Q.Let A = { a, b }, B = { a, b, c}. Is A ⊂ B ? What is A ∪ B ?

Concept: Subset relation and union of sets.

We check whether every element of $A$ belongs to $B$. The set $A = \{a, b\}$ contains exactly two elements: $a$ and $b$. Both $a$ and $b$ are present in $B = \{a, b, c\}$. Since every element of $A$ is also an element of $B$, we have $A \subset B$ (or $A \subseteq B$ if we allow equality, though here $A \neq B$).

For the union $A \cup B$, we collect all elements that appear in either set, without repetition. The elements are $a$, $b$ from $A$ and $a$, $b$, $c$ from $B$. Combining these gives $\{a, b, c\}$.

Understanding Subsets and Unions

When we ask whether $A \subset B$, we're checking if $A$ is a subset of $B$. This means every single element that lives in $A$ must also live in $B$. Think of it as asking: "Can I find everything from the first set inside the second set?"

The union $A \cup B$, on the other hand, gathers together all elements that appear in either set (or both), without repetition. It's the combined collection of distinct elements.

Checking the Subset Relationship

1. List what's in each set.

We have $A = \{a, b\}$ and $B = \{a, b, c\}$.

2. Verify membership element by element.

For $A \subset B$ to be true, we need:

• Is $a \in B$? Yes, $a$ appears in $B$.
• Is $b \in B$? Yes, $b$ appears in $B$.

Every element of $A$ is indeed in $B$, so $A \subset B$ is true.

Finding the Union

3. Collect all distinct elements from both sets.

The union $A \cup B$ includes every element that appears in $A$, in $B$, or in both:

• From $A$: $a, b$
• From $B$: $a, b, c$

Combining these and removing duplicates (since sets don't repeat elements), we get:

Notice this is exactly the set $B$ itself, which makes sense because $A$ was already contained in $B$.

Concept: Subset Listing & Union of Sets

Step 1: Check if A ⊂ B

A set $A$ is a subset of $B$ ($A \subset B$) if every element of $A$ is also in $B$.

Here, $A = \{a, b\}$ and $B = \{a, b, c\}$. Both $a$ and $b$ are in $B$, so yes, $A \subset B$.

Step 2: Find A ∪ B

The union $A \cup B$ is the set of all elements in $A$ or $B$ (or both).

List all distinct elements: $\{a, b, c\}$.

Final Answer:

• $A \subset B$ is true.
• $A \cup B = \mathbf{\{a, b, c\}}$.

🧠 The Core Idea: Subset vs. Element

Before we list mistakes, remember the two key symbols:

• $\subset$ (subset): Every element of the first set must be in the second set.
• $\in$ (element): The entire thing on the left is a single member of the set on the right.

Mixing these up is the #1 cause of errors.

✗ Common Mistake #1: Confusing $\subset$ with $\in$

Example from the list:

Statement (v): $\{ a \} \in \{ a, b, c \}$

Why it’s wrong:

• $\{a\}$ is a set containing the letter $a$.
• $\{a, b, c\}$ contains the elements $a$, $b$, and $c$ — not the set $\{a\}$.
• So $\{a\}$ is not an element of $\{a, b, c\}$.

✓ Correct thinking:

• $\{a\} \subset \{a, b, c\}$ is true (every element of $\{a\}$ is in the big set).
• $\{a\} \in \{a, b, c\}$ is false unless the big set explicitly contains a set as an element, e.g., $\{a, \{a\}, b\}$.

How to avoid:

Ask yourself: “Is the left side a single object inside the right side, or is it a collection whose members are inside?”

✗ Common Mistake #2: Forgetting that $\subset$ requires all elements

Example from the list:

Statement (iii): $\{ 1, 2, 3 \} \subset \{ 1, 3, 5 \}$

Why it’s wrong:

• The left set has $1, 2, 3$.
• The right set has $1, 3, 5$.
• $2$ is missing from the right set. So it’s false.

✓ Correct thinking:

• For $\subset$ to be true, every element of the first set must appear in the second. One missing element = false.

How to avoid:

Check each element one by one. If even one is missing, the statement is false.

✗ Common Mistake #3: Misreading “not a subset” ($\not\subset$)

Example from the list:

Statement (i): $\{ a, b \} \not\subset \{ b, c, a \}$

Why it’s wrong:

• The left set has $a$ and $b$.
• The right set has $b, c, a$ — both $a$ and $b$ are present.
• So $\{a, b\}$ is a subset. The statement says it is not a subset — that’s false.

✓ Correct thinking:

• $\{a, b\} \subset \{b, c, a\}$ is true.
• Therefore $\{a, b\} \not\subset \{b, c, a\}$ is false.

How to avoid:

First check if it is a subset. Then apply the “not” ($\not\subset$) to decide true/false.

✗ Common Mistake #4: Overlooking the definition of the set on the right

Example from the list:

Statement (ii): $\{ a, e \} \subset \{ x : x \text{ is a vowel in the English alphabet} \}$

Why it’s correct (but often marked wrong by students):

• Vowels: $a, e, i, o, u$.
• The left set has $a$ and $e$ — both are vowels.
• So it is a subset — true.

Common error: Students sometimes think “vowel” means only $a, e, i, o, u$ but then forget to check if $a$ and $e$ are actually in that list. Or they misread the set-builder notation.

How to avoid:

Write out the actual elements of the set described in words. Then compare.

✗ Common Mistake #5: Not simplifying the set before comparing

Example from the list:

Statement (vi): $\{ x : x \text{ is an even natural number less than 6} \} \subset \{ x : x \text{ is a natural number which divides 36} \}$

Step-by-step:

• Left set: even natural numbers less than 6 → $\{2, 4\}$

• Right set: natural numbers that divide 36 → $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$

• Both $2$ and $4$ are in the right set → true.

Common error: Students guess without listing. They might think “divides 36” means only $\{1,2,3,4,6\}$ or forget $4$ divides $36$.

How to avoid:

Always list the elements of both sets explicitly before comparing.

✓ Quick Summary Table

🧪 Final Exam Tip

When in doubt, write it out.

Convert set-builder to roster form. Then check element by element. Never skip this step — it’s where most marks are lost.