Mathematical-Reasoning · Unit 19 · Video 2 · Interactive Practice

Indexed Unions: Subscripts for Infinitely Many Sets

IKey Formulas

Statement	Name
$\bigcup_{j=1}^{n} A_j = \{x \in U \mid x \in A_j \text{ for some } j \text{ with } 1 \le j \le n\}$	Indexed union
$\bigcap_{j=1}^{n} A_j = \{x \in U \mid x \in A_j \text{ for all } j \text{ with } 1 \le j \le n\}$	Indexed intersection
$\bigcup_{j=1}^{\infty} B_j = \{x \in U \mid x \in B_j \text{ for some } j \ge 1\}$ , and $\bigcap_{j=1}^{\infty} B_j$ uses "for all $j \ge 1$ "	Infinite family
For $n \le m$ : $\bigcup_{j=1}^{n} A_j \subseteq \bigcup_{j=1}^{m} A_j$ and $\bigcap_{j=1}^{m} A_j \subseteq \bigcap_{j=1}^{n} A_j$	Widening the range

Key Insight: "For some" versus "for all" is the whole difference between the union and the intersection. The $\infty$ on the symbol names no set (there is no $B_\infty$ ): the index takes every value $1, 2, 3, \ldots$ with no last one. The letter $j$ is a placeholder, so $\bigcup_{j=1}^{n} A_j = \bigcup_{k=1}^{n} A_k$ .

IIThe Staircase $C_n = \{n, n+1, n+2, n+3, n+4\}$

The bounds on the symbol choose which sets $C_j$ combine: the union spans them, the intersection keeps their common overlap.

Start j 4 Stop j 8

IIIThe Family $A_n = \{1, n, n^2\}$

A number joins the union when some $A_j$ in the range contains it, and the intersection only when every one does.

Start j 1 Stop j 3

💡 Each $A_n$ has at most three elements, yet $\bigcup_{j=1}^{\infty} A_j = \mathbb{N}$ , because every natural number $n$ lies in its own set $A_n$ .

IVQuiz Questions

Problem 1 · A Finite Range

Given: for each $n \in \mathbb{N}$ , $D_n = \{n, n+1, n+2\}$ . Find $\bigcup_{j=2}^{4} D_j$ and $\bigcap_{j=2}^{4} D_j$ .

The union $\bigcup_{j=2}^{4} D_j$

The intersection $\bigcap_{j=2}^{4} D_j$

✅ Correct! The union runs from

\min D_2 = 2

\max D_4 = 6

, and only

4

lies in all three sets.

❌ Not quite.

\{2, 3, 4\}

is just

D_2

. The union collects the elements of

D_2

D_3

and

D_4

❌ That is the intersection. The union needs an element to be in

D_j

for some

j

, not for all of them.

❌ Check the stop value. The index stops at

4

, so

D_5 = \{5, 6, 7\}

is not included and

7

is in none of the sets.

❌ Check the union. List

D_2

D_3

D_4

and collect every element that appears.

❌ That is the union. The intersection keeps only elements in

D_j

for every

j

from

2

4

❌ Check the intersection. Test each candidate against all three sets:

D_2 = \{2, 3, 4\}

and

D_4 = \{4, 5, 6\}

already share only

4

Show solution

The index $j$ takes the values $2, 3, 4$ :

D_2 = \{2, 3, 4\}, \quad D_3 = \{3, 4, 5\}, \quad D_4 = \{4, 5, 6\}.

An element is in the union when it lies in at least one of these sets:

\bigcup_{j=2}^{4} D_j = \{2, 3, 4, 5, 6\}.

An element is in the intersection when it lies in all three. $D_2$ and $D_4$ share only $4$ , and $4 \in D_3$ :

\bigcap_{j=2}^{4} D_j = \{4\}.

Problem 2 · Infinitely Many Tails

Given: universal set $\mathbb{N}$ and, for each $n \in \mathbb{N}$ , $F_n = \{k \in \mathbb{N} \mid k \ge n\}$ , so $F_1 = \mathbb{N}$ , $F_2 = \{2, 3, 4, \ldots\}$ , and so on. Find $\bigcap_{j=1}^{\infty} F_j$ .

✅ Correct! Every natural number

n

is missing from

F_{n+1}

, so no number lies in all of the sets.

❌ That is the union.

\bigcup_{j=1}^{\infty} F_j = F_1 = \mathbb{N}

, but the intersection needs

x \in F_j

for all

j \ge 1

❌ Not quite.

1

belongs to

F_1

only; it is already missing from

F_2 = \{2, 3, 4, \ldots\}

❌ The infinite intersection is not the "last" finite one. There is no last set

F_\infty

: the condition is

x \in F_j

for every

j \ge 1

, and each fixed

x

fails it at

j = x + 1

Show solution

By definition, $x \in \bigcap_{j=1}^{\infty} F_j$ means $x \in F_j$ for all $j \ge 1$ .

Take any $x \in \mathbb{N}$ . Then $x < x + 1$ , so $x \notin F_{x+1}$ . The condition fails at $j = x + 1$ , so $x$ is not in the intersection. Since $x$ was arbitrary,

\bigcap_{j=1}^{\infty} F_j = \varnothing.

Each finite intersection $\bigcap_{j=1}^{n} F_j = F_n$ is nonempty, but the $\infty$ on the symbol does not mean "take the last one": the index runs through every natural number, and every candidate is eventually excluded.

Problem 3 · Union of Complements

Given: universal set $\mathbb{N}$ and $C_n = \{n, n+1, n+2, n+3, n+4\}$ as in the video. Find $\bigcup_{j=3}^{6} C_j^{\,c}$ .

✅ Correct! Only

6

and

7

lie in all four sets; every other natural number misses at least one, so

\bigcup_{j=3}^{6} C_j^{\,c} = \big(\bigcap_{j=3}^{6} C_j\big)^c

❌ That set misses every $C_j$ . It is

\bigcap_{j=3}^{6} C_j^{\,c}

. The union of complements only asks for a number to miss some

C_j

❌ The universal set is $\mathbb{N}$ . Numbers like

1

2

and

11

miss

C_j

too, so they belong to its complement.

❌ That is $\bigcap_{j=3}^{6} C_j$ itself. Its elements are exactly the numbers that miss none of the sets, so they are the ones left out.

Show solution

The sets are $C_3 = \{3, \ldots, 7\}$ , $C_4 = \{4, \ldots, 8\}$ , $C_5 = \{5, \ldots, 9\}$ , $C_6 = \{6, \ldots, 10\}$ .

$x \in \bigcup_{j=3}^{6} C_j^{\,c}$ means $x \notin C_j$ for some $j$ with $3 \le j \le 6$ : $x$ misses at least one set.

$6$ and $7$ lie in all four sets, so they miss none and are excluded.
$1, 2, 3, 4, 5$ are not in $C_6$ , and every number from $8$ on is not in $C_3$ , so all of them are included.

\bigcup_{j=3}^{6} C_j^{\,c} = \mathbb{N} - \{6, 7\} = \Big(\bigcap_{j=3}^{6} C_j\Big)^c.

Problem 4 · Widening the Range

Given: sets $X_1, X_2, \ldots, X_7$ (nothing else is known about them) with $\bigcap_{j=1}^{4} X_j = \{2, 5\}$ . Decide what must be true for the wider range $j = 1, \ldots, 7$ .

$\bigcap_{j=1}^{7} X_j$ must be…

$\bigcup_{j=1}^{7} X_j$ must be…

✅ Correct! Adding sets to the range can only shrink the intersection and only grow the union.

❌ Not necessarily. If

X_5 = \{5\}

, then

2

drops out and

\bigcap_{j=1}^{7} X_j \subseteq \{5\}

❌ Backwards. Three more conditions "

x \in X_5

x \in X_6

x \in X_7

" can only remove elements, never add them.

❌ Not necessarily. If

X_5 = X_6 = X_7 = \{2, 5\}

, the intersection is still

\{2, 5\}

❌ Check the intersection. An element of

\bigcap_{j=1}^{7} X_j

is in

X_1, \ldots, X_4

, so it lies in

\{2, 5\}

❌ Check the union.

2

and

5

lie in

X_1

, so they lie in

\bigcup_{j=1}^{7} X_j

; extra sets such as

X_5

may add more elements.

Show solution

Intersection. If $x \in \bigcap_{j=1}^{7} X_j$ , then $x \in X_j$ for all $j$ with $1 \le j \le 7$ , in particular for $1 \le j \le 4$ , so $x \in \{2, 5\}$ . Hence $\bigcap_{j=1}^{7} X_j \subseteq \{2, 5\}$ . Equality can fail ( $X_5 = \{5\}$ ) and so can emptiness ( $X_5 = X_6 = X_7 = \{2, 5\}$ ), so only the subset claim must hold.

Union. $2$ and $5$ belong to $X_1$ , so each is in $X_j$ for some $j$ with $1 \le j \le 7$ . Hence $\{2, 5\} \subseteq \bigcup_{j=1}^{7} X_j$ . The union can exceed $\bigcup_{j=1}^{4} X_j$ (take $X_5 = \{100\}$ with $100$ not in $X_1, \ldots, X_4$ ), so it need not equal it. It is never empty, and it lies inside $\{2, 5\}$ only when every $X_j$ does.

\bigcap_{j=1}^{7} X_j \subseteq \bigcap_{j=1}^{4} X_j = \{2, 5\} \subseteq \bigcup_{j=1}^{4} X_j \subseteq \bigcup_{j=1}^{7} X_j.

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Indexed Unions: Subscripts for Infinitely Many Sets

IKey Formulas

IIThe Staircase Cn={n,n+1,n+2,n+3,n+4}C_n = \{n, n+1, n+2, n+3, n+4\}Cn​={n,n+1,n+2,n+3,n+4}

IIIThe Family An={1,n,n2}A_n = \{1, n, n^2\}An​={1,n,n2}

IVQuiz Questions

IIThe Staircase $C_n = \{n, n+1, n+2, n+3, n+4\}$

IIIThe Family $A_n = \{1, n, n^2\}$