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 ?
Thank you in advance!
Best Answer
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.