# question about universal sets of real numbers

If $A = \{x \in R | x \lt -5$ or $x \ge 3\}$ and $B =\{ x \in R | -7 \lt x \le 3\}$

Find $(A \cup B)^c$

I figured it to be $\{ x \in R | x \le -7$ $or$ $x \gt 3 \}$

Did I calculate it right? Can you push me in the right direction?

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Why "universal" sets? –  1015 Feb 27 '13 at 2:59

Just to make it graphical. I hope it helps:

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Nice!${}{}{}{}$ –  Asaf Karagila Feb 27 '13 at 2:58
Okay, that does make sense now. I should have done that in the first place. So if the compliment of (A u B) is an empty set. –  Max Feb 27 '13 at 2:58
Yes, indeed, Max! –  amWhy Feb 27 '13 at 2:58
Nice picture. Too bad you forgot the complement sign...Still +1. –  1015 Feb 27 '13 at 2:58
Well then the compliment would still be an empty set... –  Max Feb 27 '13 at 2:59

Well, $x\in A\cup B$ if and only if it is in either one; and it is in the complement if and only if it is not in neither.

What does it mean that $x$ is not in $A$ and not in $B$? It means that $-5\leq x<3$, and either $x<-7$ or $3<x$. Is there such $x$?

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Note that

$$A\cup B = \{x \in \mathbb R \mid x \in A \;\text{ or }\;\; x \in B \}$$

$$A\cup B = \{x \in \mathbb R \mid x \lt -5 \;\;\text{or}\;\;x \geq 3\;\;\text{or}\;\; -7 \lt x \le 3\} = \mathbb R$$

So $$(A \cup B)^c = \mathbb R\setminus(A\cup B) = \mathbb R\setminus \mathbb R = \varnothing$$

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