Q.In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

1. False. $x \in A$ means $x$ is an element of $A$; $A \in B$ means $A$ is an element of $B$. Elements of $B$ are sets, not necessarily the elements of those sets. Counterexample: Let $A = \{1\}$, $B = \{\{1\}\}$. Then $1 \in A$ and $A \in B$, but $1 \notin B$.
2. False. $A \subset B$ means $A$ is a subset of $B$; $B \in C$ means $B$ is an element of $C$. Subsets of an element of $C$ need not themselves be elements of $C$. Counterexample: Let $A = \{1\}$, $B = \{1,2\}$, $C = \{\{1,2\}\}$. Then $A \subset B$ and $B \in C$, but $A \notin C$.
3. True. If $A \subset B$ and $B \subset C$, then every element of $A$ is in $B$, and every element of $B$ is in $C$. Hence every element of $A$ is in $C$, so $A \subset C$.
4. False. It is possible that $A$ is not a subset of $B$, and $B$ is not a subset of $C$, yet $A$ is still a subset of $C$. Counterexample: Let $A = \{1\}$, $B = \{2\}$, $C = \{1,3\}$. Then $A \not\subset B$ and $B \not\subset C$, but $A \subset C$.
5. False. $x \in A$ and $A \not\subset B$ only tells us that $A$ has some element not in $B$ — it does not forbid $x$ from being outside $B$. Counterexample: Let $A = \{1,2\}$, $B = \{1\}$. Then $2 \in A$ and $A \not\subset B$, but $2 \notin B$.
6. True. If $A \subset B$, then every element of $A$ is in $B$. So if $x \notin B$, $x$ cannot be in $A$ (otherwise it would be in $B$). Hence $x \notin A$.
   ✓Final answer

   1. False,
   2. False,
   3. True,
   4. False,
   5. False,
   6. True.

Let’s unpack each statement one by one. The core confusion here is between “is an element of” ($\in$) and “is a subset of” ($\subset$). An element is a member of a set; a subset is a set whose every element is also in another set. They are not interchangeable, and one does not imply the other.

(i) If $x \in A$ and $A \in B$, then $x \in B$

False.

Here $x$ is an element of $A$, and $A$ itself is an element of $B$. But $x$ is not necessarily an element of $B$ — $B$ contains the set $A$, not the individual elements of $A$.

Counterexample:

Let $A = \{1, 2\}$ and $B = \{A\} = \{\{1, 2\}\}$.

Then $1 \in A$ and $A \in B$, but $1 \notin B$ (the only element of $B$ is the set $\{1,2\}$, not the number 1).

(ii) If $A \subset B$ and $B \in C$, then $A \in C$

False.

$A$ is a subset of $B$, meaning every element of $A$ is in $B$. But $C$ contains $B$ as an element — it does not automatically contain subsets of $B$.

Counterexample:

Let $A = \{1\}$, $B = \{1, 2\}$, and $C = \{B\} = \{\{1, 2\}\}$.

Then $A \subset B$ and $B \in C$, but $A \notin C$ (the only element of $C$ is the set $\{1,2\}$, not $\{1\}$).

(iii) If $A \subset B$ and $B \subset C$, then $A \subset C$

True.

This is the transitivity of the subset relation. If every element of $A$ is in $B$, and every element of $B$ is in $C$, then every element of $A$ is in $C$.

Proof:

Take any $x \in A$. Since $A \subset B$, we have $x \in B$. Since $B \subset C$, we have $x \in C$. Hence every element of $A$ is in $C$, so $A \subset C$.

(iv) If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$

False.

Just because $A$ is not a subset of $B$, and $B$ is not a subset of $C$, it does not mean $A$ cannot be a subset of $C$. The two conditions are independent.

Counterexample:

Let $A = \{1\}$, $B = \{2\}$, $C = \{1, 3\}$.

• $A \not\subset B$ because $1 \notin B$.
• $B \not\subset C$ because $2 \notin C$.
• Yet $A \subset C$ because $1 \in C$.

(v) If $x \in A$ and $A \not\subset B$, then $x \in B$

False.

$A \not\subset B$ means there is at least one element of $A$ that is not in $B$. But the statement claims that any element $x$ of $A$ must be in $B$ — that contradicts the meaning of $A \not\subset B$.

Counterexample:

Let $A = \{1, 2\}$, $B = \{1\}$.

Then $2 \in A$ and $A \not\subset B$ (since $2 \notin B$), but $2 \notin B$.

(vi) If $A \subset B$ and $x \notin B$, then $x \notin A$

True.

This is the contrapositive of the definition of subset. If every element of $A$ is in $B$, then anything outside $B$ cannot be in $A$.

Proof:

Assume $A \subset B$ and $x \notin B$. If $x$ were in $A$, then by $A \subset B$, $x$ would be in $B$ — contradiction. Hence $x \notin A$.

Method: Definition of Disjoint Sets

Concept: Two sets are disjoint if they have no elements in common — i.e., their intersection is the empty set ($A \cap B = \emptyset$).

Steps to solve:

1. List the elements of each set clearly.
2. Check for common elements — look for any element that appears in both sets.
3. Conclusion:
   • If at least one element is common → False (not disjoint)
   • If no element is common → True (disjoint)

(i) $\{2, 3, 4, 5\}$ and $\{3, 6\}$

• Common element: 3 appears in both sets.
• Answer: False — they are not disjoint.

(ii) $\{a, e, i, o, u\}$ and $\{a, b, c, d\}$

• Common element: a appears in both sets.
• Answer: False — they are not disjoint.

(iii) $\{2, 6, 10, 14\}$ and $\{3, 7, 11, 15\}$

• Check: 2, 6, 10, 14 vs 3, 7, 11, 15 — no common element.
• Answer: True — they are disjoint.

(iv) $\{2, 6, 10\}$ and $\{3, 7, 11\}$

• Check: 2, 6, 10 vs 3, 7, 11 — no common element.
• Answer: True — they are disjoint.

Final Summary:

🧠 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.