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?