Assume the negation: 
$$\exists A>0 ,\quad \forall B>0, \quad n\ge B\; \text{and}\; f(n)<A$$
then pick any $B$ and
$$f\colon \left\{\lfloor B\rfloor+1,\ldots\right\}\rightarrow \left\{1,\ldots,\lfloor A\rfloor\right\}$$
is also injective which's a contradiction due to the cardinality of the two sets.