Worked examples · Sheet 1

Operations on given sets

Theory used: Set, membership, inclusion Union, intersection, difference, complement Cartesian product Intervals and their notation

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Given $A = {1, 2, 3}$ and $B = {2, 4}$ , determine $A \cup B$ , $A \cap B$ , $A ∖ B$ , $B ∖ A$ and $A \times B$ .

A \cup B = {1, 2, 3, 4}, A \cap B = {2}

A ∖ B = {1, 3}, B ∖ A = {4}

A \times B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}

A ∖ B = {1, 3}

: the elements of

A

outside

B

B ∖ A = {4}

: the elements of

B

outside

A

; the common element

2

lies in the intersection

The union collects the elements that lie in at least one of the two sets, so it contains $1, 2, 3$ coming from $A$ and $4$ coming from $B$ ; the element $2$ appears only once, because repetitions do not count in a set. The intersection collects the common elements, and the only number that belongs both to $A$ and to $B$ is $2$ . The difference $A ∖ B$ contains the elements of $A$ that are not in $B$ , that is $1$ and $3$ ; the difference $B ∖ A$ contains only the elements of $B$ outside $A$ , that is $4$ : the two differences are different, because the order of the sets matters. The Cartesian product is the set of ordered pairs with first component in $A$ and second component in $B$ : since $A$ has $3$ elements and $B$ has $2$ , there are $3 \cdot 2 = 6$ pairs.

Exercise 2

Given the intervals $A = [- 1, 3)$ and $B = (1, 5]$ , determine $A \cup B$ , $A \cap B$ , $A ∖ B$ and the complement $A^{c}$ with respect to $R$ .

A \cup B = [- 1, 5], A \cap B = (1, 3)

A ∖ B = [- 1, 1], A^{c} = (- \infty, - 1) \cup [3, + \infty)

The two intervals overlap in the region between $1$ and $3$ , so the union is a single interval that goes from the left endpoint of $A$ to the right endpoint of $B$ : the point $- 1$ is included because it belongs to $A$ , the point $5$ is included because it belongs to $B$ . The intersection is the common region: a point lies in both sets when it is greater than $1$ , by the condition of $B$ , and less than $3$ , by the condition of $A$ ; the endpoints $1$ and $3$ are excluded because $1 \in / B$ and $3 \in / A$ . The difference $A ∖ B$ contains the points of $A$ that are not in $B$ , that is the points of $[- 1, 3)$ with $x \leq 1$ : the point $1$ belongs to it, since $1 \in A$ and $1 \in / B$ . The complement of $A$ contains the reals less than $- 1$ , an endpoint included in $A$ and therefore excluded from the complement, and the reals from $3$ onward, since $3 \in / A$ .

Exercise 3

Given $A = {x \in R : x^{2} \leq 9}$ and $B = {x \in R : x > 1}$ , determine $A \cap B$ , $A \cup B$ and $B ∖ A$ .

A = [- 3, 3], B = (1, + \infty)

A \cap B = (1, 3], A \cup B = [- 3, + \infty), B ∖ A = (3, + \infty)

Before operating it is convenient to write the two sets as intervals. The inequality $x^{2} \leq 9$ is equivalent to $∣ x ∣ \leq 3$ , that is to $- 3 \leq x \leq 3$ , so $A$ is the closed interval $[- 3, 3]$ ; the set $B$ is the open half-line $(1, + \infty)$ .

The intersection is the set of points that satisfy both conditions $x > 1$ and $x \leq 3$ , that is $(1, 3]$ : the point $3$ is included because it belongs both to $A$ and to $B$ , while $1$ is excluded because $1 \in / B$ . Since the two sets overlap, the union is a single set that starts from $- 3$ , included, and is not bounded above, since $B$ contains all the reals greater than $1$ . The difference $B ∖ A$ contains the points of $B$ outside $A$ , that is the reals with $x > 1$ and $x > 3$ : the more restrictive condition is $x > 3$ , and the endpoint $3$ is excluded because $3 \in A$ .