If the closure of each $\mathcal{R}$-saturated set is also $\mathcal{R}$-saturated, is then $\mathcal{R}$ an open relation (projection is open)?

Question

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.

Answer 1

No, it does not have to be the case that every open set is saturated. That would make the quotient map just identity (if the original space is $T_0$).

Let $U ⊆ X$ be open. You cannot use that $π^{-1}(π(U)) = U$, but it is enough to show that $π^{-1}(π(U))$ is open, or equivalently that its complement is closed. Since the complement is a saturated set, you may use the hypothesis…

Answer 2

Solution:

Let $(X,T)$ be a topological space, $U\in T \ ,\mathcal{R}$ an equivalence relation in $X$ and $\pi:X\to X/R$ its canonical projection. Then $\pi^{-1}(\pi(U))$ is $\mathcal{R}$-saturated, because $\pi$ is onto. In particular:

$$\pi^{-1}(\pi(U)) \ \mathcal{R}\text{-saturated}\implies X\backslash\pi^{-1}(\pi(U)) \ \mathcal{R}\text{-saturated}\implies\overline{X\backslash\pi^{-1}(\pi(U))}=X\backslash \text{int}\big(\pi^{-1}(\pi(U))\big) \ \mathcal{R}\text{-saturated}\implies\text{int}\big(\pi^{-1}(\pi(U))\big) \mathcal{R} \ \text{-saturated}$$

On one hand, $U\subset \pi^{-1}(\pi(U))$ which implies $U=\text{int}(U)\subset\text{int}\big(\pi^{-1}(\pi(U))\big)$.

As it alwais holds that $\text{int}(S)\subset S$, we also have that $\text{int}\big(\pi^{-1}(\pi(U))\big)\subset\pi^{-1}(\pi(U))$, and therefore $\pi^{-1}(\pi(U))$ is open in $X$. As $X/R$ has the quotient topology, $\pi^{-1}(\pi(U))$ is open in $X\iff\pi(U)$ is open in $X/R$, and we can conclude that $\pi$ is an open aplication, i.e. $\mathcal{R}$ is an open relation.