Recall that a topological space is a set $X$ equipped with a topological structure $\mathcal{T}$. Well, a bitopological space is simply a set equipped with two topological structures $(X, \mathcal{T}, \mathcal{T}^*)$. Unlike with bialgebras, no compatibility condition is required between these structures.
A bicontinous map is a function between bitopological spaces that is continuous with respect to each topological structure.
Bitopological spaces and bicontinuous maps form a category $BiTop$.
Let $Cl$ denote the closure operator with respect to $\mathcal{T}$ and let $Cl^*$ denote the closure operator with respect to $\mathcal{T}^*$.
Let $(X, \mathcal{T}, \mathcal{T}^*)$ be a bitopological space. Consider the following properties of this space:
for each point $x$ there is a $\mathcal{T}$-neighborhood base consisting of $\mathcal{T}^*$-closed sets;
for all $x\in X$ and all $\mathcal{T}$-opens $U$ containing $x$ there is a $\mathcal{T}^*$-closed $\mathcal{T}$-neighborhood $V$ of $x$ such that $V \subset U$;
$Cl^*(O) \subset Cl(O)$ for each $\mathcal{T}^*$-open $O$;
for all $x\in X$ and all $\mathcal{T}$-neighborhoods $U$ of $x$ the closure $Cl^*(U)$ is a $\mathcal{T}^*$-neighborhood;
for each point $x$ and each $\mathcal{T}$-closed $\mathcal{T}$-neighborhood $V$ of $x$ in $X$ there exists a $\mathcal{T}^*$-closed $\mathcal{T}$-neighborhood $U$ of $x$ in $X$ such that $U$ is contained in $V$.
There are the following implications among these properties
Especially, all properties are equivalent if $\mathcal{T}$ is regular.
(1) $\iff$ (2): Given a neighborhood base for a point $x$ as guaranteed by the first property. When you spell out the properties of this neighborhood base, you end up with the second property. For the reverse direction start with an arbitrary $\mathcal{T}$-neighborhood base of a point $x$ consisting of open. Apply the second property to every element of this neighborhood base to the desired neighborhood base.
(3) $\iff$ (4): Suppose property (3), and let $U$ be a $\mathcal{T}$-neighborhood of an arbitrary point $x$. Then the complement $\widetilde{Cl^*(U)}$ is in $\mathcal{T}^*$, so that $Cl^*(\widetilde{Cl^*U}) \subset Cl(\widetilde{Cl^* U})$ by the first property. Hence $\widetilde{Cl(\widetilde{Cl^*U})} \subset \widetilde{Cl^*(\widetilde{Cl^* U})}$ for the complements. Since $U$ is a $\mathcal{T}$-neighborhood of $x$, $x$ does not belong to $Cl(\widetilde{Cl^*U})$. Moreover, $\widetilde{Cl^*(\widetilde{Cl^*U})}$ is $\mathcal{T}^*$-open and a subset of $Cl^*(U)$. Hence $Cl^*(U)$ is a $\mathcal{T}^*$-neighborhood of $x$.
For the converse suppose property (4). Let $O$ be a nonempty $\mathcal{T}^*$-open set and $x$ an element of $Cl^*(O)$. Then if $U$ is any $\mathcal{T}$-neighborhood of $x$, some point $y \in O$ belongs to $Cl^*(U)$ due to the second property. Hence, as $O$ is a $\mathcal{T}^*$-neighborhood of $y$, some point of $U$ belongs to $O$. Thus $x \in Cl(O)$, and therefore $Cl^*(G) \subset Cl(O)$.
(1) $\implies$ (4): Given $x\in X$ and a $\mathcal{T}$-neighborhood $U$ by property (1) there is a $\mathcal{T}^*$-open $O \subset U$ containing $x$. Hence $O \subset Cl^*(U)$, and $Cl^*(U)$ is a $\mathcal{T}^*$-neighborhood.
(3) and $\mathcal{T}$ regular $\implies$ (2): Let $x\in X$ and $U$ be a $\mathcal{T}$-open containing $x$. By regularity of $\mathcal{T}$ we can find disjoint $\mathcal{T}$-opens $V' \ni x$ and $U' \supset \tilde{U}$ ($\tilde{U}$ denotes the complement). Set $V \coloneqq Cl^*(V')$. This set is obviously a $\mathcal{T}^*$-closed $\mathcal{T}$-neighborhood of $x$. Due to property (3) $V \subset Cl(V')$. Since also $Cl(V') \subset \widetilde{U'}$, we have $V \subset U$. This is to say that $V$ is the $\mathcal{T}$-neighborhood we sought.
(5) and $\mathcal{T}$ regular $\implies$ (2): Let $x\in X$ and $U$ be a $\mathcal{T}$-open containing $x$. By regularity of $\mathcal{T}$ we can find disjoint $\mathcal{T}$-opens $V' \ni x$ and $U' \supset \tilde{U}$ ($\tilde{U}$ denotes the complement). Due to property (5) the closed set $\widetilde{U'}$ contains a $\mathcal{T}^*$-closed neighborhood of $x$. This is the neighborhood we sought.
(1) $\implies$ (5): Given some $\mathcal{T}$-closed $\mathcal{T}$-neighborhood $V$ of some point $x$ choose a neighborhood base according to property (1) and take an element $U$ therein that is contained in $V$.
Let $(X, \mathcal{T}, \mathcal{T}^*)$ be a bitopological space. The topology $\mathcal{T}$ is regular with respect to $\mathcal{T}^*$ if one of the two equivalent conditions (1) and (2) from proposition holds. A bitopological space $(X, \mathcal{T}, \mathcal{T}^*)$ is called pairwise regular if $\mathcal{T}$ is regular with respect to $\mathcal{T}^*$ and vise versa.
Let $(X, \mathcal{T}, \mathcal{T}^*)$ be a bitopological space. The topology $\mathcal{T}^*$ is coupled to $\mathcal{T}$ if one of the two equivalent conditions (3) and (4) from proposition holds.
Not that is if $\mathcal{T}^*$ is coupled to a finer topology $\mathcal{T} \supset \mathcal{T}^*$ then $\mathcal{T}^*$ is coupled to every topology coarser than $\mathcal{T}$ due to property (3). Moreover in this case also $\mathcal{T}$ is coupled to $\mathcal{T}^*$ (again a direct consequence of property (3)).
Let $(X, \mathcal{T}, \mathcal{T}^*)$ be a bitopological space. The topology $\mathcal{T}^*$ is called a cotopology of $\mathcal{T}$ if $\mathcal{T}^* \subseteq \mathcal{T}$ and property (5) from proposition holds. The space $(X, \mathcal{T}^*)$ is also called a cospace of $(X, \mathcal{T}$.
It is interesting and perhaps surprising that many advanced topological notions can be described using bitopological spaces, even when you would not naïvely think that there are two topologies around. (At least, that's my vague memory of what they were good for. I think that this was in some article by Isbell.)
Jiri Adamek, Horst Herrlich, and George Strecker, Abstract and Concrete Categories: The Joy of Cats, Dover New York 2009. (pdf) pp. 59-60, 278
B. Dvalishvili, Bitopological Spaces: Theory, Relations with Generalized Algebraic Structures and Applications, Elsevier Amsterdam 2005.
Peter Johnstone, Collapsed Toposes as Bitopological Spaces, pp. 19-35 in Categorical Topology, World Scientific Singapore 1989.
O. K. Klinke, A. Jung, A. Moshier, A bitopological point-free approach to compactications (2011). (preprint)
R. Kopperman, Asymmetry and duality in topology, Topology Appl. 66 no. 1 (1995) pp. 1-39.
The idea naturally appeared first in the context of quasi-metric spaces
The notions of separation axioms were introduced in
J. D. Weston, On the comparison of topologies 1956, Journal of the London Mathematical Society, vol. s1-32 no. 3, pp. 342-354,
J. C. Kelly, Bitopological spaces, Proc. London Math. Soc. 13 no.3 (1963) pp. 71-89.
Only Kelly introduced the concept in its nowadays formulation of a set equipped with two topologies. The Russian school contributed the following comprehensive overviews of this and related topics
A. A. Ivanov, Problems of the Theory of Bitopological Spaces, 1990, Journal of Soviet Mathematics, vol. 52, Issue 1, pp. 2759-2790. Originally published as Проблематика теории битопологических пространств in Zap. Nauchn. Sem. POMI, 1988, vol. 167 (Russian version).
A. A. Ivanov, Problems of the Theory of Bitopological Spaces 2, 1996, Journal of Math. Sciences, vol. 81, Issue 2. Originally publishes as Проблематика теории битопологических пространств. 2 in Zap. Nauchn. Sem. POMI, 1993, Volume 208, pp. 5–67 (Russian version).
A. A. Ivanov, Problems of the Theory of Bitopological Spaces 3, 1998, Journal of Math. Sciences, vol. 91, Issue 6, pp 3339–3364. Originally published as Проблематика теории битопологических пространств. 3 in Zap. Nauchn. Sem. POMI, 1995, Volume 231, pp. 9–54 (Russian version).
as well as a more introductory text book
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