🔵 Sets — Revision Notes

Walkthrough 1 — listing elements, and n(A) Basic

1. P = {4, 5, 6, 7, 8, 9, 10, 11}
   4 ≤ x means 4 is allowed in; x < 12 means 12 is not. The two ends of the inequality are read separately, and only one of them is inclusive here. Braces, commas, ascending order.
2. Q = {5, 6, 7, 8, 9, 10, 11, 12}
   "Between 5 and 12 inclusive" puts both endpoints in, so the list starts at 5 and ends at 12. Without the word "inclusive" this set would have been {6, 7, 8, 9, 10, 11}.
3. n(P) = 8 and n(Q) = 8, so n(P) = n(Q).
   Count the elements in each list. n(…) is a number, so these two answers are bare numbers with no braces. Both sets happen to be runs of eight consecutive integers, which is why the counts agree.
4. No. P ≠ Q, because 4 ∈ P but 4 ∉ Q.
   Equal sets contain exactly the same elements, so one element in P and not in Q settles it. Naming it with ∈ and ∉ is the form the mark is for; 12 ∈ Q but 12 ∉ P would serve equally well.

Walkthrough 2 — distinct elements, and true or false Basic

1. S = {B, A, L, O, N}
   BALLOON has seven letters but only five distinct ones: the second L and the second O are already in the set. The letters are kept in the order the word supplies them.
2. n(S) = 5
   Count what is actually inside the braces — five elements, so five. Counting the letters of the word instead gives 7, and that is the whole point of the question.
3. (a) O ∈ S is true. (b) T ∉ S is true.
   O appears in the list, so it is an element. T does not appear anywhere in BALLOON, so "T is not an element" is a true statement — a ∉ claim is true exactly when the letter is missing.
4. (c) {B} ∈ S is false. The elements of S are letters, not sets; the correct statement is {B} ⊆ S (or B ∈ S).
   ∈ connects an element to a set; ⊆ connects a set to a set. {B} is a set, and it is not one of the five letters inside S — but every element of {B} is in S, which is what ⊆ says.

Walkthrough 3 — empty sets and equal sets Intermediate

1. CRYPT contains no A, E, I, O or U, so D = Ø.
   Check the five vowels a, e, i, o, u one at a time against the word; Y is not one of them. Nothing qualifies, so the set is empty and the answer is the symbol Ø.
2. Every multiple of 4 is 4, 8, 12, …, and each has 2 as a factor, so none is prime. E = Ø.
   A prime has exactly two factors, 1 and itself. Any multiple of 4 is also a multiple of 2, so it has at least three factors and cannot be prime. The condition is impossible to satisfy, which is what makes the set empty.
3. 2 is even and 2 is prime, so F = {2}, and n(F) = 1.
   This is the description that looks impossible and is not. 2 is the only even prime, so F has exactly one element — a set with one element, written with braces, not the bare number 2.
4. D and E are the empty sets, and D = E.
   Both contain no elements at all, so they contain exactly the same elements — none — and that is the definition of equal sets. All empty sets are equal, however differently they were described.
5. No. E ≠ F, because E is an empty set whereas F is not: 2 ∈ F but 2 ∉ E.
   The explanation is the mark. Say which one is empty and which is not, and name the element that separates them. n(E) = 0 while n(F) = 1, so they cannot possibly be equal.

Walkthrough 4 — universal set and complement Basic

1. ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
   Write the universal set out in full before anything else. "Between 1 and 12 inclusive" keeps both endpoints, so the list runs 1 to 12. Everything later in the question — and especially A′ — is measured against this list.
2. A = {3, 6, 9, 12}
   The multiples of 3 that are in ξ. 15 and 18 are multiples of 3 but are not under consideration here, so they are not elements of A — a set is cut down to the universal set it lives in.

3. Fig. 5.2

   Draw the rectangle, put the circle inside it, and place the four multiples of 3 in the circle. The other eight elements of ξ go inside the rectangle but outside the circle, which is where the next part reads A′ from.
4. A′ = {1, 2, 4, 5, 7, 8, 10, 11}
   Read the eight elements straight off the region outside the circle. This is the answer the diagram was drawn for, and it is a set, so it needs braces.
5. n(ξ) = 12, n(A) = 4, n(A′) = 8
   Three numbers, no braces. They are worth checking against each other: 4 + 8 = 12, which is n(A) + n(A′) = n(ξ), so nothing has been dropped or double-counted.

Walkthrough 5 — subset and proper subset Intermediate

1. X = {P, O, R, T, A, B, L, E} and Y = {P, O, R, T}
   List both sets before drawing anything. PORTABLE has eight letters, all different, so n(X) = 8; PORT has four, so n(Y) = 4. Every letter of Y appears in X, which is what the picture will have to show.

2. Fig. 5.4

   Every element of Y is an element of X, so the circles are drawn one inside the other rather than overlapping. Each common letter is written once, inside Y, and the other four letters of X go in the ring.
3. Yes. Y ⊂ X, because every element of Y is an element of X and Y ≠ X.
   A proper subset needs both halves stated. The first half is the containment; the second is what makes it proper, and it is true here because A ∈ X but A ∉ Y, so the two sets are not equal.
4. Yes. X ⊈ Y, because there are elements in X — for example A — that are not elements of Y.
   One escaping element settles "is not a subset of", so naming A is enough; B, L or E would serve equally well. Note the direction: Y ⊆ X is true while X ⊆ Y is false.

Walkthrough 6 — listing every subset Basic

1. n(T) = 3, so T has 2³ = 8 subsets.
   Work out how many there should be before listing them. Each of the three elements is either in a subset or out of it, giving 2 × 2 × 2 choices. Now the list has a target to hit.
2. With no elements: Ø
   The empty set is a subset of every set, since it has no element that could fail to be in T. It is the easiest one to leave out, so write it first.
3. With one element: {2}, {5}, {9}
   One subset per element, each in its own braces. {2} is a set containing the number 2, not the number itself — the braces are the difference between an element and a subset.
4. With two elements: {2, 5}, {2, 9}, {5, 9}
   Take the elements in order and pair each with every later one, so nothing is repeated and nothing is missed. Three pairs from three elements.
5. Ø, {2}, {5}, {9}, {2, 5}, {2, 9}, {5, 9}, {2, 5, 9}
   With all three elements comes T itself, since every set is a subset of itself, and 1 + 3 + 3 + 1 = 8 matches the 2³ predicted at the start. The proper subsets would drop that last one, leaving seven.

Walkthrough 7 — intersection and union of two listed sets Basic

1. C ∩ D = {e, g}
   Go through D one element at a time and ask whether it also appears in C: e yes, g yes, i no, k no. Only the elements in both lists survive, and the answer is a set, so it is written in braces.

2. Fig. 5.9

   The overlap is filled first, with the two elements just found. Everything else in C then goes in the left-hand part and everything else in D in the right-hand part. Nothing is written twice, which is exactly what a Venn diagram is for.
3. C ∪ D = {c, d, e, f, g, h, i, k}
   Read every element inside either circle straight off the diagram, working left to right: the four in C only, then the two in the overlap, then the two in D only. The shared elements e and g appear once.
4. n(C ∪ D) = 8
   Count what is in the braces: 4 + 2 + 2 = 8. Notice that n(C) + n(D) = 10 is too big by 2, exactly n(C ∩ D), because e and g were counted twice.

Walkthrough 8 — sets given by a rule, with a universal set Intermediate

1. G = {1, 2, 3, 4, 6, 8, 12, 24} and H = {4, 8, 12, 16, 20, 24}
   List both sets inside ξ = {1, 2, 3, …, 24}. The factors of 24 come in pairs — 1×24, 2×12, 3×8, 4×6 — which guarantees none is missed. The multiples of 4 stop at 24, because 28 is not in ξ.
2. G ∩ H = {4, 8, 12, 24}
   The overlap is worked out before the diagram is drawn. These are the numbers that are both factors of 24 and multiples of 4 — and 16 and 20 are not among them, because neither divides 24 exactly.

3. Fig. 5.10

   The overlap first, then the rest of each circle, then everything left over inside the rectangle. That last group is the one the next line reads its answer from.
4. (G ∪ H)′ = {5, 7, 9, 10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23}
   The bracket says: take the union first, then complement it. So this is everything outside both circles — read straight off the bottom of the rectangle. There are 14 of them, and 10 + 14 = 24 = n(ξ) confirms nothing was dropped.
5. G ∩ H′ = {1, 2, 3, 6}
   "In G and not in H" is the part of the G circle lying outside H — the left-hand region, read off in one glance. By lists it would mean writing out H′, eighteen elements, and then intersecting.

Walkthrough 9 — when one set is inside the other, and when they share nothing Intermediate

1. J = {10, 20, 30, 40} and K = {5, 10, 15, 20, 25, 30, 35, 40}
   List them both. Every multiple of 10 is also a multiple of 5, so the whole of J is going to reappear inside K — which is the structure the question is really about.
2. J ⊂ K, so J ∩ K = J = {10, 20, 30, 40}.
   Every element of J is an element of K, and J ≠ K, so it is a proper subset. When one set sits inside the other, the overlap is all of the smaller set.
3. J ∪ K = K = {5, 10, 15, 20, 25, 30, 35, 40}
   The union adds the elements of J to K, but they are already there, so nothing changes. Together with the line above, this is the general rule: if J ⊆ K then J ∩ K = J and J ∪ K = K.
4. L = {1, 3, 5, 7, 9} and M = {2, 4, 6, 8}
   "Between 0 and 10" excludes both endpoints, so L stops at 9 and M at 8. No number is both odd and even, so the two lists have nothing in common.
5. L ∩ M = Ø — L and M are disjoint sets.
   An empty intersection is written Ø. The answer 0 belongs to a different question, n(L ∩ M), because 0 is a number. On a diagram, disjoint sets are two separate circles that do not touch.

Walkthrough 10 — shade A ∪ B′ Basic

1. Fig. 5.12

   A is the first set named, and it is made of two regions: the part of A outside B, and the overlap. Ticking the overlap matters — the elements in both sets are still in A.

2. Fig. 5.13

   B′ is everything not in B: the part of A outside B, and the region outside both circles. So the left region now carries two ticks and the part of B outside A carries none.

3. Fig. 5.14

   Everything is shaded except the part of B outside A.
   The operator is ∪, so at least one tick is enough. Three of the four regions qualify; the part of B lying outside A has no tick, so it stays blank. Rub the ticks out at the end.

Walkthrough 11 — shade A′ ∩ B Intermediate

1. Fig. 5.16

   Start with A′, the first set named: everything outside circle A. That is two regions — the part of B outside A, and the region outside both circles. The overlap is inside A, so it gets no tick.

2. Fig. 5.17

   B is the overlap and the part of B outside A. Only that second region already carried a tick, so only it now has two. The overlap has one; the outside region has one; the part of A outside B still has none.

3. Fig. 5.18

   Only the part of B that lies outside A is shaded.
   The operator is ∩, so a region needs two ticks, not one. Shading the whole of B would answer B, and shading the overlap would answer A ∩ B; what is wanted is "in B but not in A".

Walkthrough 12 — shade (A ∪ B)′, and a surprise Intermediate

1. Fig. 5.19

   The bracket has to be built first. A ∪ B is everything inside either circle, which is three of the four regions — left, overlap and right. Only the region outside both circles is left untouched.

2. Fig. 5.20

   The prime sits outside the bracket, so it complements the whole union: shade everything the union did not reach. That is the single region outside both circles — the elements that are in ξ but in neither set, which every real-world question calls "neither".

3. Fig. 5.21

   Starting again on a fresh diagram: A′ is the right region and the outside; B′ is the left region and the outside. The outside is the only region ticked twice, because it is the only one that is outside both circles.
4. Both parts shade the same region, so (A ∪ B)′ = A′ ∩ B′.

   Fig. 5.22

   ∩ needs two ticks and only the outside region has two, so part (ii) shades exactly what part (i) shaded: "not in either" and "not in A and not in B" say the same thing. Moving the prime inside the bracket flips the operator.

Walkthrough 13 — describing a region in one piece Intermediate

1. Is the shaded region inside A or outside A? — inside A.
   The whole of the shading lies within circle A, so A is one of the two ingredients. Ask about one circle at a time; trying to read the whole expression off at once is what produces guesses.
2. Is the shaded region inside B or outside B? — outside B, so B′.
   None of the shading reaches into circle B — the overlap is deliberately left white. "Outside B" is written B′, and that prime is the whole reason this region is not just A.
3. Both conditions hold at the same time, so the operator is ∩.
   Every shaded point is inside A and outside B — both at once, not one or the other. "And" is ∩. Had it been "or", the shading would have spilled into the region outside both circles.
4. The shaded region is A ∩ B′.
   Read it back as a check: "in A and not in B" is the part of circle A with the overlap taken out. A alone would have included the overlap.

Walkthrough 14 — describing a region in two pieces Advanced

1. Inside A or outside? Both appear — part of the shading is inside A and part is outside it. So this is not a single ∩.
   Question 1 has no single answer here, which is the signal that the region comes in two pieces. Rather than force one expression, name each piece separately and join them at the end.
2. Left piece: inside A, outside B → A ∩ B′.

   Fig. 5.25

   Treat the left piece as its own little question and run the three questions on it. Inside A, outside B, both at once — which is the region named in Walkthrough 13.
3. Right piece: outside A, inside B → A′ ∩ B.

   Fig. 5.26

   The same three questions on the other piece, and the answers are the mirror image. Naming the two pieces in the same order as the diagram — left then right — keeps the primes in the right places.
4. The shaded region is (A ∩ B′) ∪ (A′ ∩ B).
   A point is shaded if it is in the left piece or the right piece, so the two are joined by ∪, each keeping its brackets so its intersection is worked out first. In words: "in exactly one of the two sets".

Walkthrough 15 — filling the four regions Basic

1. Let C be the set of residents who own a bicycle and S the set of residents who own a scooter.
   Define the sets in one line before drawing. It costs five seconds and it is what turns a wordy question into a diagram — and if the question later asks for something in set notation, the letters are already there.

2. Fig. 5.27

   The middle first. "Own both" is n(C ∩ S) = 7, the only number here that belongs to a single region as it stands; every other given number is a total with the overlap buried in it.
3. Bicycle only: 19 − 7 = 12.    Scooter only: 23 − 7 = 16.
   The 19 counts everyone with a bicycle, including the 7 who also have a scooter, so the part of the circle outside the overlap holds 19 − 7. Same for the 23; writing 19 and 23 straight into the circles would double-count the 7.

4. Fig. 5.28

   Neither: 45 − 12 − 7 − 16 = 10.
   The three regions inside the circles hold 12 + 7 + 16 = 35 residents, so the rest of the group — the region inside the rectangle but outside both circles — holds 45 − 35. That region is the answer to "neither".
5. 10 residents own neither.
   Look back: 12 + 7 = 19 bicycles ✓, 7 + 16 = 23 scooters ✓, 12 + 7 + 16 + 10 = 45 residents ✓. All three agree with the question, and the answer is given in words about residents.

Walkthrough 16 — finding the overlap with x Intermediate

1. Let M be the set of members who read mysteries and B the set who read biographies. Let x be the number who read both.
   Name the sets, then name the unknown. x is put on the one region no given number reaches directly — the overlap — because every other region can then be written in terms of it.

2. Fig. 5.29

   The 31 mystery readers split into the x who read both and the 31 − x who read only mysteries; the 28 split the same way, and the 6 go outside both circles. All four regions are now labelled.
3. (31 − x) + x + (28 − x) + 6 = 52
   The four regions are disjoint and together they are the whole club, so their sum is the total membership. Written out in full, this line is where the method mark sits.
4. 65 − x = 52
   Collecting terms: 31 + 28 + 6 = 65, and −x + x − x leaves one −x. The x in the middle cancels one of the two subtractions, which is exactly the double-counting being undone.
5. x = 65 − 52 = 13 members read both.
   Check on the diagram: only mysteries 31 − 13 = 18, only biographies 28 − 13 = 15, and 18 + 13 + 15 + 6 = 52 ✓. The question asked for a number of members, so the answer says so.

Walkthrough 17 — three sets, filled from the centre Advanced

1. Fig. 5.30

   Let N, R and S be the sets of students buying noodles, rice and satay. The centre is the only region a given number names outright, so it is filled first, exactly as the overlap was in the two-set case.
2. Noodles and rice only: 22 − 7 = 15.   Noodles and satay only: 16 − 7 = 9.   Rice and satay only: 18 − 7 = 11.
   "22 bought noodles and rice" counts everyone in both circles, including the 7 who also bought satay, so the region in N and R but not S holds 22 − 7. The same subtraction handles the other two pairs.
3. Noodles only: 55 − 15 − 9 − 7 = 24.   Rice only: 51 − 15 − 11 − 7 = 18.   Satay only: 39 − 9 − 11 − 7 = 12.
   Each circle is made of four regions, three of which are now known, so the fourth is the given total minus those three. Subtracting only the centre would leave the two pair regions inside the "only" count.

4. Fig. 5.31

   Inside the circles: 24 + 15 + 18 + 9 + 7 + 11 + 12 = 96.
   Add the seven regions that lie inside at least one circle. This total is n(N ∪ R ∪ S), obtained without any formula — just by adding disjoint regions, so nothing can be double-counted.
5. None of the three: 120 − 96 = 24 students.
   The eighth region is whatever the rectangle has left. Check: 24 + 15 + 9 + 7 = 55, 18 + 15 + 11 + 7 = 51, 12 + 9 + 11 + 7 = 39, and all eight regions total 120 ✓.

Walkthrough 18 — largest and smallest possible values Advanced

1. n(A) + n(B) = 12 + 21 = 33, and 33 > 30, so A and B cannot be disjoint.
   Test this comparison first. With no common element the two sets would need 33 different elements between them, and ξ has only 30 to give, so they overlap in at least 3.
2. n((A ∪ B)′) is largest when n(A ∪ B) is smallest.
   The complement and the union share out the 30 elements of ξ, so one is big exactly when the other is small. Writing this line turns the question into one about the union, which the two extremes describe directly.
3. The union is smallest when A ⊆ B, giving n(A ∪ B) = n(B) = 21.
   Pushing the smaller circle wholly inside the larger wastes no room: every one of A's 12 elements is already one of B's 21, so the union is just B. This is allowed because 12 ≤ 21.
4. Largest possible value of n((A ∪ B)′) = 30 − 21 = 9
   n((A ∪ B)′) = n(ξ) − n(A ∪ B), the complement rule from 5.2. Putting the smallest union into it gives the largest complement.
5. The union is largest when A ∪ B = ξ, so n(A ∪ B) = 30 and the smallest possible value of n((A ∪ B)′) is 30 − 30 = 0.
   Since the two sets cannot be disjoint, the largest the union can get is the whole of ξ: 12 + 21 = 33 covers 30 elements with 3 to spare in the overlap, so nothing is left outside and the complement is empty.