Q.Decide, among the following sets, which sets are subsets of one and another: A = { x : x ∈ R and x satisfy x2 – 8x + 12 = 0 }, B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }.

Concept: Set Membership and Subset Relations

First, identify each set explicitly.

Set $A$ contains real solutions to $x^2 - 8x + 12 = 0$. Factoring: $(x-2)(x-6) = 0$, so $x = 2$ or $x = 6$. Thus $A = \{2, 6\}$.

Set $B = \{2, 4, 6\}$ is given.

Set $C = \{2, 4, 6, 8, \ldots\}$ is the set of all positive even integers.

Set $D = \{6\}$ is a singleton.

Now check subset relations ($X \subseteq Y$ means every element of $X$ is in $Y$):

• $D = \{6\} \subseteq A = \{2, 6\}$ ✓
• $D = \{6\} \subseteq B = \{2, 4, 6\}$ ✓
• $D = \{6\} \subseteq C$ (since $6$ is even) ✓
• $A = \{2, 6\} \subseteq B = \{2, 4, 6\}$ ✓
• $A = \{2, 6\} \subseteq C$ (both are even) ✓
• $B = \{2, 4, 6\} \subseteq C$ (all are even) ✓

The question asks us to identify all subset relationships among four sets. A set $X$ is a subset of $Y$ (written $X \subseteq Y$) when every element of $X$ also belongs to $Y$. The strategy is straightforward: first determine what each set actually contains, then systematically compare them.

Finding set $A$

Set $A$ is defined by a condition: $x \in \mathbb{R}$ satisfying $x^2 - 8x + 12 = 0$.

Factoring the quadratic:

So $x = 2$ or $x = 6$, giving us $A = \{2, 6\}$.

Identifying the other sets

• $B = \{2, 4, 6\}$ is explicitly listed
• $C = \{2, 4, 6, 8, \ldots\}$ is the set of all positive even integers
• $D = \{6\}$ is a singleton set

Checking all subset relationships

Now we compare each pair. There are ${4 \choose 2} = 6$ pairs to check, plus we should verify if any set is a subset of itself (which is always true, but trivial).

1. Is $A \subseteq B$?

$A = \{2, 6\}$ and $B = \{2, 4, 6\}$. Both $2$ and $6$ are in $B$, so yes, $A \subseteq B$.

2. Is $A \subseteq C$?

$C$ contains all positive even integers. Since $2$ and $6$ are both positive and even, yes, $A \subseteq C$.

3. Is $A \subseteq D$?

$D = \{6\}$ contains only $6$, but $A$ contains $2$ as well. So no, $A \not\subseteq D$.

4. Is $B \subseteq A$?

$B$ contains $4$, which is not in $A = \{2, 6\}$. So no, $B \not\subseteq A$.

5. Is $B \subseteq C$?

$B = \{2, 4, 6\}$ and all three elements are positive even integers, so they're all in $C$. Yes, $B \subseteq C$.

6. Is $B \subseteq D$?

$B$ has three elements but $D$ has only one. No, $B \not\subseteq D$.

7. Is $C \subseteq A$?

$C$ is infinite while $A$ has only two elements. No, $C \not\subseteq A$.

8. Is $C \subseteq B$?

$C$ contains $8, 10, 12, \ldots$ which are not in $B$. No, $C \not\subseteq B$.

9. Is $C \subseteq D$?

$C$ is much larger than the singleton $D$. No, $C \not\subseteq D$.

10. Is $D \subseteq A$?

$D = \{6\}$ and $6 \in A$. Yes, $D \subseteq A$.

11. Is $D \subseteq B$?

$6 \in B$, so yes, $D \subseteq B$.

12. Is $D \subseteq C$?

$6$ is a positive even integer, so yes, $D \subseteq C$.

Summary of all subset relationships

All other potential subset relations (like $B \subseteq A$, $C \subseteq B$, etc.) are false.

Method: Direct Set Listing and Subset Testing

Step 1: List each set explicitly

Set A — Solve $x^2 - 8x + 12 = 0$

Factor: $(x - 2)(x - 6) = 0$

So $x = 2$ or $x = 6$

Thus A = { 2, 6 }

Set B = { 2, 4, 6 }

Set C = { 2, 4, 6, 8, … } (all positive even integers)

Set D = { 6 }

Step 2: Test subset relationships

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

• D ⊆ A?

  D = {6}, A = {2, 6} → 6 ∈ A → Yes

• D ⊆ B?

  6 ∈ B → Yes

• D ⊆ C?

  6 ∈ C → Yes

• A ⊆ B?

  A = {2, 6}, B = {2, 4, 6} → both 2 and 6 are in B → Yes

• A ⊆ C?

  2 and 6 are both even positive integers → Yes

• B ⊆ C?

  B = {2, 4, 6}, C = {2, 4, 6, 8, …} → all three are in C → Yes

• B ⊆ A?

  4 ∉ A → No

• C ⊆ B?

  8 ∉ B → No

Step 3: Summarise all subset relations

D ⊆ A ⊆ B ⊆ C

Also:

• D ⊆ B, D ⊆ C
• A ⊆ B, A ⊆ C
• B ⊆ C

Final Answer:

The chain of subsets is:

Common Mistakes on Set Membership & Subset Questions

Mistake 1: Solving the Quadratic Incorrectly

The error: Students often factor $x^2 - 8x + 12 = 0$ as $(x-6)(x-2) = 0$ but then write the solution set as $\{6, -2\}$ or $\{2, -6\}$.

Why it happens: Rushing through factorization or sign errors.

How to avoid: Always double-check:

• $x^2 - 8x + 12 = 0$
• Factors: $(x-2)(x-6) = 0$
• Solutions: $x = 2$ or $x = 6$
• So A = {2, 6}

Pro tip: Expand your factors mentally to verify: $(x-2)(x-6) = x^2 - 8x + 12$ ✓

Mistake 2: Confusing Set C's Definition

The error: Students think C = {2, 4, 6, 8, ...} means "all even numbers" and incorrectly include 0 or negative numbers.

Why it happens: The "..." notation is ambiguous — students assume it starts from 2 and continues indefinitely.

How to avoid: Read the pattern carefully:

• C starts at 2 and increases by 2 each step
• So C = {2, 4, 6, 8, 10, ...} — only positive even numbers
• C does NOT contain 0, -2, -4, etc.

Mistake 3: Forgetting the Empty Set & Improper Subsets

The error: Students list subsets but forget that every set is a subset of itself.

Why it happens: Focusing only on "proper subsets" when the question asks for all subsets.

How to avoid: Remember the definition:

• Set X is a subset of set Y if every element of X is also in Y
• This is always true when X = Y

For this problem:

• A ⊆ A ✓ (always)
• B ⊆ B ✓
• C ⊆ C ✓
• D ⊆ D ✓

Mistake 4: Missing the D ⊆ A Relationship

The error: Students see D = {6} and A = {2, 6} but say "D is not a subset of A because A has an extra element."

Why it happens: Misunderstanding that subsets can be smaller — a subset doesn't need to contain all elements of the superset.

How to avoid: Check element-by-element:

• D = {6}
• Is 6 ∈ A? Yes (A = {2, 6})
• Therefore D ⊆ A ✓

Key insight: A subset can be "smaller" — it just needs every element to belong to the larger set.

Mistake 5: Incorrectly Claiming A ⊆ B or B ⊆ A

The error: Students say A ⊆ B because both contain 2 and 6, forgetting that A also contains 2.

Why it happens: Not checking every element systematically.

How to avoid: Use the element test:

• A = {2, 6}, B = {2, 4, 6}
• Is every element of A in B? 2 ∈ B ✓, 6 ∈ B ✓ → A ⊆ B ✓
• Is every element of B in A? 4 ∈ A? No → B ⊈ A ✗

Mistake 6: Confusing "∈" with "⊆"

The error: Writing "6 ⊆ A" instead of "6 ∈ A" or "D ⊆ A".

Why it happens: Mixing up element membership vs. subset notation.

How to avoid: Remember:

• ∈ = "is an element of" (for individual elements)
• ⊆ = "is a subset of" (for sets)

Correct usage:

• 6 ∈ A ✓ (6 is an element of set A)
• D ⊆ A ✓ (set D is a subset of set A)
• {6} ⊆ A ✓ (the set containing 6 is a subset of A)

Final Correct Answer

A = {2, 6}, B = {2, 4, 6}, C = {2, 4, 6, 8, ...}, D = {6}

Subset relationships:

• A ⊆ B ✓ (2 and 6 are both in B)
• D ⊆ A ✓ (6 is in A)
• D ⊆ B ✓ (6 is in B)
• D ⊆ C ✓ (6 is in C)
• A ⊆ C ✓ (2 and 6 are both in C)
• B ⊆ C ✓ (2, 4, 6 are all in C)
• Every set is a subset of itself