Q.Consider the sets φ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol ⊂ or ⊄ between each of the following pair of sets:

$X \subset Y$ means every element of $X$ is also in $Y$; if even one element of $X$ is missing from $Y$, write $X \not\subset Y$.

(i) $\phi$ has no elements, so it is a subset of every set $\implies \phi \subset B$.

(ii) $A = \{1,3\}$, $B = \{1,5,9\}$: $3 \in A$ but $3 \notin B$ $\implies A \not\subset B$.

(iii) $A = \{1,3\}$, $C = \{1,3,5,7,9\}$: both $1$ and $3$ are in $C$ $\implies A \subset C$.

(iv) $B = \{1,5,9\}$, $C = \{1,3,5,7,9\}$: every element of $B$ is in $C$ $\implies B \subset C$.

Understanding the subset relation

$X \subset Y$ ("$X$ is a subset of $Y$") means every element of $X$ is also an element of $Y$. To decide the symbol, check each element of the first set against the second: if all of them are found in the second set, use $\subset$; if even one is missing, use $\not\subset$.

Given: $\phi$ (empty set), $A = \{1, 3\}$, $B = \{1, 5, 9\}$, $C = \{1, 3, 5, 7, 9\}$.

(i) $\phi$ and $B$

The empty set contains no elements at all, so there is nothing in it that could fail to be in $B$. By definition, the empty set is a subset of every set.

$\phi \subset B$

(ii) $A$ and $B$

$A = \{1, 3\}$, $B = \{1, 5, 9\}$.

• $1 \in A$ and $1 \in B$ — passes.
• $3 \in A$ but $3 \notin B$ — fails.

Since not every element of $A$ is in $B$, $A$ is not a subset of $B$.

$A \not\subset B$

(iii) $A$ and $C$

$A = \{1, 3\}$, $C = \{1, 3, 5, 7, 9\}$.

• $1 \in A$ and $1 \in C$ — passes.
• $3 \in A$ and $3 \in C$ — passes.

Every element of $A$ is in $C$, so $A$ is a subset of $C$.

$A \subset C$

(iv) $B$ and $C$

$B = \{1, 5, 9\}$, $C = \{1, 3, 5, 7, 9\}$.

• $1 \in B$ and $1 \in C$ — passes.
• $5 \in B$ and $5 \in C$ — passes.
• $9 \in B$ and $9 \in C$ — passes.

Every element of $B$ is in $C$, so $B$ is a subset of $C$.

$B \subset C$

Method: Subset Test by Element Checking

This method checks whether every element of the first set is also present in the second set.

• If yes → the first set is a subset ($\subset$) of the second.
• If even one element is missing → it is not a subset ($\not\subset$).

Steps

1. List all elements of the first set.
2. Check each element — is it present in the second set?
3. Decision:
   • All found → $\subset$
   • Any missing → $\not\subset$

Applying to each pair

(i) $\phi \dots B$

• $\phi$ has no elements.
• There is no element to check — the condition "every element is present" is vacuously true.
• Result: $\phi \subset B$

(ii) $A \dots B$

• $A = \{1, 3\}$, $B = \{1, 5, 9\}$
• Check: $1 \in B$ ✓, $3 \in B$ ✗ (3 is not in B)
• Result: $A \not\subset B$

(iii) $A \dots C$

• $A = \{1, 3\}$, $C = \{1, 3, 5, 7, 9\}$
• Check: $1 \in C$ ✓, $3 \in C$ ✓
• Result: $A \subset C$

(iv) $B \dots C$

• $B = \{1, 5, 9\}$, $C = \{1, 3, 5, 7, 9\}$
• Check: $1 \in C$ ✓, $5 \in C$ ✓, $9 \in C$ ✓
• Result: $B \subset C$

Final Answer

🧠 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.