The Limit of a Sequence Is Not Unique in a General Topological Space
Theorem
In general, the limit of a sequence in a topological space is not unique.
Explanation
You might wonder what on earth this means, but surprisingly it is true. Until now, in analysis and the like, we have pictured an image where the interval containing a sequence gets narrower and narrower, converging to a single point. However, according to the notion of convergence defined in topology, depending on the topological space there is no reason at all for it to converge to a single point.
To guarantee the uniqueness of the limit, a Hausdorff space is usually assumed.
Disproof
It suffices to present a counterexample where multiple limits exist.
In the finite complement space $\left( \mathbb{R} , \mathscr{T}_{f} \right)$, consider a sequence $\left\{ x_{n} \right\}$ consisting of distinct points. First, let the point to which $\left\{ x_{n} \right\}$ converges be an arbitrary $x \in \mathbb{R}$. Here there exists an open set $U \in \mathscr{T}_{f}$ containing $x$, and $\mathbb{R} \setminus U$ is a finite set. Since $\left\{ x_{n} \right\}$ consists of distinct points, there cannot exist $n_{0} \in \mathbb{N}$ satisfying $x_{n} \notin \mathbb{R} \setminus U$ for all $n > n_{0}$. However, since $\left\{ x_{n} \right\}$ does converge, there must exist $n_{0} \in \mathbb{N}$ satisfying $x_{n} \in U$ for all $n > n_{0}$. By the definition of convergence, $\left\{ x_{n} \right\}$ does converge to $x$, but here $x$ can be anything without it mattering.
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If this is hard to understand, it is a good idea to think about how the definition of convergence has changed and what the open sets of a finite complement space are.
In the traditional metric space, an open set referred to the set of points within a given distance centered on a point. Therefore, although we say “all open sets,” in practice the key was to satisfy the condition in the neighborhood of that point. If the condition continued to be satisfied no matter how small we chose it, that was as good as finishing the check over “all open sets.”
However, if we think about $U$ and yet another open space in a finite complement space, something like $U \setminus \left\{ a \right\}$ also becomes an open set. Whether the whole space is the reals or whatever, an open set remains an open set even if we just remove points one by one, and here it is meaningless to think about “distance.” Therefore, even if we check over all open sets, there is no reason for them to get smaller and smaller as in a metric space, and the limit is not determined.