Conditions for the uniqueness of the limit

Let $(𝑋, T)$ be a topological space. If $𝑋$ is Hausdorff then every convergent sequence $(𝑎 𝑛) \infty 𝑛 = 1$ has a unique limit, #m/thm/topology i.e. every sequence has at most one limit point. Moreover, if $𝑋$ is first-countable, then it is hausdorff iff all limits are unique.

Proof

First assume $𝑋$ is Hausdorff. Let $(𝑎 𝑛) \to 𝑎$ in $𝑋$ , and some $𝑏 \neq 𝑎$ . Then there exist open neighbourhood $𝐴 \in T (𝑎)$ and $𝐵 \in T (𝑏)$ such that $𝐴 \cap 𝐵 = \emptyset$ . Moreover there exists $𝑁$ , such that $𝑎 𝑛 \in 𝐴$ for all $𝑛 > 𝑁$ . Then $𝑎 𝑛 \notin 𝐵$ for all such $𝑛$ and hence $(𝑎 𝑛) ↛ 𝑏$ . Thus limits are unique for any hausdorff space, without invoking first-countability.

Now assume $𝑋$ is first-countable with unique limit, and let $𝑎, 𝑏 \in 𝑋$ such that $𝑎 \neq 𝑏$ . Let $(𝐴 𝑛) 𝑛 \in ℕ$ and $(𝐵 𝑛) 𝑛 \in ℕ$ be nested open neighbourhood bases of $𝑎$ and $𝑏$ respectively. Assume $𝑋$ is not hausdorff, i.e. $𝐴 𝑛 \cap 𝐵 𝑛 \neq \emptyset$ for all $𝑛 \in ℕ$ . Then we can construct a sequence $(𝑥 𝑛) \infty 𝑛 = 1$ in $𝑋$ such that $𝑥 𝑛 \in 𝐴 𝑛 \cap 𝐵 𝑛$ for all $𝑛 \in ℕ$ , in which case $(𝑥 𝑛) \to 𝑎$ and $(𝑥 𝑛) \to 𝑏$ violating the uniqueness of limits. Therefore $𝑋$ must be hausdorff.

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