A B-topological ring that is not semi-topological? 
A topological space $A$ that is also a ring with operations '+' and '.'   where  $ A\times A$  carries product topology, is a semi-topological ring if the mappings $$a_1: A\times A \to A\ \text{such that}\ (x,y)\to x+y$$ , $$a_2: A \to A\ \text{such that}\ x\to -x$$ and $$a_3: A \to A\ \text{such that}\ (x,y) \to x.y$$ are semi-continuous.


A topological space $A$ that is also a ring with operations '+' and '.'   where  $ A\times A$  carries product topology, is a B-topological ring if the mappings $$a_1: A\times A \to A\ \text{such that}\ (x,y)\to x+y$$ , $$a_2: A \to A\ \text{such that}\ x\to -x$$ and $$a_3: A \to A\ \text{such that}\ (x,y) \to x.y$$ are b-continuous.

where,

A function $f:X \to Y$ from topological space $X$ to topological space $Y$ is said to be b-continuous (semi continuous) if for each $x\in X$ and
each open set $U \subset Y$ containing $f(x)$ there exists a b-open (semi-open) set $V \subset X$ containing $x$ such that $f(V) \subset U$.


A set $B$ is called b-open if $B \subset Int(cl(B)) \cup cl(Int(B))$ and $S$ is semi open if $S \subset cl(Int(S))$

Clearly, Topological rings $\implies$ semi-topological rings $\implies$ B-topological rings but not conversely.
After all these definitions I seek a B-topological ring that is not semi-topological. I already have other counterexamples but this one I have no idea about.
 A: It is straightforward to adapt Sebastian Spindler's answer to your other question to this one.  Informally, redefine multiplication on $\mathbb{R}$ so that the product of any two irrational numbers is $0$, but otherwise multiplication is the same.
Formally, let $\{1\}\sqcup T$ be a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space.  Let $X$ be a set of indeterminates equinumerous with $T$ and $S=\mathbb{Q}[X]$.  Finally, choose $R=S/(ab:a,b\in X)$; then the bijection between $T$ and $X$ extends by linearity to an isomorphism of $\mathbb{Q}$-vector spaces between $R$ and $(1\sqcup T)\mathbb{Q}=\mathbb{R}$.  Give $R$ the topology induced by this bijection, and then abuse notation by identifying $T$ and $X$.
Then $({\times}_R)^{-1}((1,2))\subseteq\mathbb{R}^2\setminus(\mathbb{Q}T)^2$, which has empty interior.  But $R$ is $B$-topological, for:

*

*If $0\in U$, then $({\times}_R)^{-1}(U)\supseteq(\mathbb{Q}T)^2$, which is dense in $\mathbb{R}^2$.

*Otherwise, $({\times}_R)^{-1}(U)=(\times_{\mathbb{R}})^{-1}(U)\cap(\mathbb{R}^2\setminus(\mathbb{Q}T)^2)$, which is dense in $(\times_{\mathbb{R}})^{-1}(U)$.

