Show that a first countable space $X$ is Hausdorff if and only if every sequence converging in $X$ has a unique limit.
My Proof
$X$ is Hausdorff, then it is trivial that every convergent sequence has a unique limit…
Now assume that $X$ is not Hausdorff then $\exists x \neq y$ such that they cannot be seperated by disjoint open neighborhoods of $x$ and $y$
Now because $X$ is first-countable then there exists a sequence of decreasing neighborhoods $U_i$ of $x$ and $V_i$ of $y$ such that $U_i \cap V_j \neq \phi$, in particular $U_i \cap V_i \neq \phi$, choose $z_i\in U_i \cap V_i$, then the sequence $( z_i) $ converges to both $x$ snd $y$.
Is it correct? (note that the fact that $U_i$ and $V_i$ is decreasing sequence of neighborhoods is necessary to prove the convergence)
When $X$ is not Hausdorff, then there is a sequence whose limit is not unique.
Is this proof correct.
Best Answer
It's better stated thus: $x$ has a decreasing countable neighbourhood base $B_n(x)$ and $y$ also has one, say $B_n(y)$.
Because we assume $x$ and $y$ are witnesses of the failure of $X$ to be Hausdorff, we know that for all $n$: $B_n(x) \cap B_n(y) \neq \emptyset$. (In fact any neighbourhood of $x$ intersects any neighbourhood of $y$ in that case.)
Picking $x_n \in B_n(x) \cap B_n(y)$ for all $n$ gives us a sequence that converges to $x$ and to $y$ simultaneously (this follows from the decreasingness of the local base as usual), contradicting the assumption on $X$ on unique sequential convergence.
So the idea is indeed correct; the write-up could be better, e.g. you have to explicitly take decreasing bases, etc.