Sets | CBSE Class 11 Maths Notes

This Mathematics note covers sets and their representation, empty and finite sets, equality, subsets, real-number intervals, universal sets, Venn diagrams, union, intersection, difference, complements, De Morgan’s laws and proofs using membership.

What is a set, and how is membership decided?

Definition: A set is a well-defined collection of objects. The words object, element and member refer to the objects belonging to that collection.

Well-defined means that membership can be decided definitely. The vowels in the English alphabet form a set because there is a clear rule for deciding whether a letter belongs. A collection described as the five most renowned mathematicians does not give an unambiguous criterion.

The criterion matters more than whether the collection is familiar or easy to count. Different people may choose different members for a collection based on personal judgements such as “most renowned”. A mathematical set requires a definite membership decision.

How do we read the membership symbols?

Sets are usually represented by capital letters, and their elements by small letters. The statement $a\in A$ means that the object $a$ belongs to the set $A$. The statement $b\notin A$ means that $b$ does not belong to it.

If $V$ is the set of vowels, then $a\in V$, whereas $b\notin V$. If $P$ is the set of prime factors of $30$, then $3\in P$, but $15\notin P$. A factor must also satisfy the stated requirement of being prime.

A membership statement compares an object with a set. Before choosing a symbol, read the complete description of the set. “Prime factors” and “factors” do not describe the same collection, even when the number being considered is unchanged.

How are sets written in roster and set-builder forms?

In roster form, list the distinct elements inside braces and separate them with commas. The even positive integers less than $7$ form $\{2,4,6\}$. Listing the members in another order does not change the set.

Repetition also does not change membership. The set of letters in SCHOOL is $\{S,C,H,O,L\}$. Although the letter O occurs twice in the word, it is listed once as an element. A set records which elements occur, rather than their number of occurrences.

In set-builder form, specify a property that identifies precisely the required elements. The braces mean “the set of all”, and the colon means “such that”. The set of vowels may be written as $V=\{x:x\text{ is a vowel in the English alphabet}\}$.

How does the domain control a roster?

The description $A=\{x:x\in\mathbb N,\ 3\lt x\lt10\}$ gives $A=\{4,5,6,7,8,9\}$. Both the natural-number restriction and the strict inequalities matter. Real numbers between the same limits would not give this finite list.

Worked example 1. Write the solution set of $x^{2}+x-2=0$ in roster form.

1. Factorise the polynomial: $$x^{2}+x-2=(x-1)(x+2).$$
2. Use the zero-product condition; the symbol $\implies$ means “implies”: $$(x-1)(x+2)=0\implies x-1=0\text{ or }x+2=0.$$
3. Solve the two linear equations: $$x=1\text{ or }x=-2.$$
4. Check both substitutions: $$1^{2}+1-2=0,\qquad (-2)^{2}+(-2)-2=4-2-2=0.$$

Answer: The solution set is $\{1,-2\}$. Both elements satisfy the equation, and the factorisation gives all its roots.

Worked example 2. Write $\{x:x\text{ is a positive integer and }x^{2}\lt40\}$ in roster form.

1. Use the domain: $x\in\mathbb Z^{+}$, so start with the positive integers.
2. Check the qualifying squares: $$1^{2}=1,\quad2^{2}=4,\quad3^{2}=9,\quad4^{2}=16,\quad5^{2}=25,\quad6^{2}=36.$$
3. Check the next positive integer: $7^{2}=49\gt40$. Larger positive integers have still larger squares.
4. Collect exactly the integers whose squares meet the condition: $$\{1,2,3,4,5,6\}.$$

Answer: $\{1,2,3,4,5,6\}$. Zero and negative integers are excluded by the domain before the inequality is applied.

The reverse conversion asks for a rule that describes the full list. For example, $\{1,4,9,16,25,\ldots\}=\{x:x=n^{2},\ n\in\mathbb N\}$. The ellipsis indicates that the displayed pattern continues; it does not remove the need for a clear defining property.

How do empty, singleton, finite and infinite sets differ?

An empty set contains no element. It is also called the null set or void set, and is written as $\varnothing$ or $\{\}$. A description can define a set even when no object satisfies its conditions.

For instance, $\{x:x\in\mathbb N,\ 1\lt x\lt2\}=\varnothing$, because no natural number lies strictly between those bounds. Similarly, there is no even prime number greater than $2$. The requirement is clear, but it has no qualifying member.

What is counted when a set is finite?

A finite set is empty or consists of a definite number of elements. The notation $n(S)$ denotes the number of distinct elements in a finite set $S$. A singleton set contains exactly one element. An infinite set is not finite.

Worked example 3. Classify $S=\{x:x\in\mathbb N,\ 2x-1=0\}$.

1. Solve the equation by adding $1$ to both sides: $$2x=1.$$
2. Divide both sides by $2$: $$x=\frac12.$$
3. Apply the natural-number restriction: $$\frac12\notin\mathbb N.$$
4. Reject the only algebraic candidate: $$S=\varnothing.$$

Answer: The set is empty and therefore finite. The candidate $\frac12$ solves the equation but does not belong to the required domain.

Some infinite sets can be indicated by a roster with dots, such as the integers $\{\ldots,-2,-1,0,1,2,\ldots\}$. This method cannot describe every infinite set. In particular, the real numbers cannot all be listed by the roster pattern used for natural numbers.

When are two sets equal?

Two sets are equal when they have exactly the same elements. We write $A=B$ for equal sets and $A\ne B$ for unequal sets. Equality concerns membership, so neither the order of a roster nor repeated entries changes it.

For example, $\{1,2,3,4\}=\{3,1,4,2\}$. Also, $\{1,2,3\}=\{2,2,1,3,3\}$. In each comparison, removing repetitions and checking the distinct elements gives the same collection.

How can equality be checked?

Check membership in both directions. Every element of the first set must occur in the second, and every element of the second must occur in the first. Finding one element present in just one of the sets is enough to show inequality.

Worked example 4. Compare $A=\{n:n\in\mathbb Z,\ n^{2}\le4\}$ with $B=\{x:x\in\mathbb R,\ x^{2}-3x+2=0\}$.

1. Apply the square condition to integers: $$n^{2}\le4\implies -2\le n\le2.$$
2. List the integer solutions: $$A=\{-2,-1,0,1,2\}.$$
3. Factorise the equation defining the other set: $$x^{2}-3x+2=(x-1)(x-2)=0.$$
4. Solve and check its roots: $$B=\{1,2\},\qquad1-3+2=0,\quad4-6+2=0.$$
5. Identify a membership difference: $$0\in A,\qquad0\notin B.$$

Answer: $A\ne B$, because the first set contains $0$ and the second does not.

Letters provide another useful illustration. ALLOY and LOYAL give the same set $\{A,L,O,Y\}$, despite a different order in the words. When converting words into sets, keep each distinct letter and then compare the resulting collections.

Note: A set containing zero is not empty. In $\{0\}$, the number $0$ is an element; in $\varnothing$, there is no element to list.

How do subsets differ from elements and proper subsets?

A set $A$ is a subset of $B$ when every element of $A$ is also an element of $B$. We use $A\subset B$ for this relationship, allowing equality. The condition is $a\in A\implies a\in B$.

To disprove a subset statement, identify an element of the first set that is absent from the second. To prove it, the membership requirement must hold for every element of the first set, rather than for just some shared members.

Property: Mutual inclusion gives equality

The two inclusion statements together express equality: $A=B\iff A\subset B\text{ and }B\subset A$. The symbol $\iff$ means “if and only if”, so the statement holds in both directions. Every set is a subset of itself, and the empty set is a subset of every set.

A proper subset is a subset that is unequal to the containing set. Thus $\{1,2,3\}$ is a proper subset of $\{1,2,3,4\}$. The larger containing set is a superset of the smaller set.

Worked example 5. For $A=\{1,3\}$, $B=\{1,5,9\}$ and $C=\{1,3,5,7,9\}$, decide whether the first set is a subset of the second in each ordered pair: (i) $\varnothing,B$, (ii) $A,B$, (iii) $A,C$, (iv) $B,C$.

1. Use the empty-set property: $$\varnothing\subset B.$$
2. Compare the members of the first pair: $$3\in A,\quad3\notin B\implies A\not\subset B.$$
3. Check both members against the larger set: $$1\in C,\quad3\in C\implies A\subset C.$$
4. Check all three members of the other set: $$1\in C,\quad5\in C,\quad9\in C\implies B\subset C.$$

Answer: $\varnothing\subset B$, $A\not\subset B$, $A\subset C$ and $B\subset C$. The missing element $3$ settles the failed inclusion.

Why must membership and inclusion be kept separate?

A set can itself be an element of another set. Consider $A=\{1\}$, $B=\{\{1\},2\}$ and $C=\{\{1\},2,3\}$. Here $A\in B$ and $B\subset C$, but $A\not\subset C$: the element $1$ is absent from $C$, although $\{1\}$ is present.

The braces are therefore essential. Membership asks whether a particular object is listed; inclusion asks whether all the elements of one set occur in another. Replacing one symbol with the other changes the claim being made.

To list all subsets of $\{-1,0,1\}$, organise them by size: $\varnothing$; then $\{-1\},\{0\},\{1\}$; then $\{-1,0\},\{-1,1\},\{0,1\}$; finally $\{-1,0,1\}$. This includes both the empty set and the original set.

How do intervals represent subsets of the real numbers?

The familiar number systems are related by inclusion: $\mathbb N\subset\mathbb Z\subset\mathbb Q\subset\mathbb R$. A rational number can be written as $\frac pq$, where $p,q\in\mathbb Z$ and $q\ne0$. Integers belong to the rational numbers as well as to the real numbers.

The irrational numbers are the real numbers that are not rational. Writing their set as $T$, we have $T=\{x:x\in\mathbb R,\ x\notin\mathbb Q\}$. Examples include $\sqrt2$, $\sqrt5$ and $\pi$.

How are interval endpoints included or excluded?

For real numbers $a\lt b$, an interval includes all real numbers between the stated endpoints. Round brackets exclude an endpoint; square brackets include it. Changing a bracket changes membership at that endpoint.

Four number-line segments are labelled $(a,b)$, $[a,b]$, $[a,b)$ and $(a,b]$. Open circles mark excluded endpoints, while filled circles mark included endpoints.

See Fig. 1.1 in your NCERT textbook

These intervals contain infinitely many points, even though their endpoints are finite real numbers. Their length is $b-a$, whether neither, both or just one of the endpoints belongs to the interval.

Worked example 6. Convert $\{x:x\in\mathbb R,\ -5\lt x\le7\}$ to interval notation and $[-3,5)$ to set-builder form.

1. Read the first lower bound: $-5\lt x$, so exclude $-5$ with a round bracket.
2. Read its upper bound: $x\le7$, so include $7$ with a square bracket.
3. Combine the endpoint decisions: $$\{x:x\in\mathbb R,\ -5\lt x\le7\}=(-5,7].$$
4. Reverse the process for the second interval: $$[-3,5)=\{x:x\in\mathbb R,\ -3\le x\lt5\}.$$

Answer: $(-5,7]$ and $\{x:x\in\mathbb R,\ -3\le x\lt5\}$, respectively. Each bracket agrees with its inequality.

Unbounded intervals also describe number sets. Here $\infty$ (positive infinity) indicates that there is no upper bound, and $-\infty$ (negative infinity) indicates that there is no lower bound; neither is a real-number endpoint. The non-negative real numbers are $[0,\infty)$, the negative real numbers are $(-\infty,0)$, and all real numbers are $(-\infty,\infty)$.

What do a universal set and a Venn diagram show?

A universal set is the basic set relevant to the particular discussion. It is usually denoted by $U$. The other sets under consideration are treated as its subsets, so their elements must belong to that chosen universe.

The universe depends on context. When studying natural numbers and collections such as primes or even natural numbers, the natural numbers provide a suitable basic set. For a discussion involving the integers, the rational numbers or the real numbers can serve as a containing universal set.

How should the regions be read?

A Venn diagram usually represents the universal set by a rectangle and subsets by closed curves, commonly circles. Elements are written inside the regions to which they belong. A circle inside another circle represents a subset relationship.

A rectangle labelled $U$ contains the numbers from $1$ to $10$. The circle $A$ contains $2,4,6,8,10$; the remaining displayed numbers lie outside that circle but inside the rectangle.

See Fig. 1.2 in your NCERT textbook

Inside the rectangle $U$, circle $B$ lies wholly inside circle $A$. The elements $4$ and $6$ lie in $B$, while $2,8,10$ lie in the part of $A$ outside $B$.

See Fig. 1.3 in your NCERT textbook

The nested diagram represents $B\subset A$. Every element in the inner circle belongs to the outer circle as well. The separate locations of the other elements show that inclusion in the outer circle does not require membership of the inner one.

When interpreting a diagram, distinguish the universe from an individual circle. An element outside a circle may still be inside the universal set. That distinction becomes essential for complements, which select elements outside a specified subset while remaining within the rectangle.

How do union and intersection combine sets?

The union of two sets contains elements belonging to either set, including those belonging to both. Common elements appear only once. Its set-builder definition is $A\cup B=\{x:x\in A\text{ or }x\in B\}$.

The intersection contains only elements belonging to both sets. Its definition is $A\cap B=\{x:x\in A\text{ and }x\in B\}$. The difference between “or” and “and” determines which elements qualify.

Worked example 7. Find the union and intersection of $A=\{2,4,6,8\}$ and $B=\{6,8,10,12\}$.

1. Start the union with all members of the first set: $$2,4,6,8.$$
2. Inspect the second set. Its new members are $10,12$; its members $6,8$ are already present.
3. Collect every distinct member: $$A\cup B=\{2,4,6,8,10,12\}.$$
4. For the intersection, retain only the members that occur in both lists: $$A\cap B=\{6,8\}.$$

Answer: $A\cup B=\{2,4,6,8,10,12\}$ and $A\cap B=\{6,8\}$. The common members are counted once in each resulting set.

Two overlapping circles labelled $A$ and $B$ lie within the rectangle $U$. Both circles, including their common region, are shaded to represent $A\cup B$.

See Fig. 1.4 in your NCERT textbook

Figure 1.5 shades only the lens-shaped overlap of two circles. Figure 1.6 shows two separate, non-overlapping circles inside the universal rectangle.

See Figs. 1.5 and 1.6 in your NCERT textbook

Property: Union and intersection obey algebraic laws

The commutative laws permit the sets to exchange places. The associative laws permit regrouping when the operation stays the same. Identity and idempotent laws describe what happens with the empty set, the universal set and repeated sets.

For disjoint sets, $A\cap B=\varnothing$. For nested sets with $B\subset A$, the union is the containing set and the intersection is the contained set: $A\cup B=A$ and $A\cap B=B$.

The distributive law connects the operations: $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$. It selects members of $A$ that also belong to at least one of the other two sets.

How is the difference of two sets found?

The difference of sets $A$ and $B$, in that order, consists of elements belonging to $A$ but not to $B$. Its definition is $A-B=\{x:x\in A\text{ and }x\notin B\}$.

The order matters because the first set supplies the candidates. Elements that belong only to the second set cannot enter the difference. Removing shared elements from the first set leaves exactly the members required by the definition.

Worked example 8. Find both differences for $A=\{1,2,3,4,5,6\}$ and $B=\{2,4,6,8\}$.

1. Identify the shared elements: $$A\cap B=\{2,4,6\}.$$
2. For the first difference, start with $A$ and remove those shared members: $$A-B=\{1,3,5\}.$$
3. Reverse the starting set and remove the same shared members from $B$: $$B-A=\{8\}.$$
4. Compare the outputs: $$\{1,3,5\}\ne\{8\}\implies A-B\ne B-A.$$

Answer: $A-B=\{1,3,5\}$, whereas $B-A=\{8\}$. Each retained member occurs only in the corresponding starting set.

How do the three parts of overlapping sets differ?

The regions $A-B$, $A\cap B$ and $B-A$ are mutually disjoint. No element belongs to any two of these regions. The middle region contains common members; the two outer regions contain members exclusive to the respective sets.

Figure 1.8 shades the portion of circle $A$ outside circle $B$. Figure 1.9 labels the left-only region $A-B$, the overlap $A\cap B$, and the right-only region $B-A$.

See Figs. 1.8 and 1.9 in your NCERT textbook

To avoid reversing the answer, read a difference from left to right: first choose the starting set, then exclude members of the following set. This is a membership operation; subtracting the written numerical values of elements would not find a set difference.

How are complements and De Morgan’s laws used?

The complement of $A$, relative to a universal set $U$, contains elements of $U$ that are not in $A$. It is denoted by $A'$, with $A'=\{x:x\in U\text{ and }x\notin A\}=U-A$.

Keep the same universal set throughout a calculation involving complements. Both the original set and its complement are subsets of that universe. The complement concerns membership within the chosen universe, rather than every conceivable object outside the set.

Worked example 9. Find the complement of $A=\{1,3,5,7,9\}$ in $U=\{1,2,3,4,5,6,7,8,9,10\}$.

1. Use the complement rule: $$A'=U-A.$$
2. Remove the listed members of $A$ from $U$: exclude $1,3,5,7,9$.
3. Retain the remaining members of the universe: $$A'=\{2,4,6,8,10\}.$$
4. Check the separation and coverage: $$A\cap A'=\varnothing,\qquad A\cup A'=U.$$

Answer: $A'=\{2,4,6,8,10\}$. Every retained number belongs to the universe and none belongs to the original set.

The rectangle $U$ contains a circle labelled $A$. The area inside the rectangle but outside the circle is shaded and labelled $A'$; the circle itself is unshaded.

See Fig. 1.10 in your NCERT textbook

Property: Complement laws and double complementation

The fundamental complement laws are $A\cup A'=U$ and $A\cap A'=\varnothing$. Taking the complement twice restores the original set: $(A')'=A$. Also, $\varnothing'=U$ and $U'=\varnothing$.

Identity: De Morgan’s laws

The complement of a union equals the intersection of the complements: $(A\cup B)'=A'\cap B'$. The complement of an intersection equals the union of the complements: $(A\cap B)'=A'\cup B'$. Both identities refer to a common universal set.

Worked example 10. Verify the first De Morgan law for $U=\{1,2,3,4,5,6\}$, $A=\{2,3\}$ and $B=\{3,4,5\}$.

1. Remove the first set from the universe: $$A'=\{1,4,5,6\}.$$
2. Remove the second set from the universe: $$B'=\{1,2,6\}.$$
3. Find the members shared by these complements: $$A'\cap B'=\{1,6\}.$$
4. Independently combine the original sets: $$A\cup B=\{2,3,4,5\}.$$
5. Remove that union from the universe: $$(A\cup B)'=\{1,6\}.$$
6. Compare the independently obtained sets: $$(A\cup B)'=\{1,6\}=A'\cap B'.$$

Answer: Both sides give $\{1,6\}$, verifying the law for these sets. The calculation checks this example; the identity applies generally to subsets of a common universe.

How can membership arguments prove a set identity?

A membership proof starts with an arbitrary element and follows what the definitions require. For equality, prove inclusion in both directions. Checking a particular list of numbers verifies one example, whereas an arbitrary-element argument establishes the result for all sets satisfying the assumptions.

Keep track of the operation used at each step. A member of a union need belong to only one of the sets. A member of an intersection must belong to both. These definitions justify the changes in membership statements.

Result: Equal union and intersection force equal sets

Suppose $A\cup B=A\cap B$. The required conclusion is $A=B$. The proof uses the given equality twice, once to establish each subset relationship.

1. Choose any element $a\in A$. By the definition of union, $$a\in A\implies a\in A\cup B.$$
2. Use the assumed equality and then the definition of intersection: $$a\in A\cup B=A\cap B\implies a\in B.$$
3. Since the choice was arbitrary, record the first inclusion: $$A\subset B.$$
4. Now choose any element $b\in B$ and repeat in the reverse direction: $$b\in B\implies b\in A\cup B=A\cap B\implies b\in A.$$
5. Record the reverse inclusion: $$B\subset A.$$
6. Apply mutual inclusion: $$A\subset B\text{ and }B\subset A\implies A=B.$$

Conclusion: $A\cup B=A\cap B\implies A=B$. The proof compares the elements of the sets rather than the appearance of a particular diagram.

The two-direction method also explains why proving only one subset relationship is insufficient for equality. A proper subset satisfies one inclusion without containing every element of the larger set. Each direction contributes a separate part of the equality claim.

Glossary

• Set — A well-defined collection of objects for which membership can be decided definitely.
• Element — An object belonging to a set, also called a member of that set.
• Roster form — A representation listing distinct elements within braces, separated from one another by commas.
• Set-builder form — A representation specifying the common property that identifies precisely the elements of a set.
• Empty set — A set containing no elements, also called the null set or void set.
• Finite set — A set that is empty or consists of a definite number of elements.
• Infinite set — A set that does not contain a finite number of elements.
• Equal sets — Sets containing exactly the same elements, irrespective of their listed order or repetition.
• Subset — A set whose every element also belongs to another specified set.
• Proper subset — A subset that is unequal to the set which contains it.
• Universal set — The basic set relevant to a discussion, containing the other sets being considered.
• Union — The set containing members of either of two sets, including their common members.
• Intersection — The set containing precisely those elements that belong to both of two sets.
• Difference — The set of elements belonging to the first set but not to the second.
• Complement — The set of elements in the chosen universe that do not belong to the specified subset.

Common errors and misconceptions

• Misconception: Changing the order of a roster changes the set. Correct: Membership determines the set; $\{1,2,3,4\}=\{3,1,4,2\}$.
• Misconception: Repeated letters become different elements. Correct: List distinct letters once; SCHOOL gives $\{S,C,H,O,L\}$.
• Misconception: A set containing zero is empty. Correct: $\{0\}$ contains an element, whereas $\varnothing$ contains none.
• Misconception: Every algebraic root belongs to the requested solution set. Correct: Apply the domain restriction; $\frac12$ is not a natural number.
• Misconception: Membership and inclusion mean the same thing. Correct: $A\in B$ treats $A$ as an element; $A\subset B$ compares its members with those of $B$.
• Misconception: A union omits common elements. Correct: It includes common elements once; the intersection selects precisely those common elements.
• Misconception: Set difference can be reversed without changing the answer. Correct: In general, $A-B\ne B-A$; the first set supplies the candidates.
• Misconception: De Morgan’s laws preserve the operation inside the brackets. Correct: Complementing a union gives an intersection of complements, while complementing an intersection gives their union.

Exam-style questions with model answers

Q1. Define a set and explain why the collection of five most renowned mathematicians is not well-defined. [2 marks]

1. A set is a well-defined collection of objects whose membership can be decided definitely.
2. “Most renowned” depends on the criterion used by the person making the selection, so it does not determine an unambiguous collection.

Q2. Write the positive integers satisfying $x^{2}\lt40$ in roster form. [3 marks]

1. The domain requires positive integers, so zero and negative integers are excluded before checking the inequality.
2. Compute the squares of the candidates: $1^{2}=1$, $2^{2}=4$, $3^{2}=9$, $4^{2}=16$, $5^{2}=25$ and $6^{2}=36$. Each is less than the stated bound.
3. The next square is $7^{2}=49\gt40$, and subsequent positive integers have larger squares. Therefore the complete roster is $\{1,2,3,4,5,6\}$.

Q3. Explain the difference between membership and inclusion using $A=\{1\}$, $B=\{\{1\},2\}$ and $C=\{\{1\},2,3\}$. [3 marks]

1. The complete set $A=\{1\}$ is listed as an element of $B$, so $A\in B$. This is a membership statement.
2. Both elements of $B$, namely $\{1\}$ and $2$, also belong to $C$, so $B\subset C$.
3. However, $1\in A$ and $1\notin C$. Hence $A\not\subset C$. The set containing the number is present in $C$, but the number itself is not.

Q4. Convert $\{x:x\in\mathbb R,\ -5\lt x\le7\}$ to interval notation and explain both endpoints. [2 marks]

1. The strict lower inequality excludes $-5$, giving a round bracket.
2. The upper inequality includes $7$, giving a square bracket. The interval is therefore $(-5,7]$, containing every real number satisfying those bounds.

Q5. For $U=\{1,2,3,4,5,6\}$, $A=\{2,3\}$ and $B=\{3,4,5\}$, verify $(A\cup B)'=A'\cap B'$. [5 marks]

1. Use the same universal set for every complement. Removing the members of the first set from the universe gives $A'=U-A=\{1,4,5,6\}$.
2. Remove the members of the second set from the universe in the same way: $B'=U-B=\{1,2,6\}$.
3. Select the elements occurring in both complement lists. Their intersection is $A'\cap B'=\{1,6\}$.
4. Calculate the other side independently. Combining all distinct members of the original sets gives $A\cup B=\{2,3,4,5\}$.
5. Remove this union from the universe to obtain $(A\cup B)'=\{1,6\}$. Both sides contain exactly the same elements, so the required equality is verified for these sets.

Q6. Prove that $A\cup B=A\cap B$ implies $A=B$. [5 marks]

1. Assume the given equality. Choose an arbitrary element $a\in A$. The definition of union gives $a\in A\cup B$.
2. Substitute the equal intersection: $a\in A\cap B$. By the definition of intersection, $a\in B$. Since this holds for every element chosen from the first set, $A\subset B$.
3. For the reverse direction, choose an arbitrary element $b\in B$. Then $b\in A\cup B$, and the assumed equality gives $b\in A\cap B$.
4. Intersection membership requires $b\in A$. Hence every element of the second set belongs to the first, giving $B\subset A$.
5. Both inclusions have now been proved. The sets contain exactly the same elements, and therefore $A=B$.

Key takeaways

• A set needs a definite membership rule; subjective descriptions do not necessarily specify a well-defined collection.
• Roster form lists distinct members, while set-builder form states the property that identifies exactly those members.
• The empty set is finite, and a singleton contains one element; a set containing zero is not empty.
• Equal sets contain exactly the same elements, and mutual inclusion establishes equality in both directions.
• Membership compares an object with a set, while inclusion requires every member of one set to belong to another.
• Interval brackets determine endpoint membership, and the interval includes every real number between its stated bounds.
• Union collects members of either set, intersection retains common members, and difference keeps members exclusive to the first set.
• Complements depend on a chosen universe; De Morgan’s laws exchange union and intersection when taking complements.

Test yourself

Why is the collection of vowels a set?

Its membership is well-defined: a letter either is or is not a vowel in the English alphabet.

Does repeating elements change a set?

No. A set depends on its distinct members, so repetition does not change which elements belong.

Is an empty set finite?

Yes. The definition of a finite set includes the empty set as well as sets with a definite number of elements.

What distinguishes a proper subset from a subset?

A proper subset is contained in the other set and is unequal to it; subset inclusion allows equality.

What does a filled endpoint on an interval diagram indicate?

It indicates that the endpoint belongs to the interval, corresponding to a square bracket in interval notation.

When are two sets disjoint?

They are disjoint when they have no common elements, so their intersection is the empty set.

How is a complement related to set difference?

The complement is $A'=U-A$: retain elements of the chosen universal set that do not belong to the subset.

What happens when a complement is taken twice?

The original set is restored: $(A')'=A$, with the same universal set used for both operations.