At first let me recall that if There are two topology $\tau_1$and $\tau_2$ on a set $X$, the triple $(X,\tau_1,\tau_2)$ is called a bitopological space.

There are many definitions and properties which have been proved for bitopological spaces. The reason which I wrote this note for, was the difficulties of defining a special continuity on bitopological spaces. As you Know there are a lot of definitions for defining bicontinuous functions on bitopological spaces. But there is no suitable definition for bicontinuous functions which the collection $C(X,\tau_1,\tau_2)$ of all bicontinuous real funcutions on bitopological space $(X,\tau_1,\tau_2)$ into real numbers $\mathbb{R}$, becomes a ring.

I thought about the following definition:

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: $f\in C(X,\tau_1,\tau_2) $ is bicontinuous at $x\in X$, if for all $\epsilon>0$ there are $U\in \tau_1$ and $V\in \tau_2$ so that $$x\in U\cap V, f(U\cap V)\subset (f(x)-\epsilon, f(x)+\epsilon)$$ and obviously $f$ is bicontinuous on $X$, if $f$ is bicontinuous at all $x\in X$.Def

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In this definition $ C(X,\tau_1,\tau_2) $ is a ring but with a closer look at this, we have found nothing new, because with this definition $C(X,\tau_1,\tau_2)$ is exactly the ring $C(X,\tau_1 \vee \tau_2 )$ of all continuous real valued functions on topological space $(X,\tau_1 \vee \tau_2 )$. Now here is my question:

: Is there a nontrivial definition of bicontinuous real valued functions so that the collection $C(X, \tau_1, \tau_2)$ would be a ring which is not in general isomorphic to $C(Y,\tau)$ of all continuous real valued functions on some topological space$(Y,\tau)$?Question