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Union of Sets

The union of two (or three, or four, or more) sets contains the elements common to any of the sets. The union of sets is a set, itself!

The union of sets $A$ and $B$ (denoted as $A \cup B$) is the set containing the elements that are in $A$ or $B!$

Finding the union of sets is similar to solving OR compound inequalities!

Here is a graphic that visually represents the union of sets using Venn diagrams:

$A \cup B$ is just a set with every element of $A$ and every element of $B.$

Example

Let's start off with an example with two sets. Consider the following sets: $A = \{a, b, c, d\}$ and $B = \{c, d, e\}.$

The union of sets $A$ and $B$, $A \cup B$, is the set of elements that are in either $A$ or $B$. The union includes all of the elements of $A$ and all the elements of $B.$ $$A = \{a, b, c, d\} \text{ and } B = \{c, d, e\}$$ We know $A \cup B$ includes all the elements of $A,$ so let's write those down first: $$A \cup B = \{a, b, c, d, \phantom{0000}\}$$ We also know that $A \cup B$ includes all the elements of $B,$ so let's write those as well: $$A \cup B = \{a, b, c, d, \phantom{00}c, d, e\}$$ Sets don't have repeat elements (unless it is otherwise noted that the set is a multiset), so we can rewrite this as: $$A \cup B = \{a, b, c, d, e\}$$

Unions in Set-Builder Notation

When you're learning how to find the union of two sets (let's call them $P$ and $Q$ to differentiate them from our previous example) you might encounter this set-builder notation: $P \cup Q = \{x \: | \: x \in P \text{ or } x \in Q\}.$ Don't let this notation scare you! It simply means that the union of sets $P$ and $Q$ is equal to the set of all elements $x$ that are in $P$ or $Q.$

Unions and Subsets

Say $P$ is a subset of $Q.$ If we take the union of $P$ and $Q,$ the result is just $Q!$ Why? Think about it like this: $$\begin{align} P \cup Q &= \{\text{the elements that are either in P or Q\}}\\ &= \{\text{the elements that are in P and the elements that are in Q\}}\\ \end{align}$$ But since $P$ is a subset of $Q,$ all the elements of $P$ are also in $Q.$ The union is just $Q!$

Union of Three Sets

You can also find the union of more than two sets! Consider sets $X = \{1, 2, 5, 7, 8\}, Y = \{2, 3, 5, 7, 9\},$ and $Z = \{4, 5, 7, 8, 9\}.$

To find $X \cup Y,$ the union of sets $X$ and $Y,$ we find the set of elements that are in either $X$ or $Y.$ The union includes all of the elements of $X$ and all the elements of $Y.$ Forget about $Z$ for now. $$X = \{1, 2, 5, 7, 8\} \text{ and } Y = \{2, 3, 5, 7, 9\}$$ $X \cup Y$ includes all the elements of $X,$ so we we write those down first: $$X \cup Y = \{1, 2, 5, 7, 8, \phantom{00000}\}$$ $X \cup Y$ also includes all the elements of $Y,$ so we write those as well: $$X \cup Y = \{1, 2, 5, 7, 8, \phantom{00}2, 3, 5, 7, 9\}$$ Rearranging the elements and deleting repeats, we get: $$X \cup Y = \{1, 2, 3, 5, 7, 8, 9\}$$

To find $X \cup Y \cup Z,$ the union of sets $X, Y$ and $Z,$ we find the set of elements that are in either $X$ or $Y$ or $Z.$ The union includes all of the elements of $X,$ all the elements of $Y,$ and all the elements of $Z.$ $$X = \{1, 2, 5, 7, 8\}, Y = \{2, 3, 5, 7, 9\}, \text{ and } Z = \{4, 5, 7, 8, 9\}$$ $X \cup Y \cup Z$ includes all the elements of $X,$ so we we write those down first: $$X \cup Y \cup Z = \{1, 2, 5, 7, 8, \phantom{0000}\}$$ $X \cup Y \cup Z$ also includes all the elements of $Y,$ so we write those as well: $$X \cup Y = \{1, 2, 5, 7, 8, \phantom{00}2, 3, 5, 7, 9\}$$ $X \cup Y \cup Z$ also includes all the elements of $Z,$ so we write those, too: $$X \cup Y \cup Z = \{1, 2, 5, 7, 8, \phantom{00}2, 3, 5, 7, 9 \phantom{00}4, 5, 7, 8, 9\}$$ Rearranging the elements and deleting repeats, we get: $$X \cup Y \cup Z = \{1, 2, 3, 4, 5, 7, 8, 9\}$$ Notice that, to find $X \cup Y \cup Z,$ we could have just started with $X \cup Y$ and added the elements of $Z.$

The union of two sets, $A$ and $B$, consists of all those elements that are in at least one of those sets. The union is written as $A \cup B$.

Here's a simple example: $$\begin{align} A &= \{1,2,3\}\\ B &=\{-7,0,2\} \end{align}$$ On your own, try to work out what $A \cup B$ would be.

$$A\cup B = \{1,2,3,0,-7\}$$

Notice that this new set has five elements, not six. That's because $2$ was already in both sets, and every element in a set has to be unique.

Let's try to formalize this notion:

The notation here might be unfamiliar, so let's look at each of these possible answers individually.

$$A \cup B = \left\{x \, | \, x \in A \text{ and } x \in B\right\}$$

Based on what we already know the about the union of $A$ and $B$, this can't be true. This line means an element is in the union of $A$ and $B$ when it's already in both $A$ and $B$. If this were the case, the element $1$ would not satisfy this definition.

$$A \cup B = \left\{x \, | \, x \in A \text{ or }x \in B\right\}$$

This is the real definition! This line means that elements are in the union of $A$ and $B$ if they were already in $A$ or $B$, which works!

$$A \cup B = \left\{x \, | \,(x \in A \text{ and } x \notin B) \text{ or } (x \notin A \text{ and }x \in B) \right\}$$

Nope, this can't be true either. This line means that elements are in the union of $A$ and $B$ if they were only in one of $A$ or $B$. If this were the case, $2$ couldn't be in the union.

$$A \cup B = \left\{x \, | \, x \notin A \text{ or } x \notin B\right\}$$

Yikes, this one is the worst of all of them! This line means the union of $A$ and $B$ is any element that isn't in $A$ or isn't in $B$. What that means is that any element that wasn't in both $A$ and $B$ could be in the union of $A$ and $B$, so every possible number but $2$ would be in the union and the set would be infinitely big.

Here are some other examples:

It's so close to being true, but not quite! Neither of $\mathbb Z^+$ or $\mathbb Z^-$ contain $0$, but $0$ is still an integer. What we COULD say is: $$\mathbb{Z} = \mathbb{Z}^+ \cup \mathbb{Z}^- \cup \left\{0\right\}$$

Yep, the union of all rational numbers and all irrational numbers is the set of all the reals.

Now let's work out some general properties. Try drawing out the Venn diagrams of each of these expressions:

1. $A \cup B$
2. $A \cup \emptyset$
3. $A\cup (A \cap B)$
4. $(A \cup B) \cap A$

Note: The $\cap$ symbol means "intersection".

$A\cup B$

$A\cup \emptyset$

$A\cup(A\cap B)$

In words, we say that the set union operator distributes over set intersection and vice versa.

Together, these two basic operators give you a lot of power when it comes to combining sets in various ways. Together with the notion of "not" (or set complement) many logical operators can be expressed in terms of these two notions.

(Video) Sets: Union and Intersection

by PatrickJMT

This video by PatrickJMT covers set union and set intersection.

$\large{\text{Summary}}$

The most important thing to know with unions and intersections of sets are the symbols, and how they relate to each other. So here they are!

$\cup$: The union symbol. It means that you combine the sets, so each element in each set in the union is included. This symbol also formally means "or", so you might see it described that way other places!

$\cap$: The intersection symbol. It means that you take the similarities between the sets, so only the elements that are in every set are included. This symbol also means "and."

Let's try these out using two sets:

$$\begin{align} A &= \{2, 3, 5, 9, 17, 18\}\\ B &= \{2, 3, 6, 9, 33, 51, 81, 123\} \end{align}$$

What is $A \cup B$?

$$A \cup B = \{2, 3, 5, 6, 9, 17, 18, 33, 51, 81, 123\}$$

Notice that each element is only in the answer once, and every single element from both $A$ and $B$ are somewhere in the set.

What is $A \cap B$?

$$A \cap B = \{2, 3 ,9\}$$

This set is much smaller than $A \cup B$! That's because instead of combining everything, we sort of used $A$ as a filter for $B$. All the numbers that had a match got caught, and everything else fell away.