Mathematical-Reasoning · Unit 19 · Video 3 · Interactive Practice

Families Indexed by Any Set: One Interval per Real Number

IKey Formulas

Statement	Meaning
$\{A_\alpha \mid \alpha \in \Lambda\}$ , with $\Lambda \neq \emptyset$	Indexed family: one set $A_\alpha$ for each index $\alpha$ in the indexing set $\Lambda$
$\displaystyle\bigcup_{\alpha \in \Lambda} A_\alpha = \{x \in U \mid \exists\, \alpha \in \Lambda,\ x \in A_\alpha\}$	Union: $x$ lies in at least one set
$\displaystyle\bigcap_{\alpha \in \Lambda} A_\alpha = \{x \in U \mid \forall\, \alpha \in \Lambda,\ x \in A_\alpha\}$	Intersection: $x$ lies in every set
$A_\alpha = (-1, \alpha]$ , $\alpha \in \mathbb{R}^+$ : $\ \bigcup A_\alpha = (-1, \infty)$ , $\ \bigcap A_\alpha = (-1, 0]$	The video's family of intervals

Key Insight: Neither definition needs the labels to come in an order, so they work for $\Lambda = \{1, 2, 3, 4\}$ , $\mathbb{N}$ , $\mathbb{R}^+$ or $\mathbb{R}$ alike — and with infinitely many sets, the intersection need not be one of the sets.

IIDifferent Labels, Same Set

An indexed family assigns one set to each label, but different labels may name the same set.

Family

IIIOne Interval per Positive Real

For $A_\alpha = (-1, \alpha]$ , a single witness label decides whether $y$ is in the union or out of the intersection.

y 2

💡 Zero is not a positive number, so $(-1, 0]$ is not one of the sets $A_\alpha$ : every set in the family reaches past $0$ .

IVComplements: The De Morgan Pattern

With universal set $\mathbb{R}$ , the complement of the union equals the intersection of the complements, and the complement of the intersection equals the union of the complements.

Pair y 2

💡 The next video proves that this pattern, De Morgan's laws, holds for every indexed family.

VQuiz Questions

Problem 1 · Counting the Different Sets

Given: $\Lambda = \{1, 2, 3, 4, 5, 6\}$ and $A_n = \{n, 7 - n\}$ for each $n \in \Lambda$ . How many different sets does the family $\{A_n \mid n \in \Lambda\}$ contain?

✅ Correct! Six labels name only three sets:

\{1, 6\}

\{2, 5\}

and

\{3, 4\}

❌ Not quite. Different labels need not give different sets:

A_1 = \{1, 6\}

and

A_6 = \{6, 1\}

are the same set, since listing order does not matter.

❌ Not quite. Compute all six sets and compare them as sets:

A_n

and

A_{7-n}

always coincide.

Show solution

Compute each set:

A_1 = \{1, 6\},\quad A_2 = \{2, 5\},\quad A_3 = \{3, 4\},\quad A_4 = \{4, 3\},\quad A_5 = \{5, 2\},\quad A_6 = \{6, 1\}.

Order does not matter, so $A_4 = A_3$ , $A_5 = A_2$ and $A_6 = A_1$ . In general $A_{7-n} = \{7 - n, n\} = A_n$ . The family contains exactly 3 different sets, and the statement "for all $m, n \in \Lambda$ , if $m \neq n$ then $A_m \neq A_n$ " is false.

Problem 2 · Union and Intersection over $\mathbb{R}^+$

Given: for each $\alpha \in \mathbb{R}^+$ , let $C_\alpha = [-\alpha, 3)$ . Find $\bigcup_{\alpha \in \mathbb{R}^+} C_\alpha$ and $\bigcap_{\alpha \in \mathbb{R}^+} C_\alpha$ .

What is the union?

What is the intersection?

✅ Correct! The union is

(-\infty, 3)

and the intersection is

[0, 3)

— which is not itself a

C_\alpha

, because

0

is not a positive label.

❌ Check the union. Every

C_\alpha

leaves out

3

, so the union does too; and any

y < 3

lies in

C_\alpha

once

\alpha \geq |y|

❌ Close, but check $0$ . Every

C_\alpha

starts at

-\alpha < 0

, so

0 \in C_\alpha

for all

\alpha

— the intersection keeps

0

❌ Check the intersection. For

y < 0

, the label

\beta = -y/2

gives

C_\beta = [y/2, 3)

, which leaves

y

out; every

y

with

0 \leq y < 3

lies in every

C_\alpha

Show solution

Union. No $C_\alpha$ contains a number $y \geq 3$ , so neither does the union. If $y < 3$ , take $\alpha = |y| + 1 > 0$ ; then $-\alpha < y < 3$ , so $y \in C_\alpha$ . Hence $\bigcup_{\alpha \in \mathbb{R}^+} C_\alpha = (-\infty, 3)$ .

Intersection. Sort $y$ into cases.

$y \geq 3$ : $y$ is in no $C_\alpha$ .
$0 \leq y < 3$ : since $-\alpha < 0 \leq y$ , $y \in C_\alpha$ for every $\alpha$ .
$y < 0$ : take $\beta = -y/2 > 0$ . Then $-\beta = y/2 > y$ , so $y \notin C_\beta$ , and one set leaving $y$ out keeps it out of the intersection.

Hence $\bigcap_{\alpha \in \mathbb{R}^+} C_\alpha = [0, 3)$ . It is not a member of the family: $[0, 3) = [-\alpha, 3)$ would need $\alpha = 0 \notin \mathbb{R}^+$ .

Problem 3 · Which Statement Is True?

Given: for each $x \in \mathbb{R}$ , let $B_x = \{0, x^2, x^4\}$ . Which statement is true?

✅ Correct!

(-x)^2 = x^2

and

(-x)^4 = x^4

, so the labels

x

and

-x

always name the same set.

❌ Not quite. The labels

1

and

-1

are different, yet

B_1 = \{0, 1\} = B_{-1}

❌ Not quite. Repeated entries collapse:

B_0 = \{0\}

and

B_1 = \{0, 1, 1\} = \{0, 1\}

❌ Not quite. Sharing an element is not equality:

B_2 = \{0, 4, 16\}

and

B_4 = \{0, 16, 256\}

16

, but

4 \notin B_4

Show solution

(a) is false: $1 \neq -1$ but $B_1 = \{0, 1\} = B_{-1}$ .

(b) is false: $B_0 = \{0, 0, 0\} = \{0\}$ has one element, and $B_{\pm 1} = \{0, 1\}$ has two.

(d) is true: for every real $x$ , $B_{-x} = \{0, (-x)^2, (-x)^4\} = \{0, x^2, x^4\} = B_x$ .

Problem 4 · The Union of the Complements

Given: $C_\alpha = [-\alpha, 3)$ for $\alpha \in \mathbb{R}^+$ (as in Problem 2), with universal set $\mathbb{R}$ . Find $\bigcup_{\alpha \in \mathbb{R}^+} C_\alpha^{\,c}$ .

✅ Correct! This is the complement of

\bigcap C_\alpha = [0, 3)

— the De Morgan pattern.

❌ Close, but check $0$ .

0 \in C_\alpha

for every

\alpha

, so

0

lies in no complement and cannot be in their union.

❌ Not quite. Each complement

C_\alpha^{\,c} = (-\infty, -\alpha) \cup [3, \infty)

also has a left piece; a negative

y

lies in

C_{-y/2}^{\,c}

❌ Not quite. That is just

C_1^{\,c}

. Smaller labels push the left piece toward

0

y = -\tfrac{1}{2}

lies in

C_{1/4}^{\,c}

Show solution

Each complement is $C_\alpha^{\,c} = (-\infty, -\alpha) \cup [3, \infty)$ . Sort $y$ into cases.

$y \geq 3$ : $y$ lies in every $C_\alpha^{\,c}$ , so in their union.
$0 \leq y < 3$ : $y \in C_\alpha$ for every $\alpha$ , so $y$ lies in no complement.
$y < 0$ : take $\beta = -y/2 > 0$ . Then $y < y/2 = -\beta$ , so $y \in C_\beta^{\,c}$ , and one complement containing $y$ is enough.

Hence $\bigcup_{\alpha \in \mathbb{R}^+} C_\alpha^{\,c} = (-\infty, 0) \cup [3, \infty)$ , which equals $\big([0, 3)\big)^c = \big(\bigcap_{\alpha \in \mathbb{R}^+} C_\alpha\big)^c$ .

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