Animated Solution for Mathematics - Sets and Relations: Let A, B and C be sets such that $\varphi \ne A\cap B\subseteq C$. Then which of the following statements is not true?

A Detective Story

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey of logical deduction.

Set theory is the bedrock of mathematics, and in JEE Advanced, it is often tested not through complex calculations, but through the elegance of logical relationships.

Let us dissect the problem: we have three sets, $A$, $B$, and $C$, bound by the condition $\phi eq A \cap B \subseteq C$. This is our 'master rule.' Everything we do must respect this constraint.

Visualizing the Reality

Before we touch a single variable, let us paint a picture. Imagine two circles, $A$ and $B$, overlapping.

The region where they overlap, $A\cap B$, is our 'common ground.' The problem tells us this region is not empty—there is at least one element living there.

Now, introduce set $C$. The condition $A\cap B\subseteq C$ tells us that this common ground is entirely contained within $C$. Think of $C$ as a larger container that has swallowed the intersection of $A$ and $B$.

Evaluating the Truths

Let us test the options. We start with Option 4: $B \cap C eq \phi$.

We know $A\cap B\subseteq B$ (by definition) and $A\cap B\subseteq C$ (by our master rule). Therefore, the region $A\cap B$ is common to both $B$ and $C$.

Since $A\cap B$ is not empty, $B\cap C$ must also contain those elements. Thus, $B \cap C eq \phi$ is undeniably true.

Next, look at Option 2: $(C\cup A)\cap (C\cup B)=C$. This looks intimidating, but let us apply the Distributive Law of sets in reverse.

The expression simplifies to:

Now, recall our master rule: $A\cap B\subseteq C$. When you take the union of a set with its own subset, you simply get the larger set back. Thus, $C\cup (A\cap B)=C$. This statement is also true!

The Decomposition Strategy

Now, consider Option 3: If $(A-B)\subseteq C$, then $A\subseteq C$. This is a classic JEE trap.

To solve this, we use the decomposition of a set. Any set $A$ can be written as the union of two disjoint parts: the part it shares with $B$ (which is $A\cap B$) and the part it keeps to itself (which is $A-B$).

So, we have:

We already know $A\cap B\subseteq C$. If we assume $(A-B)\subseteq C$, then both parts of $A$ are inside $C$. If both parts are inside $C$, then the whole set $A$ must be inside $C$. This statement is also true.

The Counter-Example

Finally, we arrive at Option 1: If $(A-C)\subseteq B$, then $A\subseteq B$. This is the statement that feels suspicious.

Let us test it with a counter-example. Let $A=\{1,2\}$, $C=\{1,2\}$, and $B=\{1\}$.

First, check the master rule: $A\cap B=\{1\}$, which is not empty and is a subset of $C$. The rule holds.

Now, check the premise of Option 1: $A-C=\{1,2\}-\{1,2\}=\varphi$. Since $\varphi \subseteq B$ is always true, the premise is satisfied.

But look at the conclusion: is $A\subseteq B$? No! $A$ contains the element $2$, which is not in $B$. We have found our culprit. Option 1 is the statement that is not true.

Mathematics is not just about following formulas; it is about testing boundaries. By using a counter-example, we proved that a logical implication does not hold in all cases.

This is the mindset of a true problem solver. Keep questioning, keep visualizing, and keep pushing the boundaries of your understanding. You are doing great.