Convergence of Sequence in Discrete Space

Theorem

Let $T = \left({S, \tau}\right)$ be a discrete topological space.

Let $H = \left \langle{x_n}\right \rangle_{n \in \N}$ be a sequence in $S$.

Then $H$ converges in $T$ 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 $S$ and "stays there".

Corollary

Let $\left \langle {x'_n}\right \rangle_{n \in \N}$ be a sequence of distinct terms in $S$.

Then $\left \langle {x'_n}\right \rangle_{n \in \N}$ is not convergent in $T$.

Proof

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

As $T$ is a discrete space, $\left\{{L}\right\}$ is an open set in $T$.

Then trivially:

$\forall x \in H: x \in \left\{{L}\right\} \implies x = L$

So for $H$ to be convergent, it is necessary that:

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

It follows by definition of convergence that:

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

where $l = L$.

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 $L$.

$\blacksquare$

Proof of Corollary

By the definition of a sequence of distinct terms:

$\forall x \in \left \langle {x'_n}\right \rangle_{n \in \N}: r \ne s \implies x_r \ne x_s$

Hence trivially:

$\neg \exists k \in \N: \forall m \in \N: m > k: x'_m = x'_k$

$\blacksquare$