# Convergent Sequence in Set of Integers

## Theorem

Let $\left \langle {x_n}\right \rangle_{n \in \N}$ be a sequence in the set $\Z$ of integers considered as a subspace of the real number line $\R$ under the Euclidean metric.

Then $\left \langle {x_n}\right \rangle_{n \in \N}$ converges in $\R$ to a limit if and only if:

$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$

That is, if and only if the sequence reaches some value of $\Z$ and "stays there".

### Corollary

Let $\left \langle {x_n}\right \rangle_{n \in \N}$ be a sequence of distinct terms in the set $\Z$.

Then $\left \langle {x_n}\right \rangle_{n \in \N}$ is not convergent.

## Proof

Suppose $\left \langle {x_n}\right \rangle_{n \in \N}$ converges to a limit $l$.

Consider the open set in $\R$:

$U := \left({l - \dfrac 1 2 \,.\,.\, l + \dfrac 1 2}\right)$

Then $\forall x \in \Z: x \in U \implies x = l$

It follows by definition of convergence that:

$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$

where $x_k = l$.

Now suppose that:

$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$

Then trivially $\left \langle {x_n}\right \rangle_{n \in \N}$ converges to the limit $x_k$.

$\blacksquare$