Suppose that $B$ is a topological ring. Suppose that $\mathcal{I}$ is a set of ideals $I$ of $B$ which form a fundamental system of open neighbourhoods of zero.
Quoting the section 1.5 of this book:
Suppose that the topology of $B$ is complete. That means that the map $$B\rightarrow \lim_{\substack{\longleftarrow\\ I \in \mathcal{I}}} B/I$$ is surjective.
Now I don't quite understand why are we saying that the topology of $B$ is complete, because the definition for $B$ being complete seems to be algebraic (although I do agree we did use the open ideals $I\in \mathcal{I}$ for it.)
I am wondering if there are any ways to reconcile the metric space notion of completeness with this algebraic one.
Here is what I mean by an example:
Suppose that $\mathcal{I}$ is in fact a decreasing filtration of ideals $$I_1\supset I_2\supset \ldots$$ Then one can define a pseudometric $d$ on $B$ by definition $d(x,y):=1/2^{m(x,y)}$ where $m(x,y)$ is the maximum $m\in \mathbb N \cup \{\infty\}$ with $x-y\in I_m$. Then $B$ is a pseudometric space and its metric topology coincides with our original topology on $B$. Just like in metric spaces, in pseudometric spaces we have the usual topological notion of completeness. It can be easily shown to be equivalent in this case to the algebraic one quoted above.
What about the general case? Or should I simply take the above as a(n algebraic) definition of completeness and be always careful to distinguish among the algebraic and topological case?
It's common to talk about “completeness in the topology”, but actually referring to the canonical uniformity induced by the topology, where a basis for the uniformity are the set of the form $U\times U$, where $U$ is a neighborhood of $0$.
Under this uniformity, a net $(x_\delta)_{\delta\in\Delta}$ (with $\Delta,\le$ a directed set) is Cauchy if and only if, for every neighborhood $U$ of $0$, there is $\bar{\delta}\in\Delta$ such that, for every $\delta_1,\delta_2\ge\bar{\delta}$, $x_{\delta_1}-x_{\delta_2}\in U$.
The topological ring is said to be complete if every Cauchy net (in the previous sense) converges.
Note that there could be different uniformities inducing the same topology, but this one is “canonically” associated to the topology.
If the topology has a countable basis of neighborhoods of $0$, for instance in your filtration case, sequences suffice for establishing completeness. Indeed, in general, if $\mathcal{B}$ is a basis of neighborhoods of $0$, ordered by reverse inclusion, we can prove that the ring is complete if and only if every Cauchy $\mathcal{B}$-net converges.