Q.Let A = { a, e, i, o, u} and B = { a, b, c, d }. Is A a subset of B ? No. (Why?). Is B a subset of A? No. (Why?)

$A \subseteq B$ requires every element of $A$ to be in $B$; one missing element is enough to disprove it.

Is $A \subseteq B$? $A = \{a,e,i,o,u\}$, $B = \{a,b,c,d\}$: $e \in A$ but $e \notin B$ (also $i, o, u$) $\implies$ $A$ is not a subset of $B$. …

Understanding why the subset test can fail both ways

$A \subseteq B$ holds only when every element of $A$ is also in $B$. To disprove it, it is enough to find just one element of $A$ that is not in $B$.

Is $A$ a subset of $B$?

$A = \{a, e, i, o, u\}$, $B = \{a, b, c, d\}$.

1. Check each element of $A$ against $B$: $a \in B$, but $e \notin B$.
2. Since $e \in A$ and $e \notin B$, the subset condition already fails on this one element ($i$, $o$, and $u$ also fail the same way, but one counterexample is enough).
3. $A$ is not a subset of $B$, because $A$ contains vowels like $e, i, o, u$ that $B$ does not have.

Is $B$ a subset of $A$?

$B = \{a, b, c, d\}$, $A = \{a, e, i, o, u\}$.

1. Check each element of $B$ against $A$: $a \in A$, but $b \notin A$. …

1. Concept First — Set Membership and Subsets

The idea being tested here is Set Membership and Subset Relationship.

• A set is a collection of distinct objects (elements).
• A set $A$ is a subset of set $B$ (written $A \subseteq B$) if every element of $A$ is also an element of $B$.
• If even one element of $A$ is not in $B$, then $A$ is not a subset of $B$.

Intuition: Think of a subset like a "smaller club within a bigger club." If you want to say "all members of Club A are also members of Club B," you must check every single member of A. If any member of A is missing from B, the statement fails.

2. Step-by-Step Reasoning

Step 1: Check if $A \subseteq B$

We have:

• $A = \{ a, e, i, o, u \}$
• $B = \{ a, b, c, d \}$

What we need to do: For each element in $A$, check if it is present in $B$.

• $a$ is in $B$? ✓ Yes (both sets contain $a$).
• $e$ is in $B$? ✗ No — $B$ has only $a, b, c, d$.
• $i$ is in $B$? ✗ No.
• $o$ is in $B$? ✗ No.
• $u$ is in $B$? ✗ No.

Conclusion: Since $e, i, o, u$ are not in $B$, $A$ is not a subset of $B$.

Why this matters: The condition for subset is all elements must match. Even one missing element breaks the relationship.

Step 2: Check if $B \subseteq A$

Now reverse the roles:

• $B = \{ a, b, c, d \}$
• $A = \{ a, e, i, o, u \}$

Check each element of $B$ against $A$:

• $a$ is in $A$? ✓ Yes.
• $b$ is in $A$? ✗ No — $A$ contains only vowels.
• $c$ is in $A$? ✗ No.
• $d$ is in $A$? ✗ No.

Conclusion: Since $b, c, d$ are not in $A$, $B$ is not a subset of $A$.

3. Final Answer

• $A \subseteq B$? No.

  Reason: $e, i, o, u \in A$ but $e, i, o, u \notin B$.

• $B \subseteq A$? No. …

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. …