Q.If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

Concept: Set Union

The union of sets collects all distinct elements that appear in any of the sets being combined. An element belongs to $A \cup B$ if it belongs to $A$ or $B$ (or both).

Solution:

Working through each union systematically by listing all distinct elements:

1. $A \cup B = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} = \{1, 2, 3, 4, 5, 6\}$
2. $A \cup C = \{1, 2, 3, 4\} \cup \{5, 6, 7, 8\} = \{1, 2, 3, 4, 5, 6, 7, 8\}$
3. $B \cup C = \{3, 4, 5, 6\} \cup \{5, 6, 7, 8\} = \{3, 4, 5, 6, 7, 8\}$
4. $B \cup D = \{3, 4, 5, 6\} \cup \{7, 8, 9, 10\} = \{3, 4, 5, 6, 7, 8, 9, 10\}$
5. $A \cup B \cup C = \{1, 2, 3, 4, 5, 6\} \cup \{5, 6, 7, 8\} = \{1, 2, 3, 4, 5, 6, 7, 8\}$
6. $A \cup B \cup D = \{1, 2, 3, 4, 5, 6\} \cup \{7, 8, 9, 10\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ …

The union operation $A \cup B$ creates a new set containing every element that belongs to $A$, or to $B$, or to both. Think of it as pooling resources: if you have one bag of marbles and another bag of marbles, the union is all the marbles you have when you pour both bags together—but you only count each unique marble once, even if it appeared in both bags.

The key principle: an element $x \in A \cup B$ if and only if $x \in A$ or $x \in B$ (or both). Duplicates are automatically eliminated because sets contain each element at most once.

Given:

• $A = \{1, 2, 3, 4\}$
• $B = \{3, 4, 5, 6\}$
• $C = \{5, 6, 7, 8\}$
• $D = \{7, 8, 9, 10\}$

Let's work through each union systematically.

(i) $A \cup B$

1. List all elements from $A$: $1, 2, 3, 4$
2. Add elements from $B$ that aren't already listed: $5, 6$ (we skip $3$ and $4$ since they're already in $A$)
3. Arrange in ascending order: $A \cup B = \{1, 2, 3, 4, 5, 6\}$

(ii) $A \cup C$

1. Start with $A$: $1, 2, 3, 4$
2. From $C = \{5, 6, 7, 8\}$, all elements are new
3. Therefore: $A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$

(iii) $B \cup C$

1. Start with $B$: $3, 4, 5, 6$
2. From $C$, we already have $5$ and $6$, so add only $7, 8$
3. Therefore: $B \cup C = \{3, 4, 5, 6, 7, 8\}$

(iv) $B \cup D$

1. Start with $B$: $3, 4, 5, 6$
2. From $D = \{7, 8, 9, 10\}$, all elements are new
3. Therefore: $B \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}$

(v) $A \cup B \cup C$

We can build this incrementally. We already know $A \cup B = \{1, 2, 3, 4, 5, 6\}$.

1. Take $A \cup B$: $1, 2, 3, 4, 5, 6$
2. From $C = \{5, 6, 7, 8\}$, we already have $5$ and $6$, so add $7, 8$
3. Therefore: $A \cup B \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$

Thorough Solution: Set Union Problems

1. Concept First — The Idea Being Tested

What is Set Membership and Union?

The core concept here is Set Union (denoted by $\cup$). When we write $A \cup B$, we are asking: "What is the set of all elements that belong to at least one of the sets A or B?"

Intuition: Think of two circles overlapping. The union is everything inside either circle — including the overlapping part. If an element appears in both sets, we list it only once (sets don't have duplicates).

Why this matters: In exams, the most common mistake is listing an element twice. Remember: sets are collections of distinct objects.

Given Sets:

• $A = \{1, 2, 3, 4\}$
• $B = \{3, 4, 5, 6\}$
• $C = \{5, 6, 7, 8\}$
• $D = \{7, 8, 9, 10\}$

2. Step-by-Step Solutions

(i) $A \cup B$

Reasoning: We take every element that appears in A or in B (or both).

• A has: 1, 2, 3, 4
• B has: 3, 4, 5, 6
• Common elements (appear in both): 3, 4 — but we list them once

Listing in order: 1, 2, 3, 4, 5, 6

Answer: $A \cup B = \{1, 2, 3, 4, 5, 6\}$

(ii) $A \cup C$

Reasoning: Elements in A or C.

• A: 1, 2, 3, 4
• C: 5, 6, 7, 8
• No common elements (A and C are disjoint — they share nothing)

Listing: 1, 2, 3, 4, 5, 6, 7, 8

Answer: $A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$

(iii) $B \cup C$

Reasoning: Elements in B or C.

• B: 3, 4, 5, 6
• C: 5, 6, 7, 8
• Common: 5, 6 (list once)

Listing: 3, 4, 5, 6, 7, 8

Answer: $B \cup C = \{3, 4, 5, 6, 7, 8\}$

(iv) $B \cup D$

Reasoning: Elements in B or D.

• B: 3, 4, 5, 6
• D: 7, 8, 9, 10
• No common elements

Listing: 3, 4, 5, 6, 7, 8, 9, 10

Answer: $B \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}$

(v) $A \cup B \cup C$

Reasoning: Elements in any of A, B, or C. We can do this stepwise: first $A \cup B$, then union with C.

• $A \cup B = \{1, 2, 3, 4, 5, 6\}$
• Now add elements from C: 7, 8 (5 and 6 already present)

Listing: 1, 2, 3, 4, 5, 6, 7, 8

Answer: $A \cup B \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$

(vi) $A \cup B \cup D$

Reasoning: Elements in any of A, B, or D.

• $A \cup B = \{1, 2, 3, 4, 5, 6\}$
• Now add elements from D: 7, 8, 9, 10 (none already present)

Listing: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Answer: $A \cup B \cup D = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

(vii) $B \cup C \cup D$

Reasoning: Elements in any of B, C, or D.

• $B \cup C = \{3, 4, 5, 6, 7, 8\}$
• Now add elements from D: 9, 10 (7 and 8 already present)

Listing: 3, 4, 5, 6, 7, 8, 9, 10

Answer: $B \cup C \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}$

--- …

The Correct Answer

If $A \subset B$, then $A \cup B = B$.

Why?

Because every element of $A$ is already inside $B$. So when you take the union (all elements in $A$ or in $B$), you don’t add anything new beyond what $B$ already has. The union just gives back $B$.

Common Mistakes & How to Avoid Them

Mistake 1: Writing $A \cup B = A$

• What students think: “Since $A$ is inside $B$, the union is just the smaller set $A$.”
• Why it’s wrong: The union must include everything from both sets. $B$ has extra elements that $A$ doesn’t have — those must be included.
• How to avoid: Draw a Venn diagram. Shade $A$ and $B$ separately, then shade the union. You’ll see the larger set $B$ is fully covered.

Mistake 2: Writing $A \cup B = A \cap B$

• What students think: “If one is inside the other, union and intersection are the same.”
• Why it’s wrong:
  • $A \cup B$ = all elements in either set = $B$ (the bigger one).
  • $A \cap B$ = only elements in both sets = $A$ (the smaller one). They are equal only if $A = B$.
• How to avoid: Memorise the difference:
  • Union → bigger set (or equal).
  • Intersection → smaller set (or equal).

Mistake 3: Forgetting the special case $A = B$

• What students think: “If $A \subset B$, then $A$ is strictly smaller.”
• Why it’s wrong: In many textbooks, $A \subset B$ allows $A = B$ (some use $\subseteq$ for that). If $A = B$, then $A \cup B = A = B$ — still correct, but students sometimes panic. …