Suppose I have a topological space $(F,\tau)$ which has some notion of convergence for elements in $F$ (denote this by e.g. $A_n \to_c A$ for $A_n, A \in F$). Furthermore, $(F,\tau)$ is metrisable and it is a Polish space (i.e. it is separable and there is a complete metric compatible with the topology).
If I show that $A_n-A_m \to_c \emptyset$ (convergence as I wrote above) as $n,m\to\infty$, does it follow that $A_n \to_c A$ for some $A$, i.e., does it follow that $A_n$ converges? Or do I actually need to show that it's Cauchy in the complete metric that it possesses?
It seems that you not only a topological space, but also a vector space (or at least an Abelian group). If the complete metric $d$ you have is translation-invariant, i.e., $$d(x+z,y+z) = d(x,y)\,,$$ then $$d(A_n,A_m) = d(A_n-A_m,0)\,,$$ and so if $A_n-A_m \to 0$, then $(A_n)$ is a Cauchy sequence, hence convergent by the completeness of the metric.
If the complete metric is not translation-invariant, then you need to show that it is Cauchy by hand.