This is somewhat of a follow up to this question.
Background: I am interested in topological rings, and particularly in unifying the construction of the $I$-adic Topology in an arbitrary ring, with the standard topology on $\mathbb{R}$. The motivation for this comes from observing the similarities between the constructions:
The $I$-adic topology is generated by the chain of ideals $\{I^n\subseteq R| n\in\mathbb{N}\}$, by taking the sub-basis to be $\{x+I^n\subseteq R|n\in \mathbb{N}\land x\in R\}$
The topology on $\mathbb{R}$ is generated by taking the chain of intervals $\{(-a,a)\subseteq \mathbb{R}| a>0 \}$, by taking the sub-basis to be $\{x+(-a,a)|a>0\land x\in \mathbb{R}\}$
The question then becomes, are there some properties satisfied by both chains
$$\{I^n|n\in\mathbb{N}\}$$ And $$\{(-a,a)|a>0\}$$
Such that an arbitrary chain of subsets in a ring with these properties generates a topological ring?
Progress
I have identified $4$ properties for an arbitrary collection of subsets of a ring that guarantees that the Topology generated by it produces a topological ring, what is interesting is that these properties can be stated as properties of elements in the collection only, without referencing elements of the ring itself, hence the name "hyper-algebraic" (perhaps this name is not well suited, but I thought it was descriptive). Here are the properties: (quantifiers run over the collection of subsets $\mathcal{C}$ of the ring $R$)
\begin{align} 1.&\forall_I\forall_J[I\subseteq J\lor J\subseteq I]\\ \\ 2. &\forall_I\exists_K[K+K \subseteq I] \\ \\ 3.&\forall_I\forall_J\exists_K[J\cdot K\subseteq I] \\ \\ 4.&\bigcup \mathcal{C}= R \\ \end{align}
Remarkably, these can be seen simply as conditions on a partially ordered set equipped with two operations, $+$ and $\cdot$. Furthermore it can easily be verified that an arbitrary chain of ideals satisfy these conditions, as do the set of intervals around $0$, so indeed it is a unification of the aforementioned constructions. Below I give a proof-sketch that shows that multiplication is continuous in rings equipped with topologies generated by collections of subsets satisfying the conditions above:
Let $(R,+,\cdot)$ be a ring, and $\mathcal{C}$ a collection of subsets of $R$ satisfying properties $(1)$-$(4)$, then \begin{align} m: R^2&\to R \\ (x,y) &\mapsto x\cdot y \end{align} Is continuous w.r.t the topology generated by $\mathcal{C}$
proof: Let $I\in\mathcal{I}$, and $x,y\in R$. We need to show that there is some $K\in\mathcal{C}$ such that $$(x+K)(y+K)\subseteq xy+I$$
First we pass to $I_1 \in\mathcal{C}$ such that $I_1+I_1\subseteq I$ (property 2).
Now since $\bigcup \mathcal{C} = R$ there is some $J\in \mathcal{C}$ such that $x+y\in J$ (property 4)
Then we can find some $K$ such that $J\cdot K\subseteq I_1$ (property 3)
Now we can assume w.l.o.g. that $K\subseteq J$, otherwise we could use $K\cap J$ instead (which is non-empty by property 1). And then note that
\begin{align} (x+K)(y+K) &= xy+ K^2 +(x+y)K \\ &\subseteq xy + I_1 + I_1 \\ &\subseteq xy+ I \end{align}
Which is what we wanted to show. $\blacksquare$
Beyond me just wanting to share this result, I have a few questions:
- Has something like this been studied before?
- Is this essentially point-free topology, but for algebraic structures?
- Does this provide a full classification of topological rings, i.e. does every topological ring have such a collection from which it's topology can be generated?
