Natural Isomorphism between Interior functor and Closure functor

Consider any partial order set $P$$P$ (more general, pre-order set), it is well known that it could form a Category.

$$\begin{array}{}\text{(1)}& {\mathrm{Ob}}_P:P{\mathrm{Mor}}_P:\le \end{array}$$

A classic example is the power set over $X$$X$

i.e.

$$\begin{array}{}\text{(2)}& (P(X),\subseteq )\end{array}$$

And Topological Space $(X,\tau )$$(X,\tau)$

i.e.

$$\begin{array}{}\text{(3)}& (\tau ,\subseteq )\end{array}$$

Will form a Category. We call it an open set category and denote it as $\mathrm{Op}(X)$$\mathrm{Op}(X)$.

Indeed, it is a functor

$$\begin{array}{}\text{(4)}& \mathrm{Op}:{\mathrm{Top}}^{\mathrm{op}}⟶\mathrm{Frm}\end{array}$$

i.e.

$$\begin{array}{}\text{(5)}& \mathrm{Op}:X⟶\mathrm{Op}(X)\mathrm{Op}:(f:X\to Y)⟶(f^{-1}:\mathrm{Op}(Y)\to \mathrm{Op}(X))\end{array}$$

Similarly, we could define the closed set category, it is a functor as well.

i.e.

$$\begin{array}{}\text{(6)}& \mathrm{Cl}:{\mathrm{Top}}^{\mathrm{op}}⟶\mathrm{Frm}\end{array}$$

 

 

$$\begin{array}{}\text{(7)}& \mathrm{Cl}:X⟶\mathrm{Cl}(X)\mathrm{Cl}:(f:X\to Y)⟶(f^{-1}:\mathrm{Cl}(Y)\to \mathrm{Cl}(X))\end{array}$$

Indeed, there exists a natural isomorphism between $\mathrm{Op}$$\mathrm{Op}$ and $\mathrm{Cl}$$\mathrm{Cl}$.

$$\begin{array}{}\text{(8)}& \begin{array}{c}\begin{array}{ccc}\mathrm{Op}(Y)& \stackrel{({)}^c}{\to }& \mathrm{Cl}(Y)\\ f^{-1}\downarrow & & f^{-1}\downarrow & \\ \mathrm{Op}(X)& \stackrel{({)}^c}{\to }& \mathrm{Cl}(X)\end{array}\end{array}\end{array}$$

Where $({)}^c$$(\,\,\,)^c$ is complement.

If we consider a topology space $X$$X$, and its power set $\mathrm{Bool}(X)$$\mathrm{Bool}(X)$, we could induce two functors as well.

$$\begin{array}{}\text{(9)}& \mathrm{Int}:\mathrm{Bool}(X)⟶\mathrm{Op}(X)\end{array}$$

and

$$\begin{array}{}\text{(10)}& \mathrm{Clo}:\mathrm{Bool}(X)⟶\mathrm{Cl}(X)\end{array}$$

They are both surjective.

Indeed, we could consider a natural transformation between $\mathrm{Int},\mathrm{Clo}:{\mathrm{Top}}^{\mathrm{op}}⟶\mathrm{Frm}$$\mathrm{Int},\mathrm{Clo}:\mathrm{Top^{op}}\longrightarrow\mathrm{Frm}$

$$\begin{array}{}\text{(11)}& \begin{array}{c}\begin{array}{ccc}\mathrm{Int}(Y)& \stackrel{({)}^c}{\to }& \mathrm{Clo}(Y)\\ f^{-1}\downarrow & & f^{-1}\downarrow & \\ \mathrm{Int}(X)& \stackrel{({)}^c}{\to }& \mathrm{Clo}(X)\end{array}\end{array}\end{array}$$

and we have this commutative diagram.

$$\begin{array}{}\text{(12)}& \begin{array}{c}\begin{array}{ccc}\mathrm{Bool}(X)& \stackrel{({)}^c}{\to }& \mathrm{Bool}(X)\\ \mathrm{Int}\downarrow & & \mathrm{Clo}\downarrow & \\ \mathrm{Bool}(X)& \stackrel{({)}^c}{\to }& \mathrm{Bool}(X)\end{array}\end{array}\end{array}$$

i.e.

$$\begin{array}{}\text{(13)}& {\mathrm{Int}(\mathrm{A})}^c=\mathrm{Clo}(A^c)\end{array}$$

In other words

$$\begin{array}{}\text{(14)}& \begin{array}{c}(A^{\circ }{)}^c=(A^c{)}^{\prime }\\ (A^c{)}^{\circ }=(A^{\prime }{)}^c\end{array}\end{array}$$

Where $\circ$$\circ$ is $\mathrm{I}\mathrm{n}\mathrm{t}$$\rm Int$ and ${}^{\prime }$$'$ is $\mathrm{C}\mathrm{l}\mathrm{o}$$\rm Clo$.

Using the fact that ${}^c$$^c$ is invertible, you will see that the conjugation between $\mathrm{I}\mathrm{n}\mathrm{t}$$\rm Int$ and $\mathrm{C}\mathrm{l}\mathrm{o}$$\rm Clo$.

The bound operator is defined by

$$\begin{array}{}\text{(15)}& \partial (A):=\mathrm{Clo}(A)-\mathrm{Int}(A)\end{array}$$

In other words

$$\begin{array}{}\text{(16)}& \partial (A):=\mathrm{Clo}(A)\cap \mathrm{Int}(A{)}^c=\mathrm{Clo}(A)\cap \mathrm{Clo}(A^c)\end{array}$$

Here is the connection with logic

Math Essays: Topology, a Logic approach (wuyulanliulongblog.blogspot.com)