# Saturated sets and topological spaces

I saw this exercise in a book of topology that I study and I would like some help to see if my solution is valid.

Definition: A subset $A$ of a topological space $(X,\tau)$ is said to be saturated if it is an intersection of open sets in $(X,\tau)$

It is easy to show that every open set is a saturated set in $(X,\tau)$ because it is the intersection of itself with the set $X$

1) Prove that in a $T_1$ space every subset of $X$ is a saturated set.

2) Give an example of a topological space that has at least one subset that it's not saturated

3) It is true that if the topological space $(X,\tau)$ is such that every subset is saturated, then $(X,\tau)$ is a $T_1$ space?

This is what I did:

1) Let $A\subseteq X$ and $a\in X \setminus A$. Then $X\setminus A=\bigcup\{\{y\}:y\in$X\A$\}$. In a $T_1$ space every singleton is closed and $A=\bigcap\{$X \ {y}$:y\in$X\A$\}$. So we proved that $A$ is an intersection of open sets.

2) Let $X=\{a,b\}$ and $\tau=\{X,\emptyset\}$. If we take the set $\{a\}$ we see that this set is not an intersection of open sets in this topological space.

3) Suppose that every subset of $X$ is saturated and and $t\in X$. A necessary condition for a space to be $T_1$ is that every singleton must be a closed set. We have that $\{t\}=\bigcap\{A_j:j \in$ $J ,A_j\in\tau \}$ and $X\setminus \{t\}=\bigcup\{X$ \ $A_j,j\in J\}$. The set $\{t\}$ has been written as a union of closed sets. But if $X,J$ are infinite ,then we know that an infinite union of closed sets is not necessarily closed.

I would prefer a counterexample in 3) but i did not come up with something. I only made the above observation.Can someone help me ?

Yes, if every subset of $$X$$ is saturated, then $$X$$ is a T$$_1$$ space. For every set $$A\subseteq X$$ we have: $$A\text{ is saturated }\iff A\text{ is an intersection of open sets}$$$$\iff X\setminus A\text{ is a union of closed sets.}$$ For any point $$x\in X,$$ the assumption that $$X\setminus\{x\}$$ is saturated means that $$\{x\}$$ is a union of closed sets, which obviously implies that $$\{x\}$$ itself is a closed set.
• For future reference: it's true that arbitrary unions of closed sets are not necessarily closed. But the argument here, I think, is that if you have any union of sets equal to $\{ x \}$, then at least one of the sets must be $\{ x \}$ itself. – Daniel Schepler Jul 3 '18 at 23:37