Space in which All Convergent Sequences have Unique Limit not necessarily Hausdorff
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Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $T$ be such that all convergent sequences have a unique limit.
Then it is not necessarily the case that $T$ is a Hausdorff space.
Proof
Let $T = \struct{\R, \tau}$ be the set of real numbers $\R$ with the countable complement topology.
From Countable Complement Space is not $T_2$, $T$ is not a Hausdorff space.
Suppose $\sequence{x_n}$ is a sequence in $\R$ which converges to $x$.
Then $C = \set{x_n: x_n \ne x}$ is closed in $T$ because it is countable.
So $X \setminus C$ is a neighborhood of $x$.
This means there is some $N \in \N$ such that:
- $\forall n > N: x_n \in X \setminus C$
That is, $x_n = x$ for large $n$.
This means that if $x_n \to y$ then $y = x$, proving limits in $T$ are unique.
$\blacksquare$