TITLE: Deriving topology from sequence convergence/limits?
QUESTION [6 upvotes]: Normally convergence of a series and the limit of a convergent series are defined relative to a topology. That is, you start with a topology, and then you define a series as convergent based on that topology, using the well known definition:

A series $(a_n)$ converges to $a$ if for every neighbourhood $N$ of $a$ there's an $m\in\mathbb N$ such that for all $n>m$, $a_n\in N$.

Now I'm thinking about whether it is possible to reverse that process: Start with a declaration which series are divergent, and what is the limit of all convergent series, and then use that to define a topology, by simply reversing the above definition:

A set $N$ is a neighbourhood of a point $a$ if for every series $(a_n)$ converging to $a$ there exists an $m\in\mathbb N$ such that for all $n>m$, $a_n\in N$.

More formally, a limit function for a set $M$ would be defined as a function
$$ l: \mathbb N^M \to M\cup \{\text{divergent}\}$$
mapping all sequences of values in $M$ either to a value in $M$ (saying that the value converges to that value) or to the special value $\text{divergent}\notin M$ (saying that the sequence is divergent, of course). That function, to be considered a limit, would of course have to fulfil some requirements to count as limit functions. I think an obvious set of conditions would be:

For the constant sequence $(a) = (a,a,a,a,a,\ldots)$, $l((a)) = a$.
If $(a_n)$ and $(b_n)$ differ in only finitely many terms, then $l((a_n))=l((b_n))$.
If $(b_n)$ is a subsequence of $(a_n)$ and $(a_n)$ converges, then $l((b_n)) = l((a_n))$.
If $c_n$ is obtained by interleaving $a_n$ and $b_n$ (where the interleaving may be in any possible way) then
$$l((c_n)) = \begin{cases}l((a_n)) & \text{if } l((a_n))=l((b_n))\\
\text{divergent} & \text{otherwise}\end{cases}$$

Of course what remains to be shown is whether the definition above indeed results in a neighbourhood topology. Looking at the axioms given on Wikipedia, all but one are easily proved. The one I don't see how to prove is the last one:

Any neighbourhood N of x contains a neighbourhood M of x such that N is a neighbourhood of each point of M.

Therefore my question is:
Does the above neighbourhood definition together with the "limit axioms" I've given above define a topology? And if not, is there some additional property of limits I've missed, adding of which to the list would be sufficient to make it a topology?

REPLY [4 votes]: I think a simpler approach would be to define $N$ is closed if every convergent sequence in $N$ has its limit in $N$. This should be equivalent to your definition. It is obvious that the entire space and the empty set are both closed, And not much harder to see that the intersection of any collection of closed sets is also closed. The final condition - that a finite union of closed sets is closed - takes a bit more work, but follows from the fact that any convergent sequence in the union must have a convergent subsequence in one of the sets.
Of all your conditions, the only one that is necessary is that every subsequence of a convergent sequence converges to the same limit. The other conditions just express additional details about the topology.