GATE CS 2003 Q37 — Sets, Relations and Functions

To find the relationship between $g(h(D))$ and $D$, let us analyze the definitions of the functions step-by-step:

1. Definition of $h(D)$ for $D rianglelefteq B$:

   $$h(D) = \{x \in A \mid f(x) \in D\}$$

   This is the preimage of $D$ under $f$, containing all elements in $A$ that map to elements in $D$.

2. Definition of $g(C)$ for $C rianglelefteq A$:

   $$g(C) = \{f(x) \mid x \in C\}$$

   This is the image of the subset $C$ under $f$.

Now, let us evaluate $g(h(D))$ by substituting $C = h(D)$:

$$g(h(D)) = \{f(x) \mid x \in h(D)\}$$

By definition of $h(D)$, any $x \in h(D)$ satisfies $f(x) \in D$. Therefore:

$$\text{If } y \in g(h(D)) \implies y = f(x) \text{ for some } x \in h(D) \implies y \in D$$

Since every element in $g(h(D))$ must belong to $D$, we have:

$$g(h(D)) \subseteq D$$

Counterexample for other options:

Let $A = \{1\}$, $B = \{a, b\}$, and $f: A \to B$ be defined as $f(1) = a$. Note that $f$ is injective.

Let $D = \{a, b\} \subseteq B$.

• $h(D) = \{x \in A \mid f(x) \in \{a, b\}\} = \{1\}$
• $g(h(D)) = g(\{1\}) = \{f(1)\} = \{a\}$

Checking the options with these sets:

• Option A: $g(h(D)) \subseteq D \implies \{a\} \subseteq \{a, b\}$ (True)
• Option B: $g(h(D)) \supseteq D \implies \{a\} \supseteq \{a, b\}$ (False)
• Option C: $g(h(D)) \cap D = \phi \implies \{a\} \cap \{a, b\} = \{a\} \ne \phi$ (False)
• Option D: $g(h(D)) \cap (B - D) \ne \phi \implies \{a\} \cap \phi = \phi$ (False)