Q.From the sets given below, select equal sets : A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2} E = {–1, 1}, F = { 0, a}, G = {1, –1}, H = { 0, 1}
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Set Membership
Set Membership
The idea in plain words
Every set is defined by exactly one question: "does this object belong to the set, or not?" That yes/no relationship between an object and a set is called membership. If an object is in the set, it is a member (or element) of the set; if it isn't, it simply is not.
A set is only "well-defined" if this question always has a clear answer for every possible object — that's what makes membership testable.
The notation
For a set A and an object x:
- x∈A reads "x belongs to A" or "x is an element of A" — TRUE membership.
- x∈/A reads "x does not belong to A" — FALSE membership.
Example. Let A={2,4,6,8}.
4∈A(4 is listed inside A)
5∈/A(5 is not listed inside A)
Testing membership: roster form vs. set-builder form
Roster form — just look for the object in the list.
B={1,3,5,7},3∈B,4∈/B
Set-builder form — plug the candidate into the defining rule and check if it's satisfied.
C={x∣x is a prime number less than 10}
Is 7∈C? Check: is 7 prime and less than 10? Yes → 7∈C.
Is 9∈C? Check: 9 is less than 10 but not prime (9=3×3) → 9∈/C.
Properties every student must know
- Each element is either in or out — never "partly in." Membership is binary, not a matter of degree.
- Repetition doesn't affect membership. {1,1,2}={1,2} — asking "is 1 a member?" gives the same YES either way.
- Order never affects membership. 2∈{1,2,3} is exactly the same fact as 2∈{3,2,1}.
- Nothing belongs to the empty set. For any x, x∈/∅ — there is nothing inside to belong to.
- A set can be an element of another set. If D={1,{2,3}}, then 1∈D is true, and {2,3}∈D is true, but 2∈D is false — 2 is not directly listed in D; it is only inside the set that is listed.
A common slip
Students often confuse x∈A (membership: is the object present?) with B⊆A (subset: is every element of B also in A?). The two symbols compare different kinds of things:
- ∈ / ∈/ compares an element to a set.
- ⊆ / ⊆ compares a set to a set. …
Why this formula?
Let's break down the definition of a set — not as a formula to memorise, but as a fundamental idea that underpins all of mathematics.
1. What is a Set? (The Core Idea)
A set is a well-defined collection of distinct objects.
The "why" here is about clarity and precision — we need to know exactly what belongs and what does not.
- Well-defined: For any object, we can say yes or no — no ambiguity.
- Distinct: No duplicates — each object appears only once.
Why? Because if we couldn't decide membership, we couldn't do any logical operations. Sets are the building blocks of all mathematical structures.
2. The Key "Formula": Set-Builder Notation
The most common way to define a set is:
S={x∣P(x)}
This reads: "S is the set of all objects x such that property P(x) is true."
Why does this work?
- x is a placeholder for any object.
- P(x) is a logical condition (a predicate) that is either true or false for each x.
- The vertical bar ∣ means "such that".
Example:
A={n∣n∈N,n is even}
Here, P(n) is "n is a natural number and n is even".
Only those n that satisfy both conditions are included.
Why this form? It avoids listing infinitely many elements. It gives a rule — a decision procedure — for membership.
3. The Two Fundamental Properties (Axioms)
Every set definition relies on two intuitive truths:
(a) Extensionality — Two sets are equal if they have the same elements.
A=B⟺(∀x)(x∈A⟺x∈B)
Why? A set is completely determined by its members. There is no other hidden property.
If you know what's inside, you know the set.
(b) Membership — The only relation is ∈ (belongs to).
x∈Sorx∈/S
Why? Because a set is just a container. The only question we can ask is: "Is this object inside?"
4. Why Can't We Just List Everything?
For small sets, listing works:
{1,2,3}
But for infinite sets (like all natural numbers), listing is impossible.
Set-builder notation solves this by giving a rule instead of a list.
Example:
N={n∣n is a positive integer}
This is not a formula to memorise — it's a definition by property.
5. The "Empty Set" — Why It Exists
The empty set ∅ (or {}) is the set with no elements. …
Concept: Set Equality
Two sets are equal if and only if they contain exactly the same elements, regardless of order or repetition.
Solution:
Compare each set by listing elements in a standard form:
- A={2,4,8,12}
- B={1,2,3,4}
- C={4,8,12,14}
- D={3,1,4,2}={1,2,3,4}
- E={−1,1}
- F={0,a}
- G={1,−1}={−1,1}
- H={0,1} …
Two sets are equal if and only if they contain exactly the same elements, regardless of order or repetition. Comparing all given sets, B = D and E = G.
Understanding Set Equality
Set equality is one of the most fundamental concepts in set theory. Two sets are equal when every element of the first set is in the second, and every element of the second is in the first. Order doesn't matter—1,2,3 and 3,1,2 are the same set. Repetition doesn't matter either—1,2,2,3 is just 1,2,3.
The key is to look at membership: do the sets have identical rosters?
Step-by-Step Comparison
Let me organize the given sets first:
| Set | Elements |
|---|---|
| A | {2,4,8,12} |
| B | {1,2,3,4} |
| C | {4,8,12,14} |
| D | {3,1,4,2} |
| E | {−1,1} |
| F | {0,a} |
| G | {1,−1} |
| H | {0,1} |
Now I'll systematically check for equality.
-
Compare sets with four elements: A, B, C, D
Set A={2,4,8,12} contains 8 and 12 but not 1 or 3.
Set B={1,2,3,4} contains 1 and 3 but not 8 or 12.
Set C={4,8,12,14} contains 14 but not 2.
Set D={3,1,4,2}—reordering gives {1,2,3,4}, which is exactly B.
Therefore B=D.
-
Compare sets with two elements: E, F, G, H
Set E={−1,1} contains both negative and positive one. …
Method: Set Equality via Element Comparison
Two sets are equal if and only if they contain exactly the same elements — order and repetition do not matter.
Steps:
- List each set in ascending order (or any consistent order) to compare easily.
- Check element-by-element — every element of one set must be in the other, and vice versa.
- Mark pairs that are identical.
Applying the method:
| Set | Elements (sorted) |
|---|---|
| A | {2, 4, 8, 12} |
| B | {1, 2, 3, 4} |
| C | {4, 8, 12, 14} |
| D | {1, 2, 3, 4} |
| E | {–1, 1} |
Here are the common mistakes students make when identifying equal sets, along with clear strategies to avoid each.
Mistake 1: Ignoring the Order of Elements
The error:
Students see sets like B={1,2,3,4} and D={3,1,4,2} and think they are different because the numbers appear in a different sequence.
Why it’s wrong:
A set is defined only by its elements, not by the order they are listed.
{1,2,3,4} and {3,1,4,2} contain exactly the same four numbers — they are equal.
How to avoid:
Always sort or mentally rearrange the elements of each set into ascending order before comparing.
- B={1,2,3,4}
- D={1,2,3,4} Now it’s obvious they match.
Mistake 2: Confusing “Equal” with “Same Number of Elements”
The error:
A student sees A={2,4,8,12} and C={4,8,12,14} — both have 4 elements — and assumes they are equal.
Why it’s wrong:
Equal sets must have exactly the same elements.
A contains 2 but not 14; C contains 14 but not 2. They are not equal.
How to avoid:
Check element-by-element after sorting. If even one element differs, the sets are not equal — regardless of size.
Mistake 3: Treating Repeated Elements as Different
The error:
A student writes B={1,2,3,4} and sees another set listed as {1,1,2,3,4} and thinks they are different because of the extra “1”.
Why it’s wrong:
In set notation, repetitions are ignored. {1,1,2,3,4}={1,2,3,4}.
(Though in this problem, no set has explicit repetitions — but the concept is tested often.)
How to avoid:
When comparing, remove duplicates mentally first. Only the distinct elements matter.
Mistake 4: Misreading Symbols or Variables
The error:
F={0,a} and H={0,1} — a student sees “a” and thinks it means “1” or ignores it.
Why it’s wrong:
a is a variable (or a letter), not the number 1. Unless told a=1, these sets are different.
Similarly, E={−1,1} and G={1,−1} are equal (same elements, different order), but F and H are not equal.
How to avoid:
Treat letters as unknown symbols unless a specific value is given. Compare elements literally:
- F has elements 0 and a …
- COMEDK 2024Set 2024-M1 markMCQQ.Express the set A={1,7,17,31,49} in set builder form (A) {x∣x=2n2−1, where n∈N and n<5} (B) {x∣x=2n2−3, where n∈N and 2≤n≤8} (C) {x∣x=2n2+1, where n∈N and n≤7} (D) {x∣x=2n2−1, where n∈N and n≤5}
›Reveal solutionSolution
The set A={1,7,17,31,49} matches the formula x=2n2−1 for n=1,2,3,4,5, so the correct option is (D).
We need to express the given set {1,7,17,31,49} in set-builder form. The key is to find a pattern: each number is one less than twice a perfect square. Let’s check:
- 1=2(1)2−1
- 7=2(2)2−1
- 17=2(3)2−1
- 31=2(4)2−1
- 49=2(5)2−1
So the pattern is x=2n2−1 with n taking natural numbers from 1 to 5. Now we examine each option.
-
Option (A): x=2n2−1, n∈N, n<5.
This gives n=1,2,3,4 → values: 1,7,17,31. Missing 49. So incorrect.
-
Option (B): x=2n2−3, n∈N, 2≤n≤8.
For n=2: 2(4)−3=5 (not in set). So incorrect.
-
Option (C): x=2n2+1, n∈N, n≤7.
For n=1: 2+1=3 (not in set). So incorrect.
-
Option (D): x=2n2−1, n∈N, n≤5. …
- COMEDK 2023Set 2023-E1 markMCQQ.Which of the following is a singleton set? (A) {x:x2=4,x∈R} (B) {x:∣x∣<4,x∈N} (C) {x:∣x∣<−4,x∈N} (D) {x:x2=4,x∈N}
›Reveal solutionSolution
{x:x2=4, x∈N}={2} has exactly one element.
- (A) {x:x2=4, x∈R}={2,−2} — two elements.
- (B) {x:∣x∣<4, x∈N}={1,2,3} — three elements.
- (C) {x:∣x∣<−4, x∈N}=∅ — empty (no modulus is negative). …
- KCET 2019Set A-11 markMCQQ.If U is the universal set with 100 elements; A and B are two sets such that n(A)=50, n(B)=60, n(A∩B)=20 then n(A′∩B′)= (A) 40 (B) 20 (C) 90 (D) 10
›Reveal solutionSolution
Use De Morgan’s law: A′∩B′=(A∪B)′. Find n(A∪B) via the inclusion-exclusion principle, then subtract from the total. The answer is 10.
The core idea here is that the complement of the union is exactly the region outside both sets. Instead of trying to count elements outside A and outside B separately, we use De Morgan’s law to turn the problem into one we already know how to solve: finding the size of the union.
When you see A′∩B′, always think: “this is everything that is not in A and not in B” — which is the same as “everything that is not in (A or B)”. That’s the complement of the union.
- Find n(A∪B) using the inclusion-exclusion formula. For any two sets,
n(A∪B)=n(A)+n(B)−n(A∩B)
Substitute the given values:
n(A∪B)=50+60−20=90
- Apply De Morgan’s law.
A′∩B′=(A∪B)′
This is a set identity — it always holds, no matter what the sets are.
- Find the size of the complement. The universal set U has 100 elements. The complement of any set X has size:
n(X′)=n(U)−n(X)
So:
n(A′∩B′)=n((A∪B)′)=n(U)−n(A∪B)=100−90=10 …
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