Let the universal set be the set $\mathbb{R}$ of all real numbers and let
$A=\{x ∈ \mathbb{R}|−3≤ x ≤0\}$, $B=\{x∈\mathbb{R}|−1<x<2\}$, $C = \{x ∈ \mathbb{R} | 6 < x ≤ 8\}$
I am supposed to find the sets $A \cup B $, $B^c$, and $(A\cap B)^c$. I did this, I am not sure if I'm solving correctly:
My solution
a) $A \cup B = \{x ∈ \mathbb R | x ∈ [−3, 0] \text { or } x ∈ (-1, 2)\} = \{x ∈\mathbb R | x ∈ [−3, 3)\} = [−3, 2)$.
f) $B^c =\{x ∈\mathbb R \mid \text{ is not the case that }x ∈(−1, 2)\} = \{x ∈ \mathbb R|\text{ it is not the case that }(−1 < x\text{ and }x < 2)\} =\{x ∈\mathbb R|x ≤−1 \text { or }x \ge 2\}=(−∞, −1]∪[2, ∞)$
i) $(A\cap B)^c= \{x ∈\mathbb R| \text{ is not the case that }x ∈[−3, 2)\}= \{x ∈ \mathbb R| \text{ it is not the case that }(−3 \le x\text{ and }x < 2)\} =\{x ∈ \mathbb R \mid x <−3\text{ or }x \ge 2\}=(−∞, −3)∪[2, ∞)$
thank you in advance!
See at the definitions of the sets you are looking for.
For example $x \in A \cup B \leftrightarrow x \in A \lor x \in B$.
$x \in A \cap C \leftrightarrow x \in A \wedge x \in C$.
So, let's find $A \cap C$.
$$A=\{ x \in \mathbb{R}| -3 \leq x \leq 0 \}$$ $$C=\{ x \in \mathbb{R}| 6<x \leq 8 \}$$
$$x \in A \cap C \leftrightarrow -3 \leq x \leq 0 \wedge 6<x \leq 8 $$
We see that we cannot find such a $x$ since it would have to be at both the intervals $[-3,0]$ and $(6,8]$, that is impossible.
Thus, $A \cap C=\varnothing$.