Q.Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

Concept: Set Membership vs. Subset — An element belongs to a set ($\in$), while a set is contained in another set ($\subset$) only if every element of the first is also in the second.

Reasoning steps:

1. Check each statement against the definition of $\in$ and $\subset$.

   $A = \{ 1, 2, \{3,4\}, 5 \}$ has four elements: the numbers 1, 2, 5 and the set $\{3,4\}$.

2. Evaluate each:

   • (i) $\{3,4\} \subset A$ — False. $\{3,4\}$ is an element of $A$, not a subset. For it to be a subset, both 3 and 4 would need to be elements of $A$, but they are not.
   • (ii) $\{3,4\} \in A$ — True. It is listed as an element.
   • (iii) $\{\{3,4\}\} \subset A$ — True. The only element of $\{\{3,4\}\}$ is $\{3,4\}$, which is in $A$.
   • (iv) $1 \in A$ — True.
   • (v) $1 \subset A$ — False. 1 is not a set; subset relation applies only between sets.
   • (vi) $\{1,2,5\} \subset A$ — True. All three numbers are elements of $A$.
   • (vii) $\{1,2,5\} \in A$ — False. The set $\{1,2,5\}$ is not listed as an element of $A$.
   • (viii) $\{1,2,3\} \subset A$ — False. 3 is not an element of $A$ (only $\{3,4\}$ is).
   • (ix) $\phi \in A$ — False. The empty set is not listed as an element.
   • (x) $\phi \subset A$ — True. The empty set is a subset of every set.
   • (xi) $\{\phi\} \subset A$ — False. The only element of $\{\phi\}$ is $\phi$, which is not in $A$.

3. Identify the incorrect statements: (i), (v), (vii), (viii), (ix), (xi).

We are given $A = \{ 1, 2, \{3, 4\}, 5 \}$.

Notice that $A$ has four elements: the numbers $1$, $2$, $5$, and the set $\{3, 4\}$.

The curly braces around $\{3,4\}$ mean that the whole thing is a single element of $A$ — it is not that $3$ and $4$ themselves are directly in $A$.

Let’s go through each statement one by one.

1. Statement (i): $\{3, 4\} \subset A$

   For $\{3,4\}$ to be a subset of $A$, every element of $\{3,4\}$ must be an element of $A$.

   The elements of $\{3,4\}$ are $3$ and $4$. Are $3$ and $4$ in $A$?

   $A$ contains $1$, $2$, $\{3,4\}$, and $5$ — no $3$ or $4$ directly. So $3 \notin A$ and $4 \notin A$.

   Hence $\{3,4\} \not\subset A$. Incorrect.

2. Statement (ii): $\{3, 4\} \in A$

   This asks: is the set $\{3,4\}$ itself an element of $A$?

   Yes — it appears as the third item inside the braces of $A$. So $\{3,4\} \in A$. Correct.

3. Statement (iii): $\{\{3, 4\}\} \subset A$

   The set $\{\{3,4\}\}$ has one element: the set $\{3,4\}$.

   For it to be a subset of $A$, that one element must belong to $A$.

   We already know $\{3,4\} \in A$, so every element of $\{\{3,4\}\}$ is in $A$.

   Therefore $\{\{3,4\}\} \subset A$. Correct.

4. Statement (iv): $1 \in A$

   $1$ is listed directly in $A$. So yes, $1 \in A$. Correct.

5. Statement (v): $1 \subset A$

   The symbol $\subset$ is for sets. $1$ is a number, not a set.

   Even if we treat $1$ as the set $\{1\}$, then $\{1\} \subset A$ would be true, but $1 \subset A$ is meaningless in standard set theory — it is false because $1$ is not a set.

   So incorrect.

6. Statement (vi): $\{1, 2, 5\} \subset A$

   Check each element: $1 \in A$, $2 \in A$, $5 \in A$. All three are in $A$.

   So $\{1,2,5\} \subset A$. Correct.

7. Statement (vii): $\{1, 2, 5\} \in A$

   This asks: is the set $\{1,2,5\}$ itself an element of $A$?

   $A$ contains $1$, $2$, $\{3,4\}$, $5$ — no $\{1,2,5\}$ as a single element.

   So incorrect.

8. Statement (viii): $\{1, 2, 3\} \subset A$

   Check elements: $1 \in A$, $2 \in A$, but $3 \notin A$.

   Since $3$ is missing, $\{1,2,3\} \not\subset A$. Incorrect.

9. Statement (ix): $\phi \in A$

   $\phi$ (the empty set) is not listed as an element of $A$. $A$ has only $1$, $2$, $\{3,4\}$, $5$.

   So $\phi \notin A$. Incorrect.

10. Statement (x): $\phi \subset A$

    The empty set is a subset of every set. This is a fundamental property: $\phi \subseteq A$ for any $A$.

    So $\phi \subset A$ is correct.

11. Statement (xi): $\{\phi\} \subset A$

    The set $\{\phi\}$ has one element: the empty set.

    For it to be a subset of $A$, that element $\phi$ must be in $A$.

    But $\phi \notin A$ (as we saw in (ix)). So $\{\phi\} \not\subset A$. Incorrect.

Method: Element vs. Subset Identification Method

This method distinguishes between membership ($\in$) and subset ($\subset$) by checking:

• Membership ($\in$): Is the item a direct element of the set?
• Subset ($\subset$): Is every element of the first set also an element of the second set?

Step 1 — List the direct elements of $A$

$A = \{ 1, 2, \{3, 4\}, 5 \}$

The elements of $A$ are:

• $1$
• $2$
• $\{3, 4\}$ (the set itself is one element)
• $5$

Step 2 — Check each statement

(i) $\{3, 4\} \subset A$

• For $\{3, 4\}$ to be a subset of $A$, both $3$ and $4$ must be elements of $A$.
• $3 \notin A$ and $4 \notin A$.
• Incorrect ✗

(ii) $\{3, 4\} \in A$

• $\{3, 4\}$ is a direct element of $A$.
• Correct ✓

(iii) $\{\{3, 4\}\} \subset A$

• The only element of $\{\{3, 4\}\}$ is $\{3, 4\}$.
• $\{3, 4\} \in A$, so every element of the left set is in $A$.
• Correct ✓

(iv) $1 \in A$

• $1$ is a direct element.
• Correct ✓

(v) $1 \subset A$

• $1$ is not a set (it's a number), so it cannot be a subset.
• Incorrect ✗

(vi) $\{1, 2, 5\} \subset A$

• $1 \in A$, $2 \in A$, $5 \in A$ — all three are elements.
• Correct ✓

(vii) $\{1, 2, 5\} \in A$

• The set $\{1, 2, 5\}$ is not listed as a direct element of $A$.
• Incorrect ✗

(viii) $\{1, 2, 3\} \subset A$

• $3 \notin A$.
• Incorrect ✗

(ix) $\phi \in A$

• $\phi$ (empty set) is not listed as an element of $A$.
• Incorrect ✗

(x) $\phi \subset A$

• The empty set is a subset of every set.
• Correct ✓

(xi) $\{\phi\} \subset A$

• The only element of $\{\phi\}$ is $\phi$.
• $\phi \notin A$, so this fails.
• Incorrect ✗

Final Answer — Incorrect Statements

Incorrect: (i), (v), (vii), (viii), (ix), (xi)

Correct: (ii), (iii), (iv), (vi), (x)

🧠 Core Concept: Elements vs. Subsets

• Element (∈): Something directly inside the set (one level only).
• Subset (⊂): Every element of the first set is also an element of the second set.

Given:

The elements of $A$ are:

• $1$
• $2$
• $\{3,4\}$ (a set as an element)
• $5$

✗ Common Mistakes & How to Avoid Them

1. Confusing ∈ and ⊂ for {3,4}

Mistake: Thinking {3,4} ⊂ A is correct because {3,4} looks like a subset.

Why it's wrong:

{3,4} is an element of $A$, not a subset. For it to be a subset, both $3$ and $4$ must be elements of $A$ — but $3$ and $4$ are not in $A$.

✓ Correct:

• {3,4} ∈ A is true (it's an element).
• {3,4} ⊂ A is false.

Avoid: Always check: "Are all members of the left set inside the right set?" If not, ⊂ is false.

2. Forgetting that {3,4} is a single element

Mistake: Treating {3,4} as two separate elements.

Why it's wrong:

In $A$, {3,4} is one object. You cannot break it apart unless you use ∈ on its contents.

✓ Correct:

• {{3,4}} ⊂ A is true because the only element of {{3,4}} is {3,4}, which is in $A$.

Avoid: Visualise the set as a bag. {3,4} is one item in the bag, not two.

3. Writing 1 ⊂ A (number as a subset)

Mistake: Thinking a number can be a subset.

Why it's wrong:

$1$ is not a set. Only sets can be subsets. 1 ∈ A is correct (element), but 1 ⊂ A is meaningless in standard set theory.

✓ Correct:

• 1 ∈ A is true.
• 1 ⊂ A is false (or invalid).

Avoid: Remember: ∈ for elements, ⊂ for sets. Numbers are elements, not sets (unless defined otherwise).

4. Thinking {1,2,5} ∈ A is correct

Mistake: Because 1,2,5 are in $A$, assuming the set {1,2,5} is also an element.

Why it's wrong:

{1,2,5} is not listed as a single element inside $A$. The elements are 1, 2, {3,4}, 5 — not {1,2,5}.

✓ Correct:

• {1,2,5} ⊂ A is true (all three are elements).
• {1,2,5} ∈ A is false.

Avoid: Check the exact list of elements. A set of elements is not automatically an element itself.

5. Confusing empty set φ with {φ}

Mistake: Thinking φ ∈ A is true because empty set is a subset of everything.

Why it's wrong:

φ is not listed as an element of $A$. φ ⊂ A is always true (empty set is subset of every set), but φ ∈ A is false unless explicitly included.

✓ Correct:

• φ ⊂ A is true.
• φ ∈ A is false.
• {φ} ⊂ A is false (because φ is not an element of $A$).

Avoid: ⊂ is about containment of members; ∈ is about direct membership. Empty set is a subset, not necessarily an element.

✓ Quick Summary Table

🧪 Final Exam Tip

Always draw the "one-level" picture: List the elements of $A$ with curly braces. Then ask:

• Is it one of these? → ∈
• Are all its pieces in this list? → ⊂

This one habit eliminates 90% of errors.