Exercise 1.3 answers: Sets

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(i) { 2, 3, 4 }

1. 2, 3 and 4 are all members of {1,2,3,4,5}.

Answer⊂

(ii) { a, b, c }

1. a is not a member of {b,c,d}.

Answer⊄

(iii) a student of Class XI of your school

1. Every Class XI student of your school is also a student of your school.

Answer⊂

(iv) a circle in the plane … radius 1 unit

1. A circle in the plane need not have radius 1 unit.

Answer⊄

(v) a triangle in a plane … a rectangle in the plane

1. No triangle is a rectangle.

Answer⊄

(vi) an equilateral triangle in a plane

1. Every equilateral triangle is, in particular, a triangle.

Answer⊂

(vii) an even natural number … an integer

1. Every even natural number is an integer.

Answer⊂

Watch this explained “One implication, about any member”, 0:53 into Containment, proper containment, and why the empty set is inside everything

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(i) { a, b } ⊄ { b, c, a }

1. a and b are both in {b,c,a}, so {a,b} IS a subset − the statement claims the opposite.

AnswerFalse

(ii) { a, e }

1. a and e are both vowels.

AnswerTrue

(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }

1. 2 is in {1,2,3} but not in {1,3,5}.

AnswerFalse

(iv) { a } ⊂ { a, b, c}

1. a is a member of {a,b,c}.

AnswerTrue

(v) { a } ∈ { a, b, c}

1. {a,b,c} has the members a, b and c − not the set {a} itself.

AnswerFalse

(vi) an even natural number less than 6

1. The even naturals less than 6 are {2,4}.
2. 36's divisors are 1,2,3,4,6,9,12,18,36, and both 2 and 4 are in that list.

AnswerTrue

Watch this explained “The thing, and the box holding it”, 6:38 into Containment, proper containment, and why the empty set is inside everything

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(i) {3, 4} ⊂ A

1. For {3, 4} ⊂ A, the numbers 3 and 4 would each have to be members of A.
2. The number 3 on its own is not one of A's four members; only the set {3, 4} is.

AnswerIncorrect: 3 ∈ {3, 4} but 3 ∉ A.

(ii) {3, 4} ∈ A

1. {3, 4} is written in A as one of its four members.

AnswerCorrect: {3, 4} ∈ A.

(iii) { {3, 4} } ⊂ A

1. { {3, 4} } is the set whose only member is {3, 4}.
2. That member, {3, 4}, is a member of A, so every member of { {3, 4} } is in A.

AnswerCorrect: { {3, 4} } ⊂ A.

(iv) 1 ∈ A

1. 1 is written in A as one of its members.

AnswerCorrect: 1 ∈ A.

(v) 1 ⊂ A

1. The symbol ⊂ is used only between two sets.
2. 1 is a number, not a set, so "1 ⊂ A" is wrongly written; the right statements are 1 ∈ A or {1} ⊂ A.

AnswerIncorrect: 1 is an element of A, so write 1 ∈ A.

(vi) {1, 2, 5} ⊂ A

1. 1, 2 and 5 are each members of A.

AnswerCorrect: {1, 2, 5} ⊂ A.

(vii) {1, 2, 5} ∈ A

1. A's four members are 1, 2, {3, 4} and 5; the set {1, 2, 5} is not one of them.
2. {1, 2, 5} is a subset of A, as (vi) shows, but not an element of A.

AnswerIncorrect: {1, 2, 5} ∉ A.

(viii) {1, 2, 3} ⊂ A

1. For {1, 2, 3} ⊂ A, the number 3 would have to be a member of A.
2. 3 on its own is not a member of A.

AnswerIncorrect: 3 ∈ {1, 2, 3} but 3 ∉ A.

(ix) φ ∈ A

1. φ is not one of A's four members 1, 2, {3, 4} and 5.

AnswerIncorrect: φ ∉ A.

(x) φ ⊂ A

1. The empty set has no members, so there is nothing in it that could fail to be in A.
2. So φ is a subset of every set, and of A in particular.

AnswerCorrect: φ ⊂ A.

(xi) {φ} ⊂ A

1. {φ} is the set whose only member is φ.
2. φ is not a member of A (see (ix)), so {φ} is not a subset of A.

AnswerIncorrect: φ ∈ {φ} but φ ∉ A.

Watch this explained “Eleven statements about one set”, 9:21 into Containment, proper containment, and why the empty set is inside everything

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(i) {a}

1. A 1-member set has 2¹ = 2 subsets: leave the member out, or keep it in.

Answerφ, {a}

(ii) {a, b}

1. A 2-member set has 2² = 4 subsets.

Answerφ, {a}, {b}, {a, b}

(iii) {1, 2, 3}

1. A 3-member set has 2³ = 8 subsets: the empty set, the three singles, the three pairs, and the whole set.

Answerφ, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}

(iv) φ

1. The empty set has exactly one subset − itself.

Answerφ

Watch this explained “All eight of them”, 11:49 into Containment, proper containment, and why the empty set is inside everything

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(i) – 4 < x ≤ 6

1. −4 is excluded (round bracket) and 6 is included (square bracket).

Answer(−4, 6]

(ii) – 12 < x < –10

1. Both ends are excluded.

Answer(−12, −10)

(iii) 0 ≤ x < 7

1. 0 is included, 7 is excluded.

Answer[0, 7)

(iv) 3 ≤ x ≤ 4

1. Both ends are included.

Answer[3, 4]

Watch this explained “Words into brackets”, 10:40 into The number systems as a chain of containments, and intervals as pieces of R

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(i) (– 3, 0)

1. Round brackets on both sides mean both ends are excluded.

Answer{x : x ∈ R, −3 < x < 0}

(ii) [6 , 12]

1. Square brackets on both sides mean both ends are included.

Answer{x : x ∈ R, 6 ≤ x ≤ 12}

(iii) (6, 12]

1. 6 is excluded (round bracket), 12 is included (square bracket).

Answer{x : x ∈ R, 6 < x ≤ 12}

(iv) [–23, 5)

1. −23 is included, 5 is excluded.

Answer{x : x ∈ R, −23 ≤ x < 5}

Watch this explained “What a bracket actually is”, 6:31 into The number systems as a chain of containments, and intervals as pieces of R

Open NCERT p. 13Checked by computerAnswers can differ: one example

(i) The set of right triangles.

1. Every right triangle is a triangle, so the set of all triangles contains every member of this set.
2. This is not the only answer: any larger set, such as the set of all polygons, would also do.

AnswerThe set of all triangles (in the plane).

(ii) The set of isosceles triangles

1. Every isosceles triangle is a triangle, so the set of all triangles again contains every member.
2. Again a larger set, such as the set of all polygons, is also correct.

AnswerThe set of all triangles (in the plane).

Watch this explained “Two families of triangle”, 8:43 into Why nothing can be complemented until the surrounding set is fixed

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(i) {0, 1, 2, 3, 4, 5, 6}

1. C contains 8, but this candidate stops at 6.

AnswerNo

(ii) φ

1. The empty set cannot hold even a single member of A, B or C.

AnswerNo

(iii) {0,1,2,3,4,5,6,7,8,9,10}

1. Every member of A, B and C (0 to 8) lies inside 0 to 10.

AnswerYes

(iv) {1,2,3,4,5,6,7,8}

1. C contains 0, but this candidate starts at 1.

AnswerNo

Watch this explained “Naming the one that escapes”, 6:17 into Why nothing can be complemented until the surrounding set is fixed