Q.Examine whether the following statements are true or false:

Concept: Set Membership and Subset Relations

We need to distinguish between $\subseteq$ (subset: every element of $A$ is in $B$), $\subset$ (proper subset: $A \subseteq B$ and $A \neq B$), and $\in$ (membership: an object belongs to a set).

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

Both elements $a$ and $b$ are in $\{b, c, a\}$, so $\{a, b\} \subseteq \{b, c, a\}$. Since the sets are not equal, $\{a, b\} \subset \{b, c, a\}$ is true. The statement claims the opposite, so it is false.

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

The vowel set is $\{a, e, i, o, u\}$. Both $a$ and $e$ are vowels, and the sets are not equal. True.

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

The element $2$ is not in $\{1, 3, 5\}$, so $\{1, 2, 3\} \not\subseteq \{1, 3, 5\}$. False.

(iv) $\{a\} \subset \{a, b, c\}$

The element $a$ is in $\{a, b, c\}$, and $\{a\} \neq \{a, b, c\}$. True.

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

The elements of $\{a, b, c\}$ are $a$, $b$, and $c$ (not the set $\{a\}$). False.

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

The first set is $\{2, 4\}$. Both $2$ and $4$ divide $36$, and the sets are not equal. True.

Understanding Set Membership vs. Subsets

The heart of this problem lies in distinguishing two fundamental relations:

• Element membership ($\in$): An object belongs to a set. For example, $a \in \{a, b, c\}$ because $a$ is one of the elements listed.
• Subset ($\subseteq$ or $\subset$): Every element of the first set is also in the second set. For example, $\{a, b\} \subseteq \{a, b, c\}$ because both $a$ and $b$ appear in the larger set.

The symbol $\subset$ sometimes means "proper subset" (excluding equality) and sometimes just "subset." In most Indian textbooks, $\subset$ is used for subset (allowing equality), while $\subsetneq$ denotes proper subset. We'll interpret $\subset$ as subset here, and $\not\subset$ as "not a subset."

Let's examine each statement systematically.

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

1. The set $\{a, b\}$ contains two elements: $a$ and $b$.
2. The set $\{b, c, a\}$ contains three elements: $b$, $c$, and $a$ (order doesn't matter in sets).
3. Check membership: Is $a \in \{b, c, a\}$? Yes. Is $b \in \{b, c, a\}$? Yes.
4. Since every element of $\{a, b\}$ belongs to $\{b, c, a\}$, we have $\{a, b\} \subseteq \{b, c, a\}$.

The statement claims $\{a, b\} \not\subset \{b, c, a\}$, which contradicts what we just found.

Statement (i) is FALSE.

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

1. The vowels in the English alphabet are: $a, e, i, o, u$.
2. So the right-hand set is $\{a, e, i, o, u\}$.
3. The left-hand set is $\{a, e\}$.
4. Check: Is $a$ a vowel? Yes. Is $e$ a vowel? Yes.
5. Both elements of $\{a, e\}$ are in $\{a, e, i, o, u\}$.

Statement (ii) is TRUE.

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

1. The left set contains $1, 2, 3$.
2. The right set contains $1, 3, 5$.
3. Check each element: Is $1 \in \{1, 3, 5\}$? Yes. Is $2 \in \{1, 3, 5\}$? No. Is $3 \in \{1, 3, 5\}$? Yes.
4. Since $2$ is in the left set but not in the right set, $\{1, 2, 3\}$ is not a subset of $\{1, 3, 5\}$.

Statement (iii) is FALSE.

Statement (iv): $\{a\} \subset \{a, b, c\}$

1. The left set contains one element: $a$.
2. The right set contains $a, b, c$.
3. Is $a \in \{a, b, c\}$? Yes.
4. Every element (just $a$) of the left set is in the right set.

Statement (iv) is TRUE.

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

1. The symbol $\in$ asks: Is the object on the left an element of the set on the right?
2. The object on the left is the set $\{a\}$.
3. The elements of $\{a, b, c\}$ are: $a$, $b$, and $c$ (three individual letters, not sets).
4. The set $\{a\}$ is not one of these elements.

Statement (v) is FALSE.

This is the classic pitfall mentioned above. We have $a \in \{a, b, c\}$ (TRUE) but $\{a\} \in \{a, b, c\}$ (FALSE).

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

1. First, list the left set. Even natural numbers less than $6$: $2, 4$. So the left set is $\{2, 4\}$.
2. Next, list the right set. Natural numbers that divide $36$: The divisors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18, 36$. So the right set is $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$.
3. Check: Is $2$ a divisor of $36$? Yes ($36 = 2 \times 18$). Is $4$ a divisor of $36$? Yes ($36 = 4 \times 9$).
4. Both elements of $\{2, 4\}$ are in the set of divisors of $36$.

Statement (vi) is TRUE.

Summary Table

Here’s a clear, concept-first solution method for listing elements of a set from its set-builder form.

Method Name: Roster Form Conversion (Set-Builder to Roster)

Core Concept:

A set defined in set-builder notation (e.g., $\{x : \text{condition on } x\}$) is a description. To list its elements, you must identify every distinct object that satisfies the given condition, within the implied universal set (natural numbers, integers, letters, etc.).

Steps for Roster Form Conversion

1. Identify the domain (what kind of objects $x$ can be: natural numbers, integers, letters, months).
2. Interpret the condition precisely. Translate any inequality or property into a clear range or list.
3. Generate candidates systematically (start from the smallest, move step-by-step).
4. Test each candidate against the condition. Include it only if it satisfies.
5. Write the final list inside curly braces $\{ \}$, separated by commas. If no elements exist, write $\{\}$ or $\emptyset$.

Applying the Method to Each Set

(i) $A = \{x : x \text{ is an odd natural number}\}$

• Domain: Natural numbers ($\mathbb{N}$). Usually $\mathbb{N} = \{1, 2, 3, \dots\}$.
• Condition: $x$ is odd.
• Generate: $1, 2, 3, 4, 5, \dots$
• Test: 1 (odd ✓), 2 (even ✗), 3 (odd ✓), 4 (even ✗), …
• Result: The set is infinite. In roster form, we show the pattern.

Answer:

(ii) $B = \{x : x \text{ is an integer, } -\frac{1}{2} < x < \frac{9}{2}\}$

• Domain: Integers ($\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}$).
• Condition: $-\frac{1}{2} < x < \frac{9}{2}$.
  • $-\frac{1}{2} = -0.5$
  • $\frac{9}{2} = 4.5$
• Generate integers between -0.5 and 4.5: $0, 1, 2, 3, 4$.
• Test: All satisfy (0 > -0.5, 4 < 4.5).

Answer:

(iii) $C = \{x : x \text{ is an integer, } x^2 \leq 4\}$

• Domain: Integers.
• Condition: $x^2 \leq 4$.
• Generate: Start from $0$, then positive and negative integers.
  • $0^2 = 0 \leq 4$ ✓
  • $1^2 = 1 \leq 4$ ✓, $(-1)^2 = 1 \leq 4$ ✓
  • $2^2 = 4 \leq 4$ ✓, $(-2)^2 = 4 \leq 4$ ✓
  • $3^2 = 9 > 4$ ✗ (stop expanding outward).
• Result: Integers whose square is at most 4.

Answer:

(iv) $D = \{x : x \text{ is a letter in the word “LOYAL”}\}$

• Domain: Letters of the English alphabet.
• Condition: $x$ appears in the word “LOYAL”.
• List letters in order: L, O, Y, A, L.
• Remove duplicates (sets contain distinct elements): L, O, Y, A.

Answer:

(v) $E = \{x : x \text{ is a month of a year not having 31 days}\}$

• Domain: The 12 months of the year.
• Condition: Month does not have 31 days.
• Recall: Months with 31 days: Jan, Mar, May, Jul, Aug, Oct, Dec.
• Remaining months (30 or fewer days): Feb (28/29), Apr (30), Jun (30), Sep (30), Nov (30).

Answer:

(vi) $F = \{x : x \text{ is a consonant in the English alphabet which precedes } k\}$

• Domain: Consonants in the English alphabet.
• Condition: The consonant comes before the letter ‘k’ in alphabetical order.
• Alphabet order: a, b, c, d, e, f, g, h, i, j, k, …
• Consonants before k: b, c, d, f, g, h, j (vowels a, e, i are excluded).

Answer:

Quick Summary Table

🧠 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.