Sets

Here is what you already know about sets from Grade 8, in one place.

Idea	What it means	Example
Set	a well-defined collection of things, its elements	the days of the week
Listing method	the elements written inside curly brackets	A = {1, 3, 5}
Set-builder method	a rule that says which things belong	A = {x:x is an odd number less than 6}
Is an element of	∈ means “belongs to”, ∉ “does not belong to”	3 ∈ A, 4 ∉ A
Empty set	a set with no elements, written {} or ∅	the months with 35 days
Universal set	everything being talked about, U	all the students of your class
Subset	every element of A is also in B	{1, 3} ⊂ {1, 3, 5}
Equal sets	exactly the same elements	{1, 3, 5} = {5, 1, 3}
Overlapping sets	some elements in common, not all	{1, 2, 3} and {3, 4}
Disjoint sets	no element in common	{1, 2} and {7, 8}

The order of the elements doesn’t matter, and an element is written only once: {b, a, b} is just {a, b}.

From two sets you can make new ones: everything in either, only what they share, what one has and the other doesn’t, and everything left out. These four ways are the operations on sets.

1.1.1 Union of sets

In Figure 1.3, the colors collected by Supriya or Anjila or both are everything inside the two circles:

S ∪ A = {red, yellow, blue, pink, white, purple, orange}

Red, yellow and blue are in both sets, but they are written once. This set is called the union of the two sets.

The union of two sets A and B, written A ∪ B and read “A union B”, is the set of all the elements that are in A or in B or in both. In set-builder form,

$$A \cup B = \{x : x \in A \text{ or } x \in B\}$$

In a Venn diagram, A ∪ B is everything inside the two circles (Figure 1.4).

The union of three sets works the same way: A ∪ B ∪ C = {x:x ∈ A or x ∈ B or x ∈ C}, everything inside any of the three circles (Figure 1.5). However many sets you join, every element is written once.

1.1.2 Intersection of sets

Some students of Grade 9 were asked which subjects they like. The students who like Science form the set S and those who like Mathematics the set M:

S = {Pemba, Rojina, Abdul, Srijana, Tenzing, Binaya}
M = {Srijana, Tenzing, Manju, Dolma, Kushal, Ajima}

Srijana and Tenzing are in both sets. The set of students who like both subjects, {Srijana, Tenzing}, is the intersection of the two sets.

The intersection of two sets A and B, written A ∩ B and read “A intersection B”, is the set of the elements that are in both A and B. In set-builder form,

$$A \cap B = \{x : x \in A \text{ and } x \in B\}$$

In a Venn diagram it is the overlap of the circles (Figure 1.11). A way to remember the symbols: ∪ is a cup that holds everything poured into it; ∩ is an arch over only the shared part. And the words: the union needs or, the intersection and.

1.1.3 Difference of sets

The students who read a newspaper but don’t watch television are P with T’s members taken away. That set is called the difference of P and T, written P − T. In the same way, the students who watch television only make T − P. The two are different sets.

If A and B are subsets of a universal set U, the difference A − B, read “A difference B”, is the set of the elements that are in A but not in B. Likewise B − A is the set of the elements in B but not in A. In set-builder form,

$$A - B = \{x : x \in A \text{ and } x \notin B\} \qquad B - A = \{x : x \in B \text{ and } x \notin A\}$$

A − B and B − A in Venn diagrams

How the difference looks depends on how the two sets sit.

A few differences are worth knowing by heart, each one clear from the pictures:

• If A and B are equal sets, A − B = B − A = ∅: nothing is left.
• Taking nothing away leaves the set as it was: A − ∅ = A.
• Nothing minus anything is nothing: ∅ − A = ∅.
• If A is a proper subset of B, then A − B = ∅: every element of A is taken away.
• If A and B are disjoint, A − B = A and B − A = B.

The union of the two differences, (A − B) ∪ (B − A), is called the symmetric difference of A and B, written $A \,\Delta\, B$: the elements in exactly one of the two sets, everything except the overlap.

1.1.4 Complement of a set

Let A be a proper subset of a universal set U. The set of all the elements of U that are not in A is called the complement of A. It is written A (read “A bar”), or A′, or $A^c$; all three are used, so recognize each.

The complement is a difference: what is left of U once A is taken away.

$$\overline{A} = U - A = \{x : x \in U \text{ and } x \notin A\}$$

In a Venn diagram, A is everything in the rectangle outside A’s circle (Figure 1.24). So a set and its complement together make up U, and the complement of the complement brings you back:

$$A \cup \overline{A} = U \qquad \overline{\overline{A}} = A$$

That is why A and A are called complements of each other. Discuss in your group: what is A ∩ A? (Example 13 answers it.)

A complement always depends on U. The complement of “the boys in your class” is “the girls in your class” if U is your class, but “everyone in the school who isn’t a boy of your class” if U is the school. So a question that asks for a complement always says what U is.

Every set made with ∪, ∩, − and bars is a choice of regions in the Venn diagram. Shade the regions first, and the listing is just the elements in them.

Back to the mountains

Now the puzzle of the opening has an exact answer. Let U be the set of the 14 mountains over 8,000 m, N those in Nepal and C those in China. A mountain on the border is in both sets.

Nepal and China share four eight-thousanders: N ∩ C = {Everest, Lhotse, Makalu, Cho Oyu}.

The other operations answer the other questions:

• N − C = {Kangchenjunga, Dhaulagiri I, Manaslu, Annapurna I}: the four that are Nepal’s and not China’s.
• C − N = {K2, Gasherbrum I, Broad Peak, Gasherbrum II, Shishapangma}: the five that are China’s and not Nepal’s. Shishapangma stands wholly in Tibet, just north of Nepal’s Langtang; the other four are on China’s border with Pakistan, in the Karakoram.
• N ∪ C is every peak in either country: 13 of the 14.
• N ∪ C = {Nanga Parbat}: the one in neither.

So 8 + 9 = 17 counted the four border peaks twice, once in each list. Section 1.2 turns that into a rule you can use without listing.

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The number of elements in a set is called its cardinality, or its cardinal number. For A = {m, a, t, h}, the cardinality is 4, written n(A) = 4 and read “n of A is 4”.

An empty set has cardinality 0. If B is the set of Grade 9 students younger than 5, there is no such student, so n(B) = 0.

Section 1.1 was about which elements; this section is about how many. Counting is often the real question: how many households, how many voters, how many peaks.

That second answer is the most useful rule in this unit. If A and B overlap,

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

because n(A) + n(B) counts the overlap twice, once in each circle, and subtracting n(A ∩ B) once leaves every element counted exactly once.

“M only” is the difference M − N, so

$$n_0(M) = n(M - N) = n(M) - n(M \cap N) \qquad n_0(N) = n(N - M) = n(N) - n(M \cap N)$$

In Activity 3: n₀(M) = 5 − 2 = 3 and n₀(N) = 6 − 2 = 4. The three regions inside the circles add up to the union: n(M ∪ N) = n₀(M) + n₀(N) + n(M ∩ N) = 3 + 4 + 2 = 9.

If one set is a proper subset of another, A ⊂ B, then A’s circle sits inside B’s, and

$$n(A \cup B) = n(B) \qquad n(A \cap B) = n(A)$$

Finding cardinality: the formulas

Compare the two Venn diagrams Figure 1.50 and Figure 1.51. For each, find n(U), n(P), n(Q), n(P ∪ Q), n(P ∩ Q) and n(P ∪ Q).

	Figure 1.50: disjoint	Figure 1.51: overlapping
n(U)	7	12
n(P)	3	5
n(Q)	4	6
n(P ∪ Q)	7	9
n(P ∩ Q)	0	2
n(P ∪ Q)	0	3
so	n(U) = n(P ∪ Q)	n(U) = n(P ∪ Q) + n(P ∪ Q)

The two tables of numbers give all the rules. For two sets A and B:

• a.If A and B are disjoint: n(A ∪ B) = n(A) + n(B).
• b.If A and B overlap: n(A ∪ B) = n(A) + n(B) − n(A ∩ B), or n(A ∪ B) = n₀(A) + n₀(B) + n(A ∩ B).
• c.If U holds nothing outside A and B: n(U) = n(A ∪ B).
• d.If U has more elements than A and B hold: n(A ∪ B) = n(U) − n(A ∪ B), that is, n(U) = n(A ∪ B) + n(A ∪ B).
• e.For overlapping A and B: n₀(A) = n(A) − n(A ∩ B) and n₀(B) = n(B) − n(A ∩ B).

Rule a is rule b with n(A ∩ B) = 0: one rule, not two to learn.

Back to the mountains, counted

The opening’s 8 + 9 = 17 is n(N) + n(C), which counts the four border peaks twice. The rule fixes it:

$$n(N \cup C) = n(N) + n(C) - n(N \cap C) = 8 + 9 - 4 = 13$$

13 of the 14 eight-thousanders are in Nepal or China or both, and n(N ∪ C) = 14 − 13 = 1: Nanga Parbat. And n₀(N) = 8 − 4 = 4 are Nepal’s alone.

The rule also works backwards. If you know only that Nepal has 8, China 9, and that 13 are in one country or the other, then n(N ∩ C) = 8 + 9 − 13 = 4: the number on the border, without naming a single peak.

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Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10 (2078). There, Sets is always one question of 6 marks, in four parts at four levels:

Part	Level	Marks	What it asks
a.	knowledge	1	a cardinal number, a symbol or a formula
b.	understanding	1	show the sets in a Venn diagram
c.	application	3	list sets to find, or prove, a relation
d.	higher ability	1	the relation between two sets, with a reason

Here is one in that shape, with how it is marked.