Animated Solution for Mathematics - Sets and Relations: If $X$ and $Y$ are two sets, then $X\cap (X\cup Y{)}^c$ equals.

A Journey into the Void

Welcome, my dear student. Today, we are not just solving a problem; we are peeling back the layers of logic that govern the very structure of mathematics.

We are looking at the expression $X\cap (X\cup Y{)}^c$. At first glance, it might look like a jumble of symbols, but I want you to see it as a landscape.

Imagine a vast, infinite plane—our Universal set, $U$. Within this plane, we have two distinct territories, set $X$ and set $Y$. They might overlap, they might be separate, but they exist within this universe.

The Visual Intuition

Let us start by visualizing the expression. We begin with the inner core: $X\cup Y$.

This is the union—the grand gathering of all elements that belong to $X$, or $Y$, or both. If you were to shade this on a Venn diagram, you would color the entire interior of both circles.

Now, we apply the complement operator: $(X\cup Y{)}^c$. This is the 'everything else' operator. It demands that we look at every single point in our universe $U$ that is NOT inside those two circles.

Imagine the entire page shaded gray, except for the two circles. That gray region is our complement.

Now, look at the first part of our expression: set $X$. This is simply the interior of the red circle.

The question asks for the intersection of the red circle and the gray region. Ask yourself: is there any point that is simultaneously inside the red circle and in the gray region outside the circles?

Of course not! The red circle is entirely contained within the union. Therefore, there is no overlap. The intersection is empty. We have arrived at our destination visually: the result is the empty set, $\varphi$.

The Algebraic Dance

While the visual method is beautiful, in the JEE Advanced arena, we need the cold, hard steel of algebra. Let us formalize our intuition using the laws of Set Theory.

We start with our expression:

Here, we invoke the legendary De Morgan's Law. It is the bridge that allows us to distribute the complement across the union. It tells us that the complement of a union is the intersection of the complements:

Watch how the union symbol $\cup$ flips into an intersection symbol $\cap$. This is a crucial moment. Now, we substitute this back into our original expression:

Because all our operations are now intersections, we are blessed with the Associative Law. We can regroup these terms however we like without changing the outcome. Let us group $X$ and $X^c$ together:

The Final Cancellation

Now, look at the term $(X\cap X^c)$. This is the intersection of a set with its own complement.

It is the mathematical equivalent of asking for a number that is both positive and negative at the same time. It cannot exist. It is the empty set, $\varphi$.

Our expression simplifies to:

Think about the nature of the empty set. If you intersect 'nothing' with 'something', you are left with 'nothing'. The empty set is the ultimate absorber in the world of intersections.

Thus, the entire expression collapses into the empty set:

My dear student, look at what we have done. We started with a complex-looking expression and, through the application of fundamental laws, reduced it to a single, elegant symbol: $\varphi$.

This is the beauty of JEE mathematics. It is not about memorizing formulas; it is about understanding the underlying logic.

Whether you use the visual power of Venn diagrams or the rigorous path of Set Algebra, the truth remains the same. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic. You are doing great.