$R \setminus (S \cup T)$ . Where is $x$? I am sorry for the messy math symbols.
If I have the set:  $R \setminus(S \cup T)$  , is it correct to assume that:
$$R \setminus(S \cup T) = \{x: x∈ \mathbb{R} \text{ and } ( x \notin S \text{ and } x \notin T) \}$$
I am confused because if I had the set $S \cup T$ I would assume that:
$$(S \cup T)= \{x: x \in S \text{ or } x \in T\}$$
Thanks!

Your statement is correct.
I think where you're getting confused is when you have to use De Morgan's Law.
This can be converted logically as follows:
$$
x \in R \setminus (S \cup T) \to x \in R \wedge \neg (x \in S \vee x \in T)\\
x \in R \wedge (\neg x \in S \wedge \neg x \in T) \\
x \in R \wedge x \notin S \wedge x \notin T
$$
De Morgan's Law is applied in the last step and it converts the or to an and.
