Sets

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 1.

Can you list "the five best cricketers"?

Ask ten friends for the five best cricketers ever and you get ten different lists. Now ask for the prime numbers below 20 and every friend gives the same eight: 2, 3, 5, 7, 11, 13, 17, 19. The difference between those two questions is the whole idea of this chapter. A set is a collection where membership is never a matter of opinion. Relations, functions, probability and even the filters on a shopping site are written in this language.

What counts as a set

A set is a well-defined collection of objects. Well-defined means that for any object you can decide, without asking anyone's opinion, whether it belongs. The objects are called elements or members. Sets get capital letters (A, B, X) and elements get small letters (a, b, x). The symbol $\in$ reads "belongs to" and $\notin$ reads "does not belong to". If V is the set of vowels, then $a \in V$ and $b \notin V$.

Some sets are used so often that they have fixed names. Learn them now, because every later chapter assumes them.

They nest inside each other: $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$ and $T \subset \mathbb{R}$. Note that $\mathbb{N}$ is not a subset of $T$.

Two ways to write a set

In roster form you list every element inside braces, separated by commas. Order does not matter and nothing is listed twice, so the letters of SCHOOL give $\{S, C, H, O, L\}$ and the letters of FOLLOW give $\{F, O, L, W\}$. Dots show an obvious pattern that goes on: $\{1, 3, 5, \dots\}$.

In set-builder form you state the property that every element has and no outsider has. The colon reads "such that" and the braces read "the set of all". The two forms describe the same set, so you must be able to move between them in both directions.

Worked example: from roster to set-builder. Write $\{2, 5, 10, 17, 26\}$ in set-builder form.

Common wrong step. Writing $\{x : x = n^2 + 1,\ n \in \mathbb{N}\}$ and stopping. Without "$n \le 5$" the rule keeps producing 37, 50, … and describes an infinite set.

Kinds of sets

Before comparing sets we need names for the common types.

Here $n(S)$ means the number of distinct elements of S, so $n(\varnothing) = 0$. Equal sets are always equivalent, but equivalent sets need not be equal.

The three symbols below look alike and are different. Picture a set as a box.

Pause and predict: how many elements does $\{\varnothing, \{\varnothing\}\}$ have?

Check your answer: two. The first element is the empty set and the second is the set that contains the empty set. They are different objects, so both count.

Subsets: boxes inside boxes

A is a subset of B, written $A \subset B$, when every element of A is also an element of B. If $A \subset B$ and $A \ne B$, then A is a proper subset and B is a superset of A. Two facts follow at once: every set is a subset of itself, and $\varnothing \subset A$ for every A.

Why the empty set is a subset of everything. A set fails to be a subset of B only if you can find one of its elements missing from B. The empty set has no elements, so it can never fail. To prove two sets equal, show both directions: $A = B$ exactly when $A \subset B$ and $B \subset A$.

The symbol $\in$ links an object to a set, while $\subset$ links two sets. Mixing them up is the most common error in this chapter. Take $A = \{1, 2, \{3, 4\}, 5\}$. Here $\{3, 4\}$ is one single element, like a bag sitting among the numbers.

The power set. The set of all subsets of A is the power set $P(A)$. For $A = \{a, b\}$ the subsets are $\varnothing, \{a\}, \{b\}, \{a, b\}$, so $P(A)$ has 4 members. To build a subset you ask each element "in or out?". That is 2 choices for each of the n elements, so there are $2 \times 2 \times \dots \times 2 = 2^n$ subsets.

$$n(P(A)) = 2^n, \qquad \text{proper subsets} = 2^n - 1$$

Worked example: counting subsets. A has 5 elements. Find the number of subsets, proper subsets and non-empty proper subsets.

Common wrong step. Writing $5^2 = 25$ or $2 \times 5 = 10$. These agree with $2^n$ only for tiny n (try n = 2, where $n^2 = 4 = 2^n$), which is why the error survives in practice. The count grows by doubling with each new element.

Intervals: subsets of the real line

Subsets of $\mathbb{R}$ that run without gaps are written as intervals. A round bracket leaves the end point out, a square bracket keeps it in. On the number line an open circle means excluded and a filled circle means included.

Each of the four has length $b - a$. The whole line is $(-\infty, \infty) = \mathbb{R}$, and infinity always gets a round bracket because it is not a number you can reach.

The universal set and the four operations

Every discussion happens inside a universal set U, the big set from which all the sets in that problem are drawn. When you talk about even numbers, $U$ might be $\mathbb{N}$. In a Venn diagram (after John Venn, 1834 to 1883) U is a rectangle and each set is a circle inside it.

"Or" in mathematics is inclusive, and common elements are counted once in a union. If $A \cap B = \varnothing$ the sets are disjoint. If $B \subset A$ then $A \cup B = A$ and $A \cap B = B$. Also $A' = U - A$, $(A')' = A$, $\varnothing' = U$ and $U' = \varnothing$.

The key picture for two overlapping sets is that U splits into four regions with no overlap. Keep this table in your head, because every survey problem is a question about one of these regions.

The first three together make up $A \cup B$, and they are pairwise disjoint. Also $A - B \ne B - A$ in general, since they are the "only A" and "only B" regions.

The laws of set algebra

Sets obey a tidy list of laws. They let you simplify an expression without drawing anything.

Notice that both distributive laws hold for sets. Numbers are not so generous: $a + bc \ne (a + b)(a + c)$ in general. So never assume that a set law fails just because its number twin does.

De Morgan's laws tell you how a complement passes through a union or an intersection:

$$(A \cup B)' = A' \cap B', \qquad (A \cap B)' = A' \cup B'$$

Why they hold. Take any element x. Then $x \in (A \cup B)'$ means x is not in "A or B", which means x is not in A and not in B, which means $x \in A' \cap B'$. In plain words, "not (A or B)" is "not A and not B". The second law is the mirror image: "not (A and B)" is "not A or not B". The connective always flips.

Pause and predict: a student is NOT (in the band AND in the choir). Which students does this describe?

Check your answer: those who are not in the band, or not in the choir, or in neither. That is $A' \cup B'$, the second De Morgan law. A student in only one of the two groups satisfies "not both", so the statement does not mean "in neither".

Worked example: De Morgan with numbers. Let $U = \{1, 2, \dots, 10\}$, $A = \{2, 4, 6, 8, 10\}$ and $B = \{3, 6, 9\}$. Find $(A \cup B)'$, $A' \cap B'$, $(A \cap B)'$ and $A' \cup B'$.

Common wrong step. Writing $(A \cup B)' = A' \cup B'$ and keeping the connective. Here that would give 9 elements instead of 3. The complement flips union to intersection, not union to union.

Worked example: simplifying with the laws. Simplify $(A \cup B) \cap (A \cup B')$.

Common wrong step. Changing the inner symbol while pulling out A, giving $A \cup (B \cup B') = A \cup U = U$. The inner operation stays $\cap$. A quick check: put $B = \varnothing$ and both sides must agree. The correct answer gives A, while the wrong one gives U.

Worked example: working with intervals. Let $A = [-2, 3)$ and $B = (0, 5]$. Find $A \cap B$, $A \cup B$, $A - B$ and $B - A$.

Common wrong step. Writing $A - B = [-2, 0)$. The point 0 belongs to A and does not belong to B, so it survives the subtraction. Differences flip the bracket at the cut: an excluded end point of B becomes an included end point of $A - B$.

Beyond the textbook. For the exam: a set must be well-defined, and "the rivers of India" passes while "the best writers" fails. In reality: the need for this rule became sharp in 1901, when Bertrand Russell asked about "the set of all sets that do not contain themselves". If it contains itself, then by its own rule it should not. If it does not, then by its own rule it should. Mathematicians resolved this by restricting how sets may be built. Why it matters: "well-defined" is not a formality. A careless collection can contradict itself.

Counting with sets

Counting is where sets earn their marks. The rule for two sets is

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

Why we subtract. Adding n(A) and n(B) counts everyone in the overlap twice, once from each set. Subtracting the overlap once puts them back to being counted once. If the sets are disjoint there is no overlap, and the sum is the answer. Two useful companions are $n(A - B) = n(A) - n(A \cap B)$ (the "only A" region) and $n(\text{neither}) = n(U) - n(A \cup B)$.

For example, in a school of 120 students, 70 take Maths, 55 take Physics and 25 take both. At least one subject: $70 + 55 - 25 = 100$ students. Neither: $120 - 100 = 20$. Exactly one subject: $100 - 25 = 75$.

For three sets the formula grows two more terms:

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

Why the last term is added. A person in all three sets is counted three times in the first sum, then subtracted three times in the pair terms, leaving zero. The final plus term brings them back exactly once.

Worked example: a three-set survey. Of 150 people, 70 read paper A, 60 read paper B and 50 read paper C. Also 25 read both A and B, 20 read both B and C, 15 read both A and C, and 8 read all three. Find how many read none, and how many read exactly one paper.

Common wrong step. Forgetting the $+n(A \cap B \cap C)$ term. That gives a union of 120 and an answer of 30 instead of 22. Another slip is reading "25 read both A and B" as "exactly A and B". The given pair counts include the three-paper readers.

Beyond the textbook. For the exam: union, intersection and complement are drawn with circles. In reality: every database filter and search engine runs on set algebra. A search for "cricket AND NOT football" returns the articles in $A \cap B'$, and a programmer's condition "not (a and b)" is the same as "(not a) or (not b)", which is the second De Morgan law. Why it matters: the same language describes events in probability, where "A or B" is a union and "not A" is a complement.

Pause and predict: if $n(A) = 30$ and $n(B) = 20$, what are the least and greatest possible values of $n(A \cup B)$?

Check your answer: the greatest is 50, when the sets are disjoint. The least is 30, when $B \subset A$, because then the union is just A. Every possible value of $n(A \cap B)$ between 0 and 20 gives a union between 30 and 50.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Next: Relations and Functions builds the Cartesian product $A \times B$ out of two sets, and a relation is simply a subset of it.
• Later in Class XI: Probability treats events as subsets of a sample space, so "A or B" is a union, "A and B" an intersection and "not A" a complement.
• Class XII link: Relations and Functions classifies relations on a set as reflexive, symmetric and transitive, using the subset language from this chapter.