Mathematical-Reasoning · Unit 18 · Video 3 · Interactive Practice

Products of Intervals Are Rectangles in the Plane

IKey Formulas

Formula	Name	What it says
$A \times B = \{(x, y) \in \mathbb{R}^2 \mid x \in A \text{ and } y \in B\}$	Product of subsets of $\mathbb{R}$	A region of the plane $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$
$[1, 3] \times [2, 5]$ : $1 \le x \le 3$ and $2 \le y \le 5$	Product of two intervals	A rectangle; here all four edges belong to it
edge $x = a$ is in $A \times B \iff a \in A$	Edge rule	Endpoint included: solid edge; excluded: dotted edge
$A \times (B \cap C) = (A \times B) \cap (A \times C)$	Product with $\cap$ , $\cup$ , $-$	The same holds with $\cup$ and with $-$ in place of $\cap$

Key Insight: An edge is solid or dotted only according to whether its points belong to the set; cutting a rectangle along a line does not make the cut edge dotted.

IIVisualization 1 — Each Endpoint Owns One Edge

In $A \times B$ , the edge at $x = a$ belongs to the set exactly when the endpoint $a$ belongs to $A$ .

A = 1, 3 B = 2, 5

IIIVisualization 2 — One Set, Two Descriptions

For $A = [0, 2]$ , $B = [2, 4)$ , $C = (3, 5]$ , both sides of each identity contain exactly the same points.

Operation

💡 Two examples do not make a proof: the next video states these identities as a theorem for any sets $A$ , $B$ , $C$ .

IVVisualization 3 — Swapping the Factors

Swapping the coordinates of every point reflects $A \times B$ across $y = x$ onto $B \times A$ , a different set.

Reflection 0%

VQuiz Questions

Problem 1 · Which Edges Are Solid?

Given: the graph of $(1, 4] \times [0, 3)$ in the plane. Which edges of the rectangle belong to the set (are drawn solid)?

✅ Correct!

4 \in (1, 4]

and

0 \in [0, 3)

, so the right edge

x = 4

and the bottom edge

y = 0

are solid;

1

and

3

are excluded, so the left and top edges are dotted.

❌ Every edge is reversed. A square bracket includes its endpoint and a parenthesis excludes it:

1 \notin (1, 4]

makes the left edge dotted, and

3 \notin [0, 3)

makes the top edge dotted.

❌ Check the left edge. The left edge is the set of points with

x = 1

, and

1 \notin (1, 4]

, so that edge is dotted; the right edge

x = 4

is solid instead.

❌ Check the top edge. The top edge is the set of points with

y = 3

, and

3 \notin [0, 3)

, so that edge is dotted; the bottom edge

y = 0

is solid instead.

Show solution

$(1, 4] \times [0, 3) = \{(x, y) \in \mathbb{R}^2 \mid 1 < x \le 4 \text{ and } 0 \le y < 3\}$ . Each edge corresponds to one endpoint:

left edge $x = 1$ : $1 \notin (1, 4]$ , dotted;
right edge $x = 4$ : $4 \in (1, 4]$ , solid;
bottom edge $y = 0$ : $0 \in [0, 3)$ , solid;
top edge $y = 3$ : $3 \notin [0, 3)$ , dotted.

So exactly the right and bottom edges are solid.

Problem 2 · The Cut Edge

Given: $A = [1, 3]$ , $B = [0, 4]$ and $C = (2, 6]$ . Which set equals $(A \times B) - (A \times C)$ ?

✅ Correct! The removed rectangle

A \times C

leaves out its bottom edge

y = 2

, because

2 \notin C

, so those points stay and the cut edge is solid.

❌ The cut edge is not dotted. A point

(x, 2)

with

1 \le x \le 3

is in

A \times B

but not in

A \times C

, since

2 \notin (2, 6]

, so it survives the subtraction and the edge

y = 2

is solid.

❌ That is the overlap.

[1, 3] \times (2, 4]

(A \times B) \cap (A \times C)

, exactly the part that the difference removes.

❌ That is the union.

[1, 3] \times [0, 6]

(A \times B) \cup (A \times C)

; the difference keeps only points of

A \times B

that are not in

A \times C

Show solution

By the difference identity, $(A \times B) - (A \times C) = A \times (B - C)$ . On the number line, $B - C$ keeps the numbers of $[0, 4]$ that are not in $(2, 6]$ :

B - C = [0, 4] - (2, 6] = [0, 2],

where $2$ stays because $2 \notin (2, 6]$ . Hence

(A \times B) - (A \times C) = [1, 3] \times [0, 2] = \{(x, y) \in \mathbb{R}^2 \mid 1 \le x \le 3 \text{ and } 0 \le y \le 2\}.

All four edges are solid, including the edge $y = 2$ along which the rectangle $A \times B$ was cut.

Problem 3 · Two Rectangles Overlapping

Find the set $\big([0, 3) \times (1, 4]\big) \cap \big((1, 5] \times [2, 6)\big)$ .

✅ Correct! The conditions on

x

combine to

1 < x < 3

and those on

y

2 \le y \le 4

, so the overlap has dotted left and right edges and solid top and bottom edges.

❌ Check the $x$ -endpoints.

1 \notin (1, 5]

and

3 \notin [0, 3)

, so neither

x = 1

nor

x = 3

satisfies both conditions on

x

❌ Check the $y$ -endpoints.

y = 2

satisfies

1 < y \le 4

and

2 \le y < 6

, and so does

y = 4

; an edge of an overlap need not be dotted.

❌ The brackets are attached to the wrong numbers. The endpoint

3

comes only from

[0, 3)

, which excludes it, and the endpoint

4

comes only from

(1, 4]

, which includes it; also

4 \in [2, 6)

Show solution

A pair $(x, y)$ is in both rectangles exactly when it meets all four conditions:

0 \le x < 3,\quad 1 < y \le 4,\quad 1 < x \le 5,\quad 2 \le y < 6.

The conditions on $x$ combine to $1 < x < 3$ , so $x \in [0, 3) \cap (1, 5] = (1, 3)$ . The conditions on $y$ combine to $2 \le y \le 4$ , so $y \in (1, 4] \cap [2, 6) = [2, 4]$ . Hence

\big([0, 3) \times (1, 4]\big) \cap \big((1, 5] \times [2, 6)\big) = (1, 3) \times [2, 4].

Its left and right edges are dotted and its top and bottom edges are solid.

Problem 4 · When There Is No Common Factor

Given: $A = B = [0, 1]$ and $C = D = [2, 3]$ . Which point belongs to $(A \cup C) \times (B \cup D)$ but not to $(A \times B) \cup (C \times D)$ ?

✅ Correct!

0.5 \in A \cup C

and

2.5 \in B \cup D

, but

(0.5, 2.5)

is in neither square

A \times B

nor

C \times D

; the two sets are not equal.

❌ This point is in both sets.

(0.5, 0.5) \in A \times B

, so it lies in the union

(A \times B) \cup (C \times D)

as well as in

(A \cup C) \times (B \cup D)

❌ This point is in both sets.

(2.5, 2.5) \in C \times D

, so it lies in the union

(A \times B) \cup (C \times D)

as well.

❌ This point is in neither set.

1.5 \notin [0, 1] \cup [2, 3]

, so

(1.5, 1.5) \notin (A \cup C) \times (B \cup D)

Show solution

$(A \cup C) \times (B \cup D) = \big([0, 1] \cup [2, 3]\big) \times \big([0, 1] \cup [2, 3]\big)$ consists of four unit squares, with lower-left corners $(0, 0)$ , $(2, 0)$ , $(0, 2)$ and $(2, 2)$ .

$(A \times B) \cup (C \times D) = [0, 1]^2 \cup [2, 3]^2$ consists of only two of them, with lower-left corners $(0, 0)$ and $(2, 2)$ .

The point $(0.5, 2.5)$ lies in the square $[0, 1] \times [2, 3]$ , which is in the first set but not the second. The identities of this video share the factor $A$ on both sides; without a common factor, a union of products need not be a product.

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