Q.If U = { a, b, c, d, e, f, g, h }, find the complements of the following sets :

Concept: Complement of a set

The complement of a set $A$ with respect to a universal set $U$, denoted $A'$ or $A^c$, consists of all elements in $U$ that are not in $A$.

Method: For each set, identify which elements from $U = \{a, b, c, d, e, f, g, h\}$ are missing.

(i) $A = \{a, b, c\}$ excludes $d, e, f, g, h$ from $U$.

So $A' = \{d, e, f, g, h\}$.

(ii) $B = \{d, e, f, g\}$ excludes $a, b, c, h$ from $U$.

So $B' = \{a, b, c, h\}$.

(iii) $C = \{a, c, e, g\}$ excludes $b, d, f, h$ from $U$.

So $C' = \{b, d, f, h\}$.

(iv) $D = \{f, g, h, a\}$ excludes $b, c, d, e$ from $U$.

So $D' = \{b, c, d, e\}$.

The complement of a set $A$ with respect to a universal set $U$, denoted $A'$ or $A^c$, is the collection of all elements that belong to $U$ but do not belong to $A$. Think of it as "everything else" in your universe of discourse.

The operation is straightforward: scan through the universal set and pick out exactly those elements that are absent from the set in question. This is a fundamental set operation that appears throughout probability, logic, and discrete mathematics.

Given $U = \{a, b, c, d, e, f, g, h\}$, we find each complement by elimination.

1. Finding $A'$ where $A = \{a, b, c\}$

The elements $a$, $b$, and $c$ are in $A$, so they cannot be in $A'$. What remains from $U$?

Looking at $U$: the elements $d, e, f, g, h$ are not in $A$.

Therefore, $A' = \{d, e, f, g, h\}$.

2. Finding $B'$ where $B = \{d, e, f, g\}$

The elements $d, e, f, g$ are in $B$. Removing these from $U$ leaves us with $a, b, c, h$.

Therefore, $B' = \{a, b, c, h\}$.

3. Finding $C'$ where $C = \{a, c, e, g\}$

The elements $a, c, e, g$ are in $C$. These are the odd-positioned letters if we think alphabetically. What's left?

From $U$, removing $a, c, e, g$ gives us $b, d, f, h$.

Therefore, $C' = \{b, d, f, h\}$.

4. Finding $D'$ where $D = \{f, g, h, a\}$

The elements $f, g, h, a$ are in $D$. Removing these from $U$ leaves $b, c, d, e$.

Therefore, $D' = \{b, c, d, e\}$.

Here's a summary table:

Concept First — The Idea of Complement of a Set

The complement of a set is one of the most intuitive ideas in set theory. It simply means:

Everything in the universal set that is not in the given set.

Think of it like this:

• You have a universal set $U$ — the “whole universe” of elements we care about.
• If you pick a subset $A$, its complement $A'$ (or $A^c$) is everything left over after removing $A$ from $U$.

Why does this matter?

• It helps us talk about “not in the set” in a precise way.
• In exams, you’ll often be asked to find complements quickly — it’s just subtraction of one set from another.

Key formula:

Step-by-Step Solution

We are given:

We need the complement of each set below.

(i) $A = \{ a, b, c \}$

Reasoning:

The complement $A'$ contains every element of $U$ that is not in $A$.

So we list all elements of $U$, then remove $a, b, c$.

Step:

$U = \{ a, b, c, d, e, f, g, h \}$

Remove $\{ a, b, c \}$ → left with $\{ d, e, f, g, h \}$

Answer:

(ii) $B = \{ d, e, f, g \}$

Reasoning:

Again, complement = everything in $U$ not in $B$.

Remove $d, e, f, g$ from $U$.

Step:

$U = \{ a, b, c, d, e, f, g, h \}$

Remove $\{ d, e, f, g \}$ → left with $\{ a, b, c, h \}$

Answer:

(iii) $C = \{ a, c, e, g \}$

Reasoning:

Remove the four elements $a, c, e, g$ from $U$.

Step:

$U = \{ a, b, c, d, e, f, g, h \}$

Remove $\{ a, c, e, g \}$ → left with $\{ b, d, f, h \}$

Answer:

(iv) $D = \{ f, g, h, a \}$

Reasoning:

Remove $f, g, h, a$ from $U$.

Notice: order doesn’t matter — sets are unordered.

Step:

$U = \{ a, b, c, d, e, f, g, h \}$

Remove $\{ a, f, g, h \}$ → left with $\{ b, c, d, e \}$

Answer:

Final Answer (Summary Table)

Why It Works & Exam Tip

Why it works:

The complement is simply the difference between the universal set and the given set.

If you understand $U$ as the “whole”, then complement is just “everything else”.

Exam tip — the common pitfall:

• Don’t forget the universal set! If $U$ changes, the complement changes.
• Don’t list elements outside $U$ — complements only contain elements from $U$.
• Order doesn’t matter — but always write elements in a clear, standard order (like alphabetical) to avoid missing any.

Quick check:

For any set $X$, $X \cup X' = U$ and $X \cap X' = \emptyset$.

Use this to verify your answer — if the union of the set and its complement doesn’t give $U$, you’ve made a mistake.

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.
• How to avoid: Check your exam board’s notation. If they use $\subset$ to mean “subset or equal”, then the answer $B$ still holds. If they use $\subsetneq$ for strict subset, the answer is still $B$.

Mistake 4: Not using a quick example to verify

• What students do: Rely only on memory.
• How to avoid: Always test with a small example: Let $A = \{1, 2\}$, $B = \{1, 2, 3, 4\}$. Then $A \cup B = \{1, 2, 3, 4\} = B$. This takes 5 seconds and eliminates doubt.

Quick Summary for Exams

Final tip: When you see “$A \subset B$”, immediately think:

“$B$ is the bigger set — union gives $B$, intersection gives $A$.”