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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?

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

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  • $\begingroup$ Thank you :). This is what I meant, yes :). $\endgroup$ – user290425 Mar 23 '16 at 11:17
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    $\begingroup$ @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. $\endgroup$ – Darío G Mar 23 '16 at 11:19
  • $\begingroup$ Haha! Yes this result does seem quite trivial and unnecessary to prove. My lecturer asked me to show this as an exercise though, so I had to explicitly prove it to verify that it is true. He wanted to see if I understand why it's true $\endgroup$ – user290425 Mar 23 '16 at 11:22
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    $\begingroup$ I think is a fair exercise to check if you understand the definition of a saturated set. $\endgroup$ – Darío G Mar 23 '16 at 12:01

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