# Prove that every open set is a saturated set

Let $(X, \mathcal T)$ be a topological space. Let $A \in \mathcal T$ be arbitrary, that is $A$ is open. Now notice that $$A = \left(\bigcup_{x \in A} \{ x \}\right)\cap X,$$ which is the intersection of open sets. That is $A$ is saturated. Since $A$ was chosen arbitrarily, we know that this holds for every $A \in \mathcal T$, that is, every open set is saturated.

Does this seem okay?

• Why is $\cup\{x\}$ open? – Shahab Mar 23 '16 at 11:14
• @Shahab - $A = \bigcup_{x \in A} \{ x \}$, and we know that $A$ is open, so $\bigcup_{x \in A} \{ x \}$ must be open – user290425 Mar 23 '16 at 11:15
• Oh of course. I missed that $A$ is open. Your proof seems okay. – Shahab Mar 23 '16 at 11:17

questionIf the definition of a saturated set is "a set that is an intersection of open subsets of $X$", then every open set is saturated because $A=A\cap X$. I assume that in your answer you meant $$A=\left(\bigcup_{x\in A}\{x\}\right)\cap X=A\cap X$$
• @user290425 Notice that it is still true that $A=\bigcup_{x\in A}(\{x\}\cap X)$, it is just not useful to prove that $A$ is saturated. – Darío G Mar 23 '16 at 11:19