Convergence in Indiscrete Space

Theorem
Let $$\left({A, \left\{{A, \varnothing}\right\}}\right)$$ be an indiscrete space.

Let $$\left \langle {x_n} \right \rangle$$ be any sequence in $A$.

Then $$\left \langle {x_n} \right \rangle$$ converges to any point $$x$$ of $$A$$.

Proof
For any open set $$U \subseteq A$$ such that $$x \in T$$, we must have $$U = A$$.

Hence $$\forall n \ge 1: x_n \in U$$.

The result follows from definition of convergence in a topological space.

Note
This demonstrates that in the general topological space it is not necessarily the case that a Sequence has One Limit at Most.