13.2 Union, Intersection, and Complement

Commonly sets interact. For example, you and a new roommate decide to have a house party, and you both invite your circle of friends. At this party, two sets are being combined, though it might turn out that there are some friends that were in both sets.

Note

Union, Interection, and Complement

The union of two sets contains all the elements contained in either set (or both sets).

The union is notated $A \cup B$

More formally, $x ∈ A union B$ if $x ∈ A$ or $x ∈ B$ (or both)

The intersection of two sets contains only the elements that are in both sets.

# The book's colour sets. EDIT them and re-run.
A = {"red", "green", "blue"}
B = {"red", "yellow", "orange"}
C = {"red", "orange", "yellow", "green", "blue", "purple"}
U = A | B | C          # a universal set has to be declared; here: the colours in play

def show(label, S):
    body = ", ".join(sorted(S)) if S else " "
    print(f"{label:<14} = {{{body}}}      n = {len(S)}")

show("A", A)
show("B", B)
show("C", C)
show("U (universe)", U)
print()
show("A union B", A | B)          # every element in either set - "red" listed once
show("A inter B", A & B)          # only what is in BOTH
show("A complement", U - A)       # everything in U that is not in A
show("Ac inter C", (U - A) & C)   # the book's part (c): orange, yellow, purple
print()
show("A union C", A | C)
show("Bc inter A", (U - B) & A)   # Try It Now 2: green, blue

print("\nwhy complements need a universe:")
print("  without U, A complement would contain fuchsia, puppies, peanut butter, ...")
print(f"  with U = {len(U)} colours, A complement is just {sorted(U - A)}")
# Try it: add "pink" to U (put it in C) and re-run - watch every complement grow.

The intersection is notated $A \cap B$

More formally, $x ∈ A intersection B$ if $x ∈ A$ and $x ∈ B$

The complement of a set A contains everything that is not in the set A.

The complement is notated $A$ , or $A^{c}$ , or sometimes $~ A$ .

Example 1

Consider the sets:

$A = {red, green, blue} B = {red, yellow, orange} C = {red, orange, yellow, green, blue, purple}$

Find $A \cup B$
Find $A \cap B$
Find $A^{c} \cap C$

Show solution

a) The union contains all the elements in either set: $A \cup B = {red, green, blue, yellow, orange}$

Notice we only list red once.

b) The intersection contains all the elements in both sets: $A \cap B = {red}$

c) Here we're looking for all the elements that are not in set $A$ and are also in $C$ .

$A^{c} \cap C = {orange, yellow, purple}$

Notice that in the example above, it would be hard to just ask for $A^{c},$ since everything from the color fuchsia to puppies and peanut butter are included in the complement of the set. For this reason, complements are usually only used with intersections, or when we have a universal set in place.

As we saw earlier with the expression $A^{c} \cap C,$ set operations can be grouped together. Grouping symbols can be used like they are with arithmetic - to force an order of operations.

# The book's animal sets (it writes F as P in a couple of spots - same set).
H = {"cat", "dog", "rabbit", "mouse"}
F = {"dog", "cow", "duck", "pig", "rabbit"}
W = {"duck", "rabbit", "deer", "frog", "mouse"}
U = H | F | W

def show(label, S):
    print(f"{label:<22} = {{{', '.join(sorted(S))}}}")

show("H", H)
show("F", F)
show("W", W)
print()
show("H inter F", H & F)
print("...then union with W:")
show("(H inter F) union W", (H & F) | W)
print()
show("F union W", F | W)
print("...then intersect with H:")
show("H inter (F union W)", H & (F | W))
print()
show("(H inter F)c inter W", (U - (H & F)) & W)

print("\nDoes the grouping matter?")
left = (H & F) | W
right = H & (F | W)
print(f"   (H inter F) union W  has {len(left)} elements")
print(f"   H inter (F union W)  has {len(right)} elements")
print(f"   same set? {left == right}")
print(f"   difference: {sorted(left ^ right)}")
# Try it: move "duck" from F into H and re-run. Which of the three answers change?

Example 4

Suppose

$H = {cat, dog, rabbit, mouse}, F = {dog, cow, duck, pig, rabbit} W = {duck, rabbit, deer, frog, mouse}$

Find $(H \cap P) \cup W$
Find $H \cap (F \cup W)$
Find $(H \cap P) \cap W$

Show solution

a) We start with the intersection: $H \cap F = {dog, rabbit}$

Now we union that result with $W : (H \cap F) \cup W = {dog, duck, rabbit, deer, frog, mouse}$

b) We start with the union: $F \cup W = {dog, cow, rabbit, duck, pig, deer, frog, mouse}$

Now we intersect that result with $H : H \cap (F \cup W) = {dog, rabbit, mouse}$

c) We start with the intersection: $H \cap F = {d o g, rabbit}$

Now we want to find the elements of $W$ that are not in $H \cap F$

$(H \cap P)^{c} \cap W = {duck, deer, frog, mouse}$

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.