Sets, Subsets and Intervals — Worked Solutions

Ques­tion 1: Decid­ing which col­lec­tions are sets#

The prob­lem#

Which of these are sets?

(a) the vow­els of Eng­lish; (b) the tall stu­dents of a class; (c) the fac­tors of $36$; (d) the dif­fi­cult chap­ters of this book.

Under­stand­ing the prob­lem#

A col­lec­tion counts as a set only if it is well defined: for any object, every­one must be able to say with­out doubt whether it belongs. Col­lec­tions described by opin­ions ("tall", "dif­fi­cult") fail this test.

The idea#

For each col­lec­tion, ask: would two dif­fer­ent peo­ple always agree on its mem­bers? If yes, it is a set.

Step-by-step solu­tion#

Step 1. (a) The Eng­lish vow­els are exactly $a, e, i, o, u$. Every­one agrees, so this is well defined: a set.

Step 2. (b) "Tall" has no fixed mean­ing. One per­son may call a $170$ cm stu­dent tall; another may not. Mem­ber­ship depends on opin­ion: not a set.

Step 3. (c) The fac­tors of $36$ are fixed num­bers:

$$\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$$

Any num­ber either divides $36$ or does not: a set.

Step 4. (d) Whether a chap­ter is "dif­fi­cult" depends on the reader: not a set.

Check­ing the answer#

The two col­lec­tions that failed both used opin­ion words; the two that passed used exact math­e­mat­i­cal or lan­guage facts.

Answer#

Only (a) and (c) are sets.

Ques­tion 2: Con­vert­ing to ros­ter form#

The prob­lem#

Write in ros­ter form:

(a) $\{x : x \in \mathbf{Z},\ -3 < x \le 2\}$;
(b) the set of let­ters of the word "ASSAS­SI­NA­TION";
(c) $\{x : x \text{ is a two-digit number whose digits add to } 4\}$.

Under­stand­ing the prob­lem#

Ros­ter form means list­ing every ele­ment inside braces. Remem­ber: order does not mat­ter, and a repeated ele­ment is writ­ten only once. Watch the signs in (a): $-3$ is excluded, $2$ is included.

The idea#

Work out exactly which objects sat­isfy each descrip­tion, then list them once each.

Step-by-step solu­tion#

Part (a)

Step 1. $x$ must be an inte­ger greater than $-3$ and at most $2$.

Step 2. Start just above $-3$: the inte­gers are $-2, -1, 0, 1, 2$. ($-3$ is not included because the sign is strict.)

$$\{-2, -1, 0, 1, 2\}$$

Part (b)

Step 1. Read the word let­ter by let­ter: A, S, S, A, S, S, I, N, A, T, I, O, N.

Step 2. Keep each dif­fer­ent let­ter once, in order of first appear­ance: A, S, I, N, T, O.

$$\{A, S, I, N, T, O\}$$

Part (c)

Step 1. Write a two-digit num­ber as tens digit $t$ and units digit $u$. Since it is two-digit, $t \ge 1$. The con­di­tion is $t + u = 4$.

Step 2. Try each allowed $t$: $t = 1, u = 3$; $t = 2, u = 2$; $t = 3, u = 1$; $t = 4, u = 0$. ($t = 0$ is not allowed, as $04$ is not a two-digit num­ber.)

$$\{13, 22, 31, 40\}$$

Check­ing the answer#

(a) has $5$ inte­gers from $-2$ to $2$. (b) The word uses only these six dis­tinct let­ters. (c) $1+3 = 2+2 = 3+1 = 4+0 = 4$.

Answer#

(a) $\{-2, -1, 0, 1, 2\}$; (b) $\{A, S, I, N, T, O\}$; (c) $\{13, 22, 31, 40\}$.

Com­mon mis­take to avoid#

For­get­ting $40$ in part (c): a units digit of $0$ is per­fectly allowed.

Ques­tion 3: Con­vert­ing to set-builder form#

The prob­lem#

Write in set-builder form: (a) $\{1, 8, 27, 64, 125\}$; (b) $\{2, 4, 8, 16, 32\}$; (c) $\{-1, 1\}$.

Under­stand­ing the prob­lem#

Set-builder form describes the ele­ments by a rule, $\{x : \text{property of } x\}$. You must spot the pat­tern in each list and state it pre­cisely, includ­ing where the list stops.

The idea#

Look for a for­mula in $n$ that pro­duces the listed num­bers for $n = 1, 2, 3, \ldots$, and state the range of $n$.

Step-by-step solu­tion#

Step 1. (a) The num­bers are $1^3, 2^3, 3^3, 4^3, 5^3$: cubes of $1$ to $5$.

$$\{x : x = n^3,\ n \in \mathbf{N},\ n \le 5\}$$

Step 2. (b) The num­bers are $2^1, 2^2, 2^3, 2^4, 2^5$: pow­ers of $2$ with expo­nent $1$ to $5$.

$$\{x : x = 2^n,\ n \in \mathbf{N},\ n \le 5\}$$

Step 3. (c) $-1$ and $1$ are exactly the real num­bers whose square is $1$.

$$\{x : x \in \mathbf{R},\ x^2 = 1\}$$

Check­ing the answer#

Sub­sti­tut­ing $n = 1, \ldots, 5$ into $n^3$ and $2^n$ gives back the lists. Solv­ing $x^2 = 1$ gives $x = \pm 1$.

Answer#

(a) $\{x : x = n^3, n \in \mathbf{N}, n \le 5\}$; (b) $\{x : x = 2^n, n \in \mathbf{N}, n \le 5\}$; (c) $\{x : x \in \mathbf{R}, x^2 = 1\}$.

Ques­tion 4: Recog­nis­ing the empty set#

The prob­lem#

Which of these are empty? (a) $\{x \in \mathbf{N} : 3 < x < 4\}$; (b) $\{0\}$; (c) $\{x : x \text{ is an even prime}\}$.

Under­stand­ing the prob­lem#

The empty set has no ele­ments at all. A set con­tain­ing the num­ber $0$ is not empty, because $0$ is an ele­ment.

The idea#

For each set, try to find at least one ele­ment. If you can, it is not empty.

Step-by-step solu­tion#

Step 1. (a) Nat­ural num­bers are $1, 2, 3, 4, \ldots$. None lies strictly between $3$ and $4$. So the set is empty.

Step 2. (b) $\{0\}$ con­tains one ele­ment, the num­ber $0$. Not empty.

Step 3. (c) $2$ is even and prime (its only fac­tors are $1$ and $2$). So the set is $\{2\}$. Not empty.

Check­ing the answer#

$n(\text{a}) = 0$, $n(\text{b}) = 1$, $n(\text{c}) = 1$.

Answer#

Only the first set, $\{x \in \mathbf{N} : 3 < x < 4\}$, is empty.

Com­mon mis­take to avoid#

Con­fus­ing $\{0\}$ with $\varnothing$. The first has one ele­ment; the sec­ond has none.

Ques­tion 5: Finite or infi­nite#

The prob­lem#

Say whether each is finite or infi­nite: (a) the cir­cles through a given point; (b) the rivers of India; (c) $\{x \in \mathbf{R} : 0 < x < 1\}$.

Under­stand­ing the prob­lem#

A set is finite if count­ing its ele­ments would come to an end, even if the num­ber is large. Oth­er­wise it is infi­nite.

The idea#

Try to see whether you can always pro­duce another ele­ment. If you can, the set is infi­nite.

Step-by-step solu­tion#

Step 1. (a) Pick any point $C$ other than the given point $P$ as a cen­tre, and use radius $CP$. That cir­cle passes through $P$. There are infi­nitely many choices of $C$, so there are infi­nitely many cir­cles: infi­nite.

Step 2. (b) India has a large but def­i­nite num­ber of rivers; count­ing them would end. Finite.

Step 3. (c) The num­bers $\displaystyle \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \tfrac{1}{5}, \ldots$ all lie between $0$ and $1$, and this list never ends. Infi­nite.

Check­ing the answer#

In (a) and (c) we pro­duced an end­less sup­ply of ele­ments; in (b) we did not need to know the exact count, only that it is a def­i­nite num­ber.

Answer#

(a) infi­nite; (b) finite; (c) infi­nite.

Ques­tion 6: Test­ing equal­ity of sets#

The prob­lem#

Are these pairs equal?

(a) $\{x : x \text{ is a letter of "FOLLOW"}\}$ and $\{x : x \text{ is a letter of "WOLF"}\}$;
(b) $\{x \in \mathbf{Z} : x^2 \le 4\}$ and $\{-2, -1, 1, 2\}$.

Under­stand­ing the prob­lem#

Two sets are equal when they have exactly the same ele­ments. Order and rep­e­ti­tion do not mat­ter.

The idea#

Write each set in ros­ter form and com­pare ele­ment by ele­ment.

Step-by-step solu­tion#

Part (a)

Step 1. FOL­LOW has let­ters F, O, L, L, O, W; dis­tinct ones: $\{F, O, L, W\}$.

Step 2. WOLF has let­ters $\{W, O, L, F\}$.

Step 3. Both con­tain F, O, L, W and noth­ing else, so the sets are equal.

Part (b)

Step 1. Find the inte­gers with $x^2 \le 4$. Test: $(\pm 2)^2 = 4$, $(\pm 1)^2 = 1$, $0^2 = 0$ all work; $(\pm 3)^2 = 9$ does not.

$$\{x \in \mathbf{Z} : x^2 \le 4\} = \{-2, -1, 0, 1, 2\}$$

Step 2. This set con­tains $0$, but $\{-2, -1, 1, 2\}$ does not. So they are not equal.

Check­ing the answer#

In (b), count­ing also shows the dif­fer­ence: $5$ ele­ments against $4$.

Answer#

(a) Yes, both are $\{F, O, L, W\}$; (b) No, the first con­tains $0$ and the sec­ond does not.

Ques­tion 7: List­ing all sub­sets#

The prob­lem#

List all sub­sets of $\{-2, 0, 2\}$. How many are there?

Under­stand­ing the prob­lem#

A sub­set is any set whose ele­ments all come from $\{-2, 0, 2\}$. This includes the empty set and the whole set itself.

The idea#

List sub­sets by size ($0$, $1$, $2$ and $3$ ele­ments) so none is missed, then check the total against the rule: a set with $n$ ele­ments has $2^n$ sub­sets.

Step-by-step solu­tion#

Step 1. Size $0$: $\varnothing$.

Step 2. Size $1$: $\{-2\}, \{0\}, \{2\}$.

Step 3. Size $2$: $\{-2, 0\}, \{-2, 2\}, \{0, 2\}$.

Step 4. Size $3$: $\{-2, 0, 2\}$.

Step 5. Count: $1 + 3 + 3 + 1 = 8$.

Check­ing the answer#

The set has $3$ ele­ments, so it should have $2^3 = 8$ sub­sets. It does.

Answer#

$\varnothing, \{-2\}, \{0\}, \{2\}, \{-2, 0\}, \{-2, 2\}, \{0, 2\}, \{-2, 0, 2\}$: $8$ sub­sets.

Ques­tion 8: Using $\in$ and $\subset$ cor­rectly#

The prob­lem#

True or false: (a) $\{3\} \in \{1, 2, 3\}$; (b) $\{3\} \subset \{1, 2, 3\}$; (c) $\varnothing \subset \{0\}$; (d) $0 \in \varnothing$.

Under­stand­ing the prob­lem#

$\in$ ("is an ele­ment of") com­pares an object with a set. $\subset$ ("is a sub­set of") com­pares a set with a set. The num­ber $3$ and the set $\{3\}$ are dif­fer­ent things.

The idea#

For $\in$, look for the object among the listed ele­ments. For $\subset$, check that every ele­ment of the left set appears in the right set.

Step-by-step solu­tion#

Step 1. (a) The ele­ments of $\{1, 2, 3\}$ are the num­bers $1$, $2$, $3$. The set $\{3\}$ is not one of them. False.

Step 2. (b) The only ele­ment of $\{3\}$ is $3$, and $3 \in \{1, 2, 3\}$. So every ele­ment of $\{3\}$ is in $\{1, 2, 3\}$. True.

Step 3. (c) The empty set is a sub­set of every set: it has no ele­ment that could fail to be in $\{0\}$. True.

Step 4. (d) The empty set has no ele­ments, so $0$ can­not be one of them. False.

Check­ing the answer#

The cor­rect state­ments are $3 \in \{1, 2, 3\}$ and $\{3\} \subset \{1, 2, 3\}$; mix­ing them up is exactly what (a) tests.

Answer#

(a) False; (b) True; (c) True; (d) False.

Ques­tion 9: Inter­vals and set-builder form#

The prob­lem#

Write as inter­vals: (a) $\{x : -4 < x \le 6\}$; (b) $\{x : 3 \le x < 8\}$. (c) Write $[0, 1)$ in set-builder form.

Under­stand­ing the prob­lem#

Here $x$ is a real num­ber. A round bracket means the end­point is excluded ($<$ or $>$); a square bracket means it is included ($\le$ or $\ge$).

The idea#

Match each inequal­ity sign at each end to the right bracket.

Step-by-step solu­tion#

Step 1. (a) Left end: $-4 < x$, so $-4$ is excluded: round bracket. Right end: $x \le 6$, so $6$ is included: square bracket.

$$(-4, 6]$$

Step 2. (b) Left end $3 \le x$: included, square. Right end $x < 8$: excluded, round.

$$[3, 8)$$

Step 3. (c) $[0, 1)$ includes $0$ and excludes $1$.

$$[0, 1) = \{x : x \in \mathbf{R},\ 0 \le x < 1\}$$

Check­ing the answer#

$6 \in (-4, 6]$ but $-4 \notin$; $3 \in [3, 8)$ but $8 \notin$; $0 \in [0, 1)$ but $1 \notin$. All match the orig­i­nal descrip­tions.

Answer#

(a) $(-4, 6]$; (b) $[3, 8)$; (c) $\{x : x \in \mathbf{R}, 0 \le x < 1\}$.

Ques­tion 10: Size of a power set#

The prob­lem#

How many ele­ments has $P(A)$ if $A$ has $5$ ele­ments? How many ele­ments has $P(P(\varnothing))$?

Under­stand­ing the prob­lem#

$P(A)$, the power set, is the set of all sub­sets of $A$. If $A$ has $n$ ele­ments, $P(A)$ has $2^n$ ele­ments, because each ele­ment of $A$ is either in or out of a sub­set. The sec­ond part applies this twice, start­ing from the empty set.

The idea#

Use $n(P(A)) = 2^{n(A)}$. For the sec­ond part, first find $P(\varnothing)$, then its power set.

Step-by-step solu­tion#

Step 1. With $n(A) = 5$:

$$n(P(A)) = 2^5 = 32$$

Step 2. $\varnothing$ has $0$ ele­ments, so $P(\varnothing)$ has $2^0 = 1$ ele­ment. That one sub­set is $\varnothing$ itself.

$$P(\varnothing) = \{\varnothing\}$$

Step 3. $\{\varnothing\}$ has $1$ ele­ment, so its power set has $2^1 = 2$ ele­ments: the empty set and the whole set.

$$P(P(\varnothing)) = \{\varnothing,\ \{\varnothing\}\}$$

Check­ing the answer#

List­ing the sub­sets of $\{\varnothing\}$ directly gives just $\varnothing$ and $\{\varnothing\}$, which agrees with $2^1 = 2$.

Answer#

$P(A)$ has $32$ ele­ments; $P(P(\varnothing))$ has $2$ ele­ments, namely $\varnothing$ and $\{\varnothing\}$.

Com­mon mis­take to avoid#

Think­ing $\{\varnothing\}$ is empty. It is a set with one ele­ment, and that ele­ment hap­pens to be the empty set.