Q.Let A = { a, e, i, o, u } and B = { a, i, u }. Show that A ∪ B = A

Concept: Set union and subset relationships

When $B \subseteq A$ (every element of $B$ is already in $A$), the union $A \cup B$ adds nothing new to $A$.

Verification:

Given $A = \{a, e, i, o, u\}$ and $B = \{a, i, u\}$.

First observe that $B \subseteq A$ since every element of $B$ (namely $a$, $i$, $u$) appears in $A$.

By definition, $A \cup B = \{x : x \in A \text{ or } x \in B\}$.

Since every element of $B$ is already in $A$, the union contains exactly the elements of $A$: $\{a, e, i, o, u\}$. …

Why the union equals the larger set

The union $A \cup B$ collects every element that belongs to at least one of the two sets. When every element of $B$ already lives inside $A$, adding $B$ to the union contributes nothing new—$A$ already contains everything $B$ has to offer.

Think of it this way: if you have a basket of all vowels and another basket containing only three of those vowels, combining both baskets still gives you just the original set of all vowels. The smaller basket adds no new letters.

Step-by-step verification

We'll show $A \cup B = A$ by proving two inclusions: every element of $A \cup B$ belongs to $A$, and every element of $A$ belongs to $A \cup B$.

1. First observe that $B \subseteq A$

Every element of $B = \{a, i, u\}$ appears in $A = \{a, e, i, o, u\}$:

• $a \in A$ ✓
• $i \in A$ ✓
• $u \in A$ ✓

So $B$ is a subset of $A$.

2. Show $A \cup B \subseteq A$

Take any element $x \in A \cup B$. By definition of union, either $x \in A$ or $x \in B$ (or both).

• If $x \in A$, we're done.
• If $x \in B$, then since $B \subseteq A$ (from step 1), we have $x \in A$.

Either way, $x \in A$. Therefore $A \cup B \subseteq A$.

3. Show $A \subseteq A \cup B$

Take any element $y \in A$. By definition of union, if $y \in A$, then automatically $y \in A \cup B$.

Therefore $A \subseteq A \cup B$.

4. Conclude equality

Since $A \cup B \subseteq A$ and $A \subseteq A \cup B$, we have $A \cup B = A$. …

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) $A \dots C$ …

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. …