Topology, a Logic approach

Consider a Topology over ${\mathbb{F}}_2$$\mathbb F_2$, $\{\mathrm{\varnothing },1,{\mathbb{F}}_2\}$$\{\empty,1,\mathbb F_2\}$

For all Topology Spaces $(X,\tau )$$(X,\tau)$, for any open set $U\in \tau$$U\in\tau$ ,

the characteristic function $\begin{array}{c}{\chi }_U(x):=\{\begin{array}{l}1,x\in U\\ 0,x\notin U\end{array}\end{array}$$\chi_U(x):=\begin{cases}1,x\in U\\0,x\notin U\end{cases}$ is continuous function

What's more, every continuous function is of the form ${\chi }_U$$\chi_U$, $f^{-1}(1)=U,f={\chi }_U$$f^{-1}(1)=U,f=\chi_U$

Thus $\mathrm{Top}(X,{\mathbb{F}}_2)$$\mathrm{Top}(X,\mathbb F_2)$ is a copy of $\tau$$\tau$

And recall the Boolean Ring and Boolean Algebra

$U\cap V\mapsto {\chi }_U\cdot {\chi }_V,U\cup V\mapsto {\chi }_U+{\chi }_V+{\chi }_U\cdot {\chi }_V$$U\cap V\mapsto \chi_U\cdot\chi_V,U\cup V\mapsto \chi_U+\chi_V+\chi_U\cdot\chi_V$

or consider $min({\chi }_U,{\chi }_V{)}^{-1}(1)=U\cap V,max({\chi }_U,{\chi }_V{)}^{-1}(1)=U\cup V$$\min(\chi_U,\chi_V)^{-1}(1)= U\cap V,\max(\chi_U,\chi_V)^{-1}(1)= U\cup V$

Consider some propositions $\mathcal{p}$$\mathscr p$ over $X$$X$, $\begin{array}{c}\mathcal{p}:X\to {\mathbb{F}}_2:=\{\begin{array}{l}1,\mathcal{p}(x)\text{is true}\\ 0,\mathcal{p}(x)\text{is false}\end{array}\end{array}$$\mathscr{p}:X\to\mathbb F_2:=\begin{cases}1,\mathscr p(x) \text{is true}\\0,\mathscr p(x)\text{is false}\end{cases}$

It can be viewed as some characteristic function.

If we consider a family $\mathcal{P}$$\mathscr P$ of propositions about $X$$X$ that can be proved to be true for $S\subseteq X$$S\subseteq X$,

$\mathcal{P}$$\mathscr P$ should have those properties

$\text{False and True}\in \mathcal{P}$$\text{False and True}\in\mathscr P$.

Any $\vee$$\vee$ of elements of $\mathcal{P}$$\mathscr P$ is an element of $\mathcal{P}$$\mathscr P$.

The reason that we can consider any $\vee$$\vee$ is we only need to prove one of them is true, then the $\vee$$\vee$ is true

Any $\wedge$$\wedge$ of finitely many elements of $\mathcal{P}$$\mathscr P$ is an element of $\mathcal{P}$$\mathscr P$.

We can not consider infinitely many ∧ because we need to prove all the proposition is true,

but we can not prove infinitely many propositions are true in finite steps.

And consider the family of ${\mathcal{p}}^{-1}(1)\subseteq X$$\mathscr p^{-1}(1)\subseteq X$, It is a topology over $X$$X$

$\{x\in X|p\vee q(x)=1\}=\{x\in X|p(x)=1\}\cup \{x\in X|q(x)\}$$\{x\in X|p\vee q(x)=1\}=\{x\in X| p(x)=1\}\cup\{x\in X| q(x)\}$

$\{x\in X|p\wedge q(x)=1\}=\{x\in X|p(x)=1\}\cap \{x\in X|q(x)\}$$\{x\in X|p\wedge q(x)=1\}=\{x\in X| p(x)=1\}\cap\{x\in X| q(x)\}$