Let be a topological space.
If is Hausdorff then every convergent sequence has a unique limit, #m/thm/topology
i.e. every sequence has at most one limit point.
Moreover, if is first-countable,
then it is hausdorff iff all limits are unique.
Proof
First assume is Hausdorff.
Let in ,
and some .
Then there exist open neighbourhood and
such that .
Moreover there exists , such that for all .
Then for all such
and hence .
Thus limits are unique for any hausdorff space,
without invoking first-countability.
Now assume is first-countable with unique limit,
and let such that .
Let and be nested open neighbourhood bases of and respectively.
Assume is not hausdorff,
i.e. for all .
Then we can construct a sequence in
such that for all ,
in which case and
violating the uniqueness of limits.
Therefore must be hausdorff.