# A convergent sequence of integers is eventually constant

Consider a sequence of integers $\{ a_i \}$, $i \in \mathbb{N}$. $\mathbb{N}$ is a metric space via the distance

\begin{equation} \forall a,b \in \mathbb{N}, d(a,b) = |b - a| \end{equation}and similarly it simply has the norm $|a|$.

As every functions we have on the integers of relevance here simply map themselves to integers, we could consider a definition of convergence based on the reals or the integers without any meaningful difference. Consider the definition of convergence using a real norm. Given a sequence $\{a_i\}$, we say that it converges to $a$ if, for every real number $\varepsilon > 0$, there exists an integer $N$ such that, for every $n > N$,

\begin{equation} |a_n - a| < \varepsilon \end{equation}If we're dealing with integers, this must be true for $\varepsilon < 1$, but the absolute value of the substraction of two integers must be an integer, and only $0$ is an integer smaller than $1$, therefore, for any $n > N$ for such an $N$, we have

\begin{equation} |a_n - a| = 0 \Leftrightarrow a_N = a \end{equation}Therefore, any integer of the sequence after the index $N$ must be equal to the limit $a$.