To give some background, if $(X,T)$ is a topological space, $C\subset X$, $\mathcal{R}$ is an equivalence relation in $X$ and $\pi:X\to X/{\mathcal R}$ is its canonical projection, then we call $C$ $\mathcal{R}$-saturated if $\pi^{-1}\big(\pi(C)\big)=C$.
I have to prove that, if for each $\mathcal{R}$-saturated subset of $X$, its closure is also $\mathcal{R}$-saturated, then $\mathcal{R}$ is an open relation, i. e. $\pi$ is an open map. However, I don't know how to proceed. I have only proved that if $C$ is $\mathcal{R}$-saturated, then $X\backslash C$ is $\mathcal{R}$-saturated.
It would be a trivial statement if we could suppose that every open subset $U$of $X$ can be expressed as:
$$U=X\backslash \bar S \quad \text{for some $\mathcal{R}$-saturated subset} \,S$$
From the hypothesis and the property I just proved, it would be clear that every open set $U$ of $X$ would then be saturated, and from the definition, we would have that $\pi^{-1}\big(\pi(U)\big)=U$. Since $X/{\mathcal R}$ has the quotient topology, $\pi(U)$ would be open in $X/{\mathcal R}$, proving that $\pi$ is an open map. However, is it possible to express every open set like the complement of a closed ${\mathcal R}$-saturated set? I don't know how to prove that from what is given.
Thanks in advance for your help.