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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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Def: $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 $x\in X$.

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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:

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$(Y,\tau)$?

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Why a ring and not a pair of rings? –  Qiaochu Yuan Aug 3 '12 at 20:46
    
We Know that if we define bicontinuous function $f:(X,\tau_1 ,\tau_2) \rightarrow \mathbb{R}$ so that $ f:(X,\tau_1) \rightarrow \mathbb{R}$ and $f:(X ,\tau_2) \rightarrow \mathbb{R}$ are continuous, then we have a pair of rings. i.e.$C(X,\tau_1)$ and $C(X,\tau_2)$. which it seems that there is no general relation between these rings which followed by bicontinuous definition of it. But it is trivial to consider the situations with which the set of all bicontinuous functions forms a ring. –  Ali Reza Aug 3 '12 at 21:07
    
What happens if you consider two different topologies on $\Bbb R$, say $t_1$ and $t_2$ and consider functions that are both $\tau_1$ to $t_1$ continuous and $\tau_2$ to $t_2$ continuous? Such functions obviously form a ring, but I don't see yet how to choose everything to get an interesting example. –  David Feldman Aug 4 '12 at 4:12
    
Consider two topologies $\tau_1$={$(a,\infty):a\in \mathbb{R}$} and $\tau_2$={$(-\infty,b):b\in \mathbb{R}$} on the real numbers. the identity function is a continuous function from $(\mathbb{R},\tau_1 ,\tau_2)\rightarrow (\mathbb{R},\tau_1 ,\tau_2)$. But its obvious that −f is not continuous. this implies that the set of all such bicontinuous functions with your definition does not form a ring. How are you sure that your definition implies that it forms a ring –  Ali Reza Aug 4 '12 at 6:09
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