Is the topological closure of a transitive relation itself transitive? Let $X$ be a topological space, and $R\subseteq X\times X$ a relation on $X$. 
Denote by $\bar{R}$ the topological closure of $R$. Suppose that $R$ is transitive. Then is $\bar R$ transitive?
Can one prove similar statements with equivalence and order relations?
Any reference would also be welcome.
 A: For convenience, let $T(X,\tau_X)$ denotes the first property - that for every $R\subset X^2$ closure preserves transitivity of relations. In general T doesn't hold. Consider $\mathbb{R}^2$ with euclidean topology and subset $R=\{(0,1)\}\cup\bigcup_{n\in \mathbb{N}}\{(1-\frac{1}{n+2},2)\}$. Then $R$ is vacuously transitive, since there is no triple $x,y,z\in \mathbb{R}$, such that both $(x,y)\in R$ and $(y,z)\in R$. Finally, $(0,1), (1,2)\in \bar R$ and $(0,2)\not\in \bar R$.
I think, that in search of topological properties that would imply $T$, one need to look at coarse topologies, so closure would be "big enough" to contain also transitive closure. There is a nice example that suggest this approach. Let $(X,cof(X))$ be arbitrary set with cofinite topology. The main idea of proof lies in simplicity of closed sets in $X^2$. Namely, $Y\subset X^2$ is closed iff it is finite sum of "lines" $\{a\}\times X$, $X\times\{b\}$ and points, or more formally, there exists $X_1\subset X$, $X_2\subset X$ and $I\subset \mathbb{N}$ all finite, such that $Y=(X_1\times X)\cup (X\times X_2)\cup(\bigcup_{i\in I}\{(x_i,x'_i)\}) $. Then it is sufficient to show by cases, that for any transitive $R\subset X^2$ and points $(x,y),(y,z)\in \bar R$ one of the following holds:

*

*both points belongs to $R$

*$\{x\}\times X_1\subset  R$ for some infinite $X_1$ so $\{x\}\times X\subset \bar R$

*analogously $X\times\{z\}\subset \bar R$

*$\bar R = X^2$.

Upgrading relation $R$ to partial order in both examples reduces to adding diagonal $\Delta_X=\{(x,x):x\in X\}$. In first example it doesn't chage much, because $\Delta_{\mathbb{R}}$ is closed, $R\cup \Delta_{\mathbb{R}}$ transitive and $(0,2)\not\in \overline {R\cup \Delta_{\mathbb{R}}}$. Second example becomes trivial: for infinite $X$, $\Delta_X$ is dense, that is, $\overline{\Delta_X}=X^2$, so closure of every reflexive relation (thus every partial order) is $X^2$ and thus transitive but not order.
Equivalence relation is similar to transitive one, as symmetry is immediate: closure of every symmetric relation in every topological space is symmetric. But my euclidean counterexample still holds, just take transitive closure of symmetric closure of $R\cup\Delta_{\mathbb{R}}$ (in that order!). Symmetric closure is just sum of relation and it's transposition, and transitive closure preserves symmetry, so $eq(R)=trans(R\cup R^T)\cup\Delta_{\mathbb{R}}$. If I'm correct, then $eq(R)=\{{0,1}\}^2\cup(\{1-\frac{1}{n+2}\}\cup\{2\})^2\cup\Delta_{\mathbb{R}}$, and $(0,2)\not\in\overline{eq(R)}$.
