Q.Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that A ⊂ C?. If not, give an example

Concept: Set Membership vs. Subset Relation

The question asks whether $A \in B$ and $B \subset C$ together imply $A \subset C$. These are fundamentally different relations: $\in$ relates an element to a set, while $\subset$ relates a set to another set.

Given $A \in B$, the set $A$ is an element of $B$. Given $B \subset C$, every element of $B$ is also in $C$. So yes, $A \in C$ follows (since $A$ is an element of $B$, and all elements of $B$ are in $C$).

But $A \subset C$ asks whether every element inside $A$ is also in $C$. There's no reason this must hold. The elements of $A$ need not be elements of $C$ just because $A$ itself sits inside $B$ and $B$ sits inside $C$. …

The heart of this question lies in understanding the distinction between two relations on sets: membership and subset.

When we write $A \in B$, we are saying that $A$ itself is an element of the set $B$. Think of $B$ as a box containing objects, and $A$ is one of those objects sitting inside the box. Crucially, $A$ might be a set, but in this relationship it is being treated as a single entity.

When we write $B \subset C$, we are saying that $B$ is a subset of $C$: every element that belongs to $B$ also belongs to $C$. This is a relationship between the contents of two sets.

Now, if $A \in B$ and $B \subset C$, what can we conclude? Since $A$ is an element of $B$, and every element of $B$ is also in $C$, we can deduce that $A \in C$ — the object $A$ is also an element of $C$. But this tells us nothing about whether $A \subset C$, which would require that every element inside $A$ is also in $C$.

Let me construct a concrete counterexample to show the statement is false.

Counterexample:

1. Let $A = \{1\}$, a set containing the single element $1$.

2. Let $B = \{\{1\}, 2\}$, a set containing two elements: the set $\{1\}$ and the number $2$.

3. Let $C = \{\{1\}, 2, 3\}$, a set containing three elements: the set $\{1\}$, the number $2$, and the number $3$.

Now verify the given conditions:

• Is $A \in B$? Yes, because $A = \{1\}$ is literally one of the two elements listed in $B = \{\{1\}, 2\}$. …

Method: Counterexample Construction

This method tests a universal claim by finding a single case where it fails. If even one valid counterexample exists, the statement is false.

Steps

1. Restate the claim

   We are asked:

   If $A \in B$ and $B \subset C$, does it follow that $A \subset C$?

2. Identify the logical trap

   • $A \in B$ means $A$ is an element of $B$ (set membership).
   • $B \subset C$ means every element of $B$ is also an element of $C$.
   • $A \subset C$ would require every element of $A$ to be an element of $C$.

   The confusion: $A$ is an element of $B$, not necessarily a subset of $B$. The condition $B \subset C$ only guarantees that elements of $B$ are in $C$, but $A$ itself (as an element of $B$) is in $C$. However, the elements inside $A$ need not be in $C$.

3. Construct a counterexample

   Choose simple sets where $A$ is an element of $B$, but the contents of $A$ are not in $C$.

   Let:

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

   Check:

   • $A \in B$? Yes, $\{1\}$ is an element of $B$.
   • $B \subset C$? Yes, every element of $B$ ($\{1\}$ and $2$) is in $C$. …

Common Mistakes on Set Membership & Subset Questions

Mistake 1: Confusing ∈ (element of) with ⊂ (subset of)

The Error: Students treat $A \in B$ as if it means $A \subset B$, leading them to incorrectly conclude $A \subset C$ is always true.

Why it's wrong:

• $A \in B$ means A is an element of set B (A is inside B as a single object)
• $B \subset C$ means every element of B is also in C
• But A is an element of B, not necessarily a subset — elements and subsets are different types of objects

How to avoid:

• Always ask: "Is A a member of B, or is A a part of B?"
• Draw a membership diagram (not a Venn diagram) — show A as a single dot inside B

Mistake 2: Assuming elements of B are automatically subsets of C

The Error: Thinking "Since $B \subset C$, everything related to B must be a subset of C."

Why it's wrong:

• $B \subset C$ only guarantees that elements of B are in C
• But A is an element of B, so $A \in C$ (A is an element of C) — this is true
• However, $A \subset C$ requires every element of A to be in C, which is not guaranteed

How to avoid:

• Separate the two statements:
  • $A \in B$ and $B \subset C$ does imply $A \in C$
  • But $A \in C$ does not imply $A \subset C$

Mistake 3: Not constructing a proper counterexample

The Error: Giving an example where the statement accidentally holds true, or giving an example that violates the given conditions.

How to avoid:

Build a counterexample step-by-step:

1. Pick A as an element of B — let $A = \{1\}$ (a set with one element)
2. Make B contain A — let $B = \{ \{1\}, 2 \}$ (A is an element here) …