Let $X$ be a set and $f:X\to\mathcal{P}(X^\omega)$. Under which circumstances is there some topology $\tau$ on $X$ such that $f$ maps each point $x$ to the set of converging sequences in $\tau$ with limit $x$? The finest topology which realizes all sequences in $f$ is $$\tau:=\{A\subseteq X:\forall x\in A.\forall s\in f(x).\exists N\ge 0.\{s_n\}_{n\ge N}\subseteq A\}.$$ But how to make sure no new converging sequences emerge in $\tau$?
There is an (open) neighbourhood base of $\tau$ whose sets $A_x$ can be constructed the following way: Modify $f$ by replacing any sequence with a tail of itself, call the result $f'$. Then define inductively $A_0:=\{x\}$, $A_{i+1}:=A_i\cup(\cup_{y\in A_i,n\ge 0}f'(y)_n)$ and take $A_x=A_{f',x}:=\cup_{i\ge 0} A_i$. Since for the construction of $f'$ one only needs to take into account the $f'(y)$ for $y$ 'reachable' from $x$, which all sit in an $\omega$-branching tree of height $\omega$, this base is countable.
So one way to make sure $\tau$ doesn't realize new sequences is to demand that for any sequence $s$ and $x\in X$ such that for any $f'$ as above there is some tail of $s$ living in $A_{f',x}$ we must have $s\in f(x)$.
But is there a simpler way to put this?
Clearly, for $s\in f(x)$ and any sequence $(k(n))_n$ of natural numbers converging to $\infty$ we need $(s_{k(n)})_n$ to be in $f(x)$, also $f(x)$ must contain the constant $x$-sequence and be closed under mergence of a finite number of sequences. Still, the case is much more subtle than this. Consider $X=\mathbf{R}$ and $f(u)$ containing all sequences $(u\pm v2^{-n})_n$ for some $v\in[1,2)$ plus their closure under the constructions just discussed. This $f$ produces the Euclidean topology, but obviously doesn't contain all convergent sequences in the Euclidean space and it's really hard to see (for me) if there is some simple, abstract condition that $f$ fails to satisfy.
One more condition I could think of, which is true whenever $\tau$ is first-countable, is that for $s\in f(x)$ and $(t^n)_n$ with $t^n\in f(s_n)$ there exists a sequence of natural numbers $(k(n))_n$ such that any $s'$ with $s'_n\in\{t^n_m\}_{m\ge k(n)}$ must be contained in $f(x)$.
But here take again $X=\mathbf{R}$ with $f(u)$ consisting of all sequences $s$ such that $s_n\in u\pm(v-w_n,v+w_n)2^{-n}$ for some $v\in[1,2)$ and $w\in\mathbf{R}^\omega_{>0}$ converging to $0$, this satisfies the last condition but doesn't contain all convergent sequences in the Euclidean space, even if we take all the closures discussed before.
Best Answer
The more usual way to go about these things is what Fréchet already did in 1906: define a limit operator $\lambda$ that assigns to all sequences of $X$ a set of limits (there can be more limits if you want to be general, and also, many sequences wil have the empty set assigned to it.) So we then consider $\lambda: X^\omega \to \mathcal{P}(X)$ and it has to be obey simple axioms:
A subtle extra condition is the following:
$$\text{if } x \in \lambda((x_n))\text{ and } \forall n : x_n \in \lambda(x^{(n)}_k)_k), \exists i_1, i_2, i_3 , \ldots, j_1, j_2, j_3, j_4., \ldots \in \mathbb{N}: x \in \lambda(x^{(i_k)}_{j_k})_k)\tag{1}$$
which ensures that the closure operation defined by the limit operator ($x$ is in the closure of $A$ iff some sequence from $A$ converges to $x$) is a topological one and defines a topology where the sequence convergence is exactly the one given by $\lambda$.
If we just have the first 3 axioms, we can already define a topology on $X$ by saying that $C$ is closed when for all sequences $(x_n)$ from $C$, all limits of that sequence lie in $C$. This is weaker and does not already garantuee that the convergence from the new topology exactly coincides with the given "convergence rule". Engelking has some exercises on this at the end of chapter 1, with references.
Of course you cannot generate all topologies in this way, just the sequential or Fréchet-Urysohn ones.