Can we find a nice definition of Congruence in Topology? According to my knowledge, quotient structure is a original structure divided by a congruence. However, quotient topology space is defined this way.
Quotient_topology
In this way, $\sim$ is only said be an equivalence relation, not need to be congruence. But what is the congruence on topology space? 
I found two ways to define:
Let $\mathfrak{X}:=(X,\mathscr{T})$ be a topology space. 


*

*First if we see open sets as unary-relations on the domain, then a equivalence relation $\sim$ is a congruence if $\forall O \in \mathscr{T}\forall x,y(Ox\land x \sim y \to Oy)$ . 

*Second if we see $\mathscr T$ as a higher-order relation with type $((0))$, then  a equivalence relation $\sim$ is a congruence if $\forall E,F\subseteq X(E \in \mathscr{T}\land E \sim F \to F \in \mathscr{T})$ . Where $E \sim F$ iff $\sim[E]=\sim[F]$.
These two method are both well-defined but seem not so nice.


*

*If $\mathfrak{X}$ is $T_0$, then for each $x,y \in X$, if $x\sim y$ and $x\ne y$. Then there is an open set $O$ which contains $x$ whereas not contains $y$. But since $\sim$ is a congruence, $x\in O$ implies $y \in O$, a contradiction. That means the only possible congruence is identity.

*If there is an open set $O$ overcasts (in this termology $O$ overcasts $E$ iff $O \cap E \ne \emptyset$) some blocks $\{E_i\}_{i \in I}$ . then every $\bigcup_{i \in I}U_i$ must be open set too where $\forall i\in I[\emptyset \ne U_i \subseteq E_i]$. Let us consider about order topology on $\mathbb R$, since every open set is uncountable, that requires every open set must overcast uncountable many blocks. Moreover,  if $E_i(i \in I)$ all contains two or more elements, then $\bigcup_{i \in I}U_i$ must not be open set where $\forall i\in I[\emptyset \ne U_i \subsetneq E_i]$ else $X\backslash \bigcup_{i \in I}U_i$ will be non-empty proper close set and there is no such non-empty clopen set in this topology space. That means $x \sim x+n$ is not a congruence.
My question is can we find a better definition?
 A: You almost got it with the first definition. The problem is you assumed congruence can be defined globally, for the whole topological space. The correct approach is to say an equivalence relation $\sim$ is a congruence relation with respect to an open set $O \in \mathscr{T}$ if $$\forall x,y(Ox\land x \sim y \to Oy)$$
Note that, if $\sim$ is a congruence relation w.r.t. the unary relation $O$ on $X$, then we can define a unary relation $O'$ on $X/{\sim}$ by
$$O'[x] \iff Ox$$
The quotient space is, then, $(X/{\sim}, \mathscr{T}')$, where $$\mathscr{T}' = \{O': O\in\mathscr{T}\text{ and } \sim \text{ is a congruence relation w.r.t. } O\}$$
Final remark: we can generalize the above definition for quotients of arbitrary structures. We throw away all relations (and operations) for which $\sim$ is not a congruence relation. We then define the quotient relations (and operations) in the natural way. This generalization works for topological quotients, whereas the usual definition doesn't. However, such generalization doesn't work for quotient graphs, whereas the usual definition does. So, as far as model theory is concerned, topological quotients and graphical quotients are different concepts.
