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

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.

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…

• Why would the complement be saturated? I just don't understand that point. – Akerbeltz Jun 9 '18 at 10:38
• Complement of a saturated set is also saturated. – user87690 Jun 9 '18 at 11:42
• Thanks for your help, now I see this clear. – Akerbeltz Jun 9 '18 at 12:13

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.