Q.Show that the following four conditions are equivalent :

The key idea is subset listing: each condition is a different way of saying that every element of $A$ is also in $B$.

Step 1 — (i) ⇒ (ii): If $A \subset B$, then every element of $A$ is in $B$, so there is no element in $A$ that is not in $B$. Hence $A - B = \phi$.

Step 2 — (ii) ⇒ (iii): $A - B = \phi$ means $A$ has no element outside $B$, so $A \subseteq B$. Then $A \cup B = B$ (union with a subset gives the larger set).

Step 3 — (iii) ⇒ (iv): $A \cup B = B$ implies $A \subseteq B$. Therefore $A \cap B = A$ (intersection with a superset gives the smaller set). …

The heart of this problem is understanding that subset ($A \subset B$) is the most fundamental idea here, and the other three conditions are just different ways of writing the same relationship. Each one captures the fact that $A$ has no elements outside $B$.

Let’s walk through the proof in a clean cycle.

1. (i) ⇒ (ii): If $A \subset B$, then $A - B = \phi$

The set difference $A - B$ means “elements in $A$ but not in $B$”. If every element of $A$ is already in $B$ (that’s what $A \subset B$ means), then there is nothing in $A$ that is missing from $B$. So $A - B$ has no elements — it’s the empty set.

2. (ii) ⇒ (iii): If $A - B = \phi$, then $A \cup B = B$

$A - B = \phi$ tells us that no element of $A$ lies outside $B$. That means every element of $A$ is actually in $B$. So $A$ is a subset of $B$ (we’ve just rediscovered (i)!). Now look at $A \cup B$: it contains everything in $A$ and everything in $B$. But since all of $A$ is already inside $B$, the union doesn’t add anything new — it’s just $B$ itself.

Formally: $A \cup B = B$ because $A \subseteq B$.

3. (iii) ⇒ (iv): If $A \cup B = B$, then $A \cap B = A$ …

Concept: Subset Listing — Proving equivalence of set conditions

We show each condition implies the next in a cycle: (i) → (ii) → (iii) → (iv) → (i).

Step 1: (i) ⇒ (ii)

If $A \subset B$, every element of $A$ is in $B$, so $A - B$ (elements in $A$ but not in $B$) has no elements. Hence $A - B = \phi$.

Step 2: (ii) ⇒ (iii)

$A - B = \phi$ means every element of $A$ is in $B$. Then $A \cup B = B$ (since $B$ already contains all of $A$).

Step 3: (iii) ⇒ (iv) …

The Correct Answer

If $A \subset B$, then $A \cup B = B$.

Why?

Because every element of $A$ is already inside $B$. So when you take the union (all elements in $A$ or in $B$), you don’t add anything new beyond what $B$ already has. The union just gives back $B$.

Common Mistakes & How to Avoid Them

Mistake 1: Writing $A \cup B = A$

• What students think: “Since $A$ is inside $B$, the union is just the smaller set $A$.”
• Why it’s wrong: The union must include everything from both sets. $B$ has extra elements that $A$ doesn’t have — those must be included.
• How to avoid: Draw a Venn diagram. Shade $A$ and $B$ separately, then shade the union. You’ll see the larger set $B$ is fully covered.

Mistake 2: Writing $A \cup B = A \cap B$

• What students think: “If one is inside the other, union and intersection are the same.”
• Why it’s wrong:
  • $A \cup B$ = all elements in either set = $B$ (the bigger one).
  • $A \cap B$ = only elements in both sets = $A$ (the smaller one). They are equal only if $A = B$.
• How to avoid: Memorise the difference:
  • Union → bigger set (or equal).
  • Intersection → smaller set (or equal).

Mistake 3: Forgetting the special case $A = B$

• What students think: “If $A \subset B$, then $A$ is strictly smaller.”
• Why it’s wrong: In many textbooks, $A \subset B$ allows $A = B$ (some use $\subseteq$ for that). If $A = B$, then $A \cup B = A = B$ — still correct, but students sometimes panic. …