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Set Unions

Precalculus Algebra Set Operations Set Unions

Set Union

Let $A$ and $B$ be sets. Then, the union of $A$ and $B$ , denoted $A\cup B$ , is the set of all elements in $A$ OR in $B$ .

The union of two sets produces a new set which consists of all the elements from both sets “dumped together” into one big set (without duplicates).

Try the following!

$A\cup B=\{1,2,3,5,7, 0,2,4,6,8\}$

Arrangement of elements in the set doesn’t matter. All we are doing is taking all the elements of $A$ and putting them together with the elements of $B$ in one big set.

$A\cup B=\{-3,-4,-5,0,1, 2, 5\}$

Notice that $A$ and $B$ have elements in common. So, when $A$ and $B$ are “dumped together” into one big set, you don’t need to list the duplicate elements twice, because sets don’t contain duplicates.

$A\cup B=\{0,1,2,3,4,5\}$

Notice that $A\subseteq B$ , so dumping the sets together and eliminating duplicate elements gives you exactly the set $B$ .

$A\cup \emptyset=\{1,2,3\}=A$ .

Think of it this way: you have a box containing 1,2 and 3, and you dump that together with a box that contains nothing. Would you get anything more than what was in the first set?

$\emptyset\cup\emptyset=\emptyset$ .

Think of it this way: if you dump two boxes that contain nothing together into one big box, what do you end up with (in that big box)?

$(0,3)\cup(2,5)=(0,5)$ .

Solution: $(0,3)\cup(2,5)=(0,5)$ . The given two intervals $(0,3)$ and $(2,5)$ overlap on the interval $(2,3)$ , so the list of elements that are in $(0,3)$ OR $(2,5)$ are all elements between 0 and 3, non-inclusive of endpoints. See number line below. Everything shaded is in the set $(0,3)\cup(2,5)$ .

$(-1,2)\cup (3,6)$ cannot be simplified any more than it is.

The two intervals listed do not overlap at all, so all numbers between 2 and 3 (including 2 and 3 themselves) are NOT in the set $(-1,2)\cup (3,6)$ . This set is drawn below on the number line. Everything that is shaded is in the set.

$(-1,4)\cup [4,6]=(-1,6]$ .

While the intervals $(-1,4)$ and $[4,6]$ do not overlap at all, the interval $(-1,4)$ ends where the interval $[4,6]$ starts. In other words, $(-1,4)$ contains all numbers greater than -1 4, and $[4,6]$ contains all numbers from 4 to 6, INCLUDING 4. Because of this “one set picking up where the other left off”, we can write this union as a single interval. NOTE: if 4 was NOT included in the second interval, we could not have written this union as a single interval; there would have been a “gap” between the two sets occurring at 4. See number line below. Everything that is shaded lives in the union.

Can’t be simplified any more, so $(-\infty, 0)\cup (0,\infty)$

Solution: Can’t be simplified any more. While the first interval stops where the second one starts, neither set contains the element 0, so one cannot write $(-\infty, 0)\cup (0,\infty)=(-\infty,\infty)$ , because the set on the right-hand side of the equation contains 0, whereas the one on the left-hand side does not!

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