Q.Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces :
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.
So for A={1,2,3}: writing 1∈A is correct, but 1⊆A is technically wrong notation (1 is an element, not a set) — although {1}⊆A is correct, because now both sides being compared are sets.
Why it matters
Membership is the single test every other set idea is built on — union, intersection, subset, and complement are all ultimately defined by asking "x∈A?" and "x∈B?" for every candidate x. Get comfortable with ∈/∈/ first, and every later operation becomes just a combination of membership questions.
Takeaway
Membership answers one question only: is this exact object listed in this exact set? Everything else in set theory is built from asking that question, over and over, about different sets.
"Set membership symbol meaning" and "element vs subset difference class 11" are common queries around this idea, which forms the very first building block of the Sets chapter in the NCERT/CBSE Class 11 Mathematics curriculum. Getting the ∈ vs ⊆ distinction right is a small but frequent trap in board and competitive exam questions.
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.
∅={x∣x=x}
Why is this allowed?
Because the condition x=x is always false — no object satisfies it.
This is a logical necessity: if we can define a set by a property, we must allow the possibility that nothing satisfies it.
Key insight: The empty set is not "nothing" — it's a set that contains nothing. It's a mathematical object.
6. Summary: The "Why" Behind the Definition
| Concept | Why it's defined this way |
|---|---|
| Set | To have a precise, unambiguous collection — no guesswork. |
| Set-builder | To define infinite or complex sets without listing. |
| Membership (∈) | The only question that matters — is it inside or not? |
| Empty set | Logical completeness — a property may have no objects. |
Final takeaway: The definition of a set is not a formula to plug numbers into. It's a logical framework for saying: "These objects, and only these, belong here." Every formula you see later (union, intersection, complement) builds on this single idea.
Concept: Set Membership — we check whether every element of the first set is also present in the second set. If yes, use ⊂; otherwise, use ⊄.
- {2, 3, 4} ⊂ {1, 2, 3, 4, 5} — all three elements are in the second set.
- {a, b, c} ⊄ {b, c, d} — a is missing from the second set.
- {Class XI students} ⊂ {all students of your school} — every Class XI student is a student of the school.
- {all circles} ⊄ {circles of radius 1} — a circle of radius 2, for example, is not in the second set.
- {triangles} ⊄ {rectangles} — no triangle is a rectangle.
- {equilateral triangles} ⊂ {all triangles} — every equilateral triangle is a triangle.
- {even natural numbers} ⊂ {integers} — every even natural number is an integer.
✓Final answer
- ⊂,
- ⊄,
- ⊂,
- ⊄,
- ⊄,
- ⊂,
- ⊂
The subset symbol ⊂ means "is a subset of" — every element of the first set must also belong to the second. We check each pair and fill ⊂ or ⊂ accordingly.
Concept first: Set membership and subsets
A set A is a subset of B (written A⊂B) if every element of A is also an element of B. If even one element of A is missing from B, then A⊂B. This is a pure "all-or-nothing" check — no partial credit in set theory.
The trick is to look at the elements, not the labels. For sets described by conditions, translate the condition into actual elements before comparing.
Let's go through each part.
-
{2,3,4}…{1,2,3,4,5}
Every element of the first set — 2, 3, and 4 — appears in the second set. So the first is a subset of the second.
⊂
-
{a,b,c}…{b,c,d}
The first set contains a, but a is not in {b,c,d}. Since one element fails, it is not a subset.
⊂
-
{x:x is a student of Class XI of your school}…{x:x is a student of your school}
Every Class XI student is, by definition, a student of the school. So the first set is entirely contained in the second.
⊂
-
{x:x is a circle in the plane}…{x:x is a circle in the same plane with radius 1 unit}
The first set includes circles of any radius — radius 2, radius 5, etc. The second set only contains circles of radius exactly 1. So a circle of radius 2 is in the first set but not in the second.
⊂
Watch outA common mistake is to reverse the direction: the set of radius‑1 circles is a subset of all circles, not the other way around. Here we are checking if all circles are radius‑1 circles — which is false.
-
{x:x is a triangle in a plane}…{x:x is a rectangle in the plane}
A triangle is never a rectangle, and a rectangle is never a triangle. The two sets have no overlap at all. So the first set is certainly not a subset of the second.
⊂
-
{x:x is an equilateral triangle in a plane}…{x:x is a triangle in the same plane}
Every equilateral triangle is a triangle. So the first set is fully contained in the second.
⊂
-
{x:x is an even natural number}…{x:x is an integer}
Even natural numbers (2, 4, 6, …) are all integers. So the first set is a subset of the second.
⊂
The correct symbols are: (i) ⊂,
(ii) ⊂,
(iii) ⊂,
(iv) ⊂,
(v) ⊂,
(vi) ⊂,
(vii) ⊂.
Method: Definition of Subset (⊂) and Not a Subset (⊄)
Concept:
A set A is a subset of set B (written A⊂B) if every element of A is also an element of B.
If there is at least one element in A that is not in B, then A⊂B.
Steps to decide ⊂ or ⊄:
- List or describe the elements of both sets clearly.
- Check each element of the first set — is it present in the second set?
- If all are present → use ⊂.
- If any one is missing → use ⊄.
Applying the method:
(i) {2,3,4}…{1,2,3,4,5}
- Every element (2, 3, 4) is in the second set.
- Answer: {2,3,4}⊂{1,2,3,4,5}
(ii) {a,b,c}…{b,c,d}
- a is not in {b,c,d}.
- Answer: {a,b,c}⊂{b,c,d}
(iii) {x:x is a student of Class XI of your school}…{x:x student of your school}
- Every Class XI student is a student of the school.
- Answer: ⊂
(iv) {x:x is a circle in the plane}…{x:x is a circle in the same plane with radius 1 unit}
- A circle of radius 2 is in the first set but not in the second.
- Answer: ⊂
(v) {x:x is a triangle in a plane}…{x:x is a rectangle in the plane}
- A triangle is never a rectangle.
- Answer: ⊂
(vi) {x:x is an equilateral triangle in a plane}…{x:x is a triangle in the same plane}
- Every equilateral triangle is a triangle.
- Answer: ⊂
(vii) {x:x is an even natural number}…{x:x is an integer}
- Every even natural number (2, 4, 6, …) is an integer.
- Answer: ⊂
Final filled statements:
- {2,3,4}⊂{1,2,3,4,5}
- {a,b,c}⊂{b,c,d}
- ⊂
- ⊂
- ⊂
- ⊂
- ⊂
Here’s a breakdown of the common mistakes students make with Set Membership and Subset (⊂ / ⊄) questions, using your examples.
🧠 Core Concept Reminder
- ⊂ (subset): Every element of the first set must also be an element of the second set.
- ⊄ (not a subset): At least one element of the first set is not in the second set.
✗ Common Mistake #1: Confusing “element of” with “subset of”
Example from (i):
{ 2, 3, 4 } . . . { 1, 2, 3, 4, 5 }
Mistake: Students write ∈ instead of ⊂.
They think: “2, 3, 4 are in the second set, so it’s a subset.” But they use the wrong symbol.
How to avoid:
∈is for single elements (e.g.,2 ∈ {1,2,3}).⊂is for whole sets (e.g.,{2,3} ⊂ {1,2,3}).- Check: Is the left side a set (with curly braces)? Then use
⊂or⊄.
✓ Correct: { 2, 3, 4 } ⊂ { 1, 2, 3, 4, 5 }
✗ Common Mistake #2: Forgetting to check every element
Example from (ii):
{ a, b, c } . . . { b, c, d }
Mistake: Student sees b and c are in the second set, but forgets a is missing. They write ⊂.
How to avoid:
- Scan each element of the first set.
- Ask: “Is this element in the second set?”
- If even one is missing →
⊄.
✓ Correct: { a, b, c } ⊄ { b, c, d }
✗ Common Mistake #3: Misreading the description (especially “your school”)
Example from (iii):
{x : x is a student of Class XI of your school} . . . {x : x student of your school}
Mistake: Student thinks “Class XI” is the whole school, or ignores “your school” and treats it as any school.
How to avoid:
- Read the condition literally.
- Class XI students are a subset of all students in the same school.
- Every Class XI student is also a student of the school → subset.
✓ Correct: ⊂
✗ Common Mistake #4: Reversing the subset direction
Example from (iv):
{x : x is a circle in the plane} . . . {x : x is a circle in the same plane with radius 1 unit}
Mistake: Student writes ⊂ thinking “circles with radius 1 are a subset of all circles” — which is true — but they put the larger set on the left.
How to avoid:
- Identify which set is bigger.
- All circles include circles of radius 1, but not vice versa.
- So the left set (all circles) is not a subset of the right set (only radius 1).
✓ Correct: ⊄
✗ Common Mistake #5: Assuming different shapes can be subsets
Example from (v):
{x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane}
Mistake: Student thinks “triangle” and “rectangle” are both shapes, so maybe one is a subset. They write ⊂.
How to avoid:
- No triangle is a rectangle.
- The two sets have no common elements.
- If they share zero elements, the first set cannot be a subset of the second.
✓ Correct: ⊄
✗ Common Mistake #6: Ignoring “special case” logic
Example from (vi):
{x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane}
Mistake: Student writes ⊄ because they think “equilateral” is different from “triangle”.
How to avoid:
- Every equilateral triangle is a triangle.
- The first set is a special case of the second.
- Special case → subset.
✓ Correct: ⊂
✗ Common Mistake #7: Confusing “natural numbers” with “integers”
Example from (vii):
{x : x is an even natural number} . . . { x : x is an integer}
Mistake: Student thinks “natural numbers” and “integers” are unrelated, or writes ⊄ because integers include negatives.
How to avoid:
- Natural numbers (usually 1,2,3,…) are a subset of integers (…,-2,-1,0,1,2,…).
- Even natural numbers are still natural numbers → they are integers.
- So every element of the first set is in the second.
✓ Correct: ⊂
📝 Quick Summary Table
| Example | Common Mistake | Correct Symbol |
|---|---|---|
| (i) | Using ∈ instead of ⊂ | ⊂ |
| (ii) | Not checking all elements | ⊄ |
| (iii) | Misreading “your school” | ⊂ |
| (iv) | Reversing subset direction | ⊄ |
| (v) | Assuming different shapes overlap | ⊄ |
| (vi) | Forgetting special case = subset | ⊂ |
| (vii) | Confusing number systems | ⊂ |
Final tip: Always ask yourself — “Is every item from the first set guaranteed to be in the second?” If yes → ⊂. If no → ⊄.
- 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.
This gives n=1,2,3,4,5 → exactly 1,7,17,31,49. Correct.
Watch outA common mistake is to misread “n<5” as including 5 — it does not. Option (A) fails because it stops at n=4, missing 49.
TipAlways test the smallest and largest elements of the set against the formula and the range of n. Here, 49 is the key: only n=5 produces it, so the range must include 5.
✓Final answerThe correct option is (D).
ANSWER: D
- 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).
- (D) {x:x2=4, x∈N}={2} — exactly one element.
Only (D) is a singleton.
✓Final answerThe correct option is (D) — {x:x2=4, x∈N}
- 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
Watch outA common mistake is to think n(A′∩B′)=n(U)−n(A)−n(B). That would give 100−50−60=−10, which is impossible. The error is double-counting the intersection — elements in both A and B get subtracted twice. Always use the union first.
TipIf you prefer a visual approach: draw a Venn diagram with three regions — only A (30), only B (40), and both (20). The union fills 30+40+20=90 elements. The remaining 10 are outside both circles, which is exactly A′∩B′.
✓Final answerThe value is 10, which corresponds to option (D).
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