Convergent Sequence in Set of Integers
Contents |
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 iff:
- $\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
That is, iff 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 $x_k$.
$\blacksquare$
Proof of Corollary
Trivially:
- $\neg \exists k \in \N: \forall m \in \N: m > k: x'_m = x'_k$
$\blacksquare$