Convergence functions Let X be a nonempty set. I define a  convergence function on X to be a partial function from the set of all sequences in X, to X, that satisfies the five additional conditions:


*

*Every constant sequence is assigned that constant.

*If a sequence converges, so does any sequence obtained from altering, inserting, or deleting finitely many terms, and to the same value.

*If a sequence converges, every subsequence converges to the same value.

*If two sequences have the same convergence value, any weave of them also converges to the same value

*This is the most complicated property. Let $(a_m)_n$ be a double sequence where every $(a_m)$ converges. And suppose the resulting sequence of convergence values also converges, say to $L$. Then there is a sequence, where the $n$-th element is in the $n$-th sequence, that converges to $L$.


The question I am trying to find out, as an independent scholar, is this: Does every convergence function on a set induce a topology such that every sequence that converges in the topological sense to a value $L$, also converges under $f$ to $L$? I have tried proving it, but I am completely stuck. Any help would be much appreciated.
 A: Yes.
A set $H$ will be closed iff it is closed under taking limit (of sequences within $H$). All you have to prove that any intersection and any finite union of closed sets is again closed.
Then define the open sets as complements of closed sets.

Note that the topology obtained this way is $T_1$: one point sets are closed by criterium 1.
Now if $x_n$ converges to $L$ according to our limit function $f$, and $U\ni L$ is any open set, then its complement, $U^\complement$ is closed, but cannot contain an infinite subsequence of $x_n$ as it doesn't contain $L$ but is closed under limit.
On the other hand, if $x_n$ converges to $L$ in the topology, then -eliminating constant subsequences- we can reduce it to the case when $x_n\ne L$ for all $n$.

Then consider $G:=\{x_n\,\mid\, n\in\Bbb N\}$. We claim that $G$ is not closed: its complement contains $L$ and if that were open, that would contain some $x_n$ because of the definition of limit. 
So, by definition of the topology, $G$ is not closed under limit, moreover by the same reasoning, $L$ must be in the closure of $G$, i.e. there is at least a subsequence $(x_{n_k})_k$ that converge to $L$, according to $f$. 
Now cut the original sequence $(x_n)$ into ${\bf A}_0:=(x_{n_k})_k$ and the rest, say $(x_{m_k})_k$ (So that $\{n_k\mid k\in\Bbb N\}\,\cup\,\{m_k\mid k\in\Bbb N\}\,=\,\Bbb N$, and repeat all with $(x_{m_k})_k$ which still converges to $L$ in topology. This yields a next subsequence ${\bf A}_1$ which already converges to $L$ according to $f$. Repeat it until all elements $x_n$ are used up (either for finite or infinite many times), then apply condition 4. or 5.
