Given a set $X$, let's call a relation $\longrightarrow \subseteq X^{\mathbb{N}} \times X$ convergence, if the following hold:
- Every constant sequence converges to the constant, i.e. $$(x, x, \dots) \longrightarrow x \quad \forall x \in X$$
- If a sequence $(x_n)_{n\in \mathbb{N}}$ converges to some $x$, i.e. $(x_n)_n \longrightarrow x$, then so does every subsequence $(x_{n_k})_{k\in \mathbb{N}}$, i.e. $(x_{n_k})_k \longrightarrow x$.
If further limits are unique, i.e. $(x_n)_n \longrightarrow x \wedge (x_n)_n \longrightarrow y \implies x = y$, the convergence is called unique.
Now, I am interested in the following question in the unique and the non-unique case:
Given a convergence $\longrightarrow$, is there always a topology $\mathcal{O}$ on $X$ such that $$s \longrightarrow x \iff \text{$s$ converges to $x$ in the topological space $(X, \mathcal{O})$}$$ i.e., $\longrightarrow$ can be expressed as topological convergence?
This is just for fun, but I have little experience in topology and only managed to show (I think :)) that the answer is no if condition 1 is omitted.