Q.Show that if A ⊂ B, then C – B ⊂ C – A.

Concept: Set difference and subset properties.

We need to prove that every element of $C - B$ also belongs to $C - A$.

Take an arbitrary element $x \in C - B$. By definition of set difference, this means $x \in C$ and $x \notin B$.

Since $A \subset B$, every element of $A$ is also in $B$. The contrapositive tells us: if $x \notin B$, then $x \notin A$. …

The set difference $C - A$ consists of all elements that belong to $C$ but not to $A$. The claim asks us to show that if $A$ is a subset of $B$, then removing $B$ from $C$ cannot give us more elements than removing $A$ from $C$.

The intuition is straightforward: if $A \subseteq B$, then $B$ contains everything $A$ does, possibly more. When we subtract $B$ from $C$, we're throwing away at least as much as we would by subtracting $A$. So $C - B$ should be "smaller" (contained in) $C - A$.

Let's prove this by showing that every element of $C - B$ must also belong to $C - A$.

1. Take an arbitrary element of $C - B$.

   Let $x \in C - B$. By the definition of set difference, this means $x \in C$ and $x \notin B$.

2. Use the hypothesis $A \subseteq B$.

   Since $A \subseteq B$, every element of $A$ is also in $B$. The contrapositive of this statement is equally useful: if an element is not in $B$, then it cannot be in $A$ either. Formally, $x \notin B \implies x \notin A$.

3. Conclude that $x \in C - A$.

   We know $x \in C$ (from step 1) and $x \notin A$ (from step 2). Therefore, by definition, $x \in C - A$.

4. Generalize. …

Concept: Set Difference

We need to show: if $A \subset B$, then $C - B \subset C - A$.

Step 1: Take any element $x \in C - B$.

By definition of set difference, $x \in C$ and $x \notin B$.

Step 2: Since $A \subset B$, every element of $A$ is also in $B$.

Thus, if $x \notin B$, then $x$ cannot be in $A$ (otherwise $x$ would be in $B$). So $x \notin A$. …

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. …