I would appreciate help answering this parenthetical question that was raised in a set of notes I am self-studying:
What assumption is necessary to make limits unique in a non-metric topology?
In that regard, I would especially appreciate an example of what can go wrong without the assumption, and how the assumption eliminates that problem.
Thanks
Hausdorff spaces have unique limits.
To see this, let $a_n \to a$, and $a_n \to b$. Then, if $a \neq b$, there exist neighbourhoods $a \in U$ and $b \in V$ such that $U \cap V = \phi$. Now, by definition of convergence, $a_n$ has to eventually, be both in $U$ and $V$, which cannot happen as they are disjoint.
However, remove Hausdorffness and you have a problem. For example, consider $\mathbb N$ with the indiscrete topology (there are only two open sets, the null set and the whole set). Then, suppose $a_n$ is a sequence of natural numbers, I claim that it can converge to any $b \in \mathbb N$.
The reason is very simple: An open neighbourhood of $b$ can only be the whole space here, and $a_n$ is already part of the whole space. Hence, $\forall a_n$, $\forall b$, $a_n \to b$, so of course limits are not unique. You can see that the Hausdorff property is very strong in that it will not allow this to happen.
All metric spaces are Hausdorff, so this would not happen in a metric space, for example.
If you want have unique limits for arbitrary nets, then the correct assumption is the $T_2$ separation axiom: every two points have disjoint neighbourhoods.
It is fairly elementary to show that $T_2$ implies unique limits: if $U$ is a neighbourhood of $x$ disjoint from a neighbourhood $V$ of $y$ and $x_i\to x$, then eventually $x_i\in U$, so almost all $x_i$ are not in $V$, and hence $x_i$ does not converge to $y$.
On the other hand, if $x$ and $y$ have no disjoint neighbourhoods, then consider a directed system $I=\{(U,V)\mid x\in U, y\in V \}$, partially ordered by reverse inclusion (so $(U,V)\leq (U',V')$ if $U'\subseteq U$ and $V'\subseteq V$), and for each $U,V$ pick an $x_{U,V}\in U\cap V$. Then it is easy to check that the net $x_{U,V}$ converges both to $x$ and to $y$.
The $T_2$ separation axiom is not equivalent, however, to all sequences having unique limits. For example, if you consider $[0,\omega_1]$ with the point $\omega_1$ doubled (so we have an additional point $\omega_1'$ with basis of neighbourhoods of the form $[\alpha,\omega_1)\cup \{\omega_1'\}$ ), then the resulting space is $T_1$, any two points except $\omega_1$ and $\omega_1'$ can be separated, and any sequence convergent to $\omega_1$ or $\omega_1'$ must be eventually constant. Using these facts, it's easy to check that limits of sequences are unique, even though the space is not $T_2$.