$R \setminus (S \cup T)$ . 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!
 A: Yes. It might be more enlightening if you rewrite the first one as
$$R\setminus (S\cup T) = \{x\in R \mbox{ and not } (x\in S \mbox{ or } x\in T)\}$$
Now you're using de Morgan's law, which is a rule from logic that says
not(X or Y) = not X and not Y
(it should be easy to convince yourself that this is true).
A: You're right. If $x$ is not in $(S$ or $T)$, then $x$ is not in $S$ and not in $T$. This sort of "flip" from or to and is referred to as DeMorgan's laws.
A: $R \backslash (S \cup T)$ is the set of all $x \in R$, $x \notin (S \cup T)$. We can write that as $R \backslash (S \cup T) =  \{x \in R | \text{x not in S and x not in T}\}$. What you wrote is correct.
Think about it: $x \in S \cup T$ if $x \in S$ or $x \in T$. But the only way for $x$ not to be in $S \cup T$ is if it is not in $S$ and not in $T$. Otherwise, we could suppose that $x \in S$, and $x \notin T$. Then $x \in S \cup T$. So we must have $x \notin S$ and $x \notin T$.
A: 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.
