In an general topology exercise I have to prove that if $(X,\tau)$ is an $T_1$- space, then every subset of $X$ is a saturated set (i.e. it is an intersection of open sets).
My approach:
Because $(X,\tau)$ is a $T_1$- space, then every singleton is closed.
Let $A \in \tau$, then we can write: $A=\bigcup \limits_{x\in A} \{x\}$. Therefore: $X \setminus A= X \setminus \bigcup \limits_{x\in A} \{x\}= \bigcap \limits_{x \in A} X\setminus \{x\}$.
If we define a set $P\in\tau, P:= X \setminus A$, then we end up with:
$$P=\bigcap \limits_{x\in X\setminus P} X\setminus\{x\}$$
And because because every singleton is closed, every $X\setminus\{x\}$ is opened, this $P$ is the intersection of open intervals.
After writing the proof I realized that I assumed that $\forall P \in \tau,\exists A \in \tau: P=X\setminus A$. Is this true in this specific case? If so how can I prove it? In : $P=\bigcap \limits_{x\in X\setminus P} X\setminus\{x\}$, the intersection of those sets can either be finite or infinite. Is this a problem? Because for a set to be a topology it has to be closed under only a finite number of intersections. In the definition of a saturated set is it specified if the number of intersections is finite or if it can be an infinite number of intersections?
You want to show every set is saturated, so you can't restrict to $P \in \tau.$ In fact, in what you have shown, is not restricted to $P \in \tau.$
Let $P \subseteq X.$ Then
$$P=X \setminus (X\setminus P)=X\setminus\left( \bigcup_{x \in X\setminus P}\{x\}\right)=\bigcap_{x \in X\setminus P}(X \setminus \{x\}).$$
Since $X$ is $T_1,$ therefore, $P$ is an intersection of open sets and hence saturated. This is essentially what you did but there is no need to first start with $A \in \tau$ or in fact, take $P \in \tau.$