GATE CS 2015 Set 3 Q.33 (CS Q.23): Sets, Relations, Functions, Partial Orders and Lattices

Suppose $U$ is the power set of the set $S = \{1, 2, 3, 4, 5, 6\}$ . For any $T \in U$ , let $|T|$ denote the number of elements in $T$ and $T'$ denote the complement of $T$ . For any $T, R \in U$ , let $T \setminus R$ be the set of all elements in $T$ which are not in $R$ . Which one of the following is true?

$\forall X \in U \; (|X| = |X'|)$

$\exists X \in U \; \exists Y \in U \; (|X| = 5, |Y| = 5 \text{ and } X \cap Y = \emptyset)$

$\forall X \in U \; \forall Y \in U \; (|X| = 2, |Y| = 3 \text{ and } X \setminus Y = \emptyset)$

$\forall X \in U \; \forall Y \in U \; (X \setminus Y = Y' \setminus X')$

Official answer and solution

Official answer: D
From: Official GATE 2015 answer key

Solution

Test standard set identities and verify whether set difference preserves equality under complementation.

Let $S = \{1, 2, 3, 4, 5, 6\}$ and $U = \mathcal{P}(S)$ .

Examine the options:

Option A: $\forall X \in U \; (|X| = |X'|)$ . Counterexample: $X = \emptyset \implies |X| = 0, |X'| = 6 \ne 0$ . False.
Option B: $\exists X, Y \in U$ with $|X| = 5, |Y| = 5$ and $X \cap Y = \emptyset$ . By Pigeonhole Principle, $|X \cap Y| = |X| + |Y| - |X \cup Y| \ge 5 + 5 - 6 = 4 > 0$ . Disjointness is impossible. False.
Option C: $\forall X, Y \in U$ with $|X| = 2, |Y| = 3$ , $X \setminus Y = \emptyset$ . Counterexample: $X = \{1, 2\}, Y = \{3, 4, 5\} \implies X \setminus Y = \{1, 2\} \ne \emptyset$ . False.
Option D: $\forall X, Y \in U \; (X \setminus Y = Y' \setminus X')$ $\forall X, Y \in U (X ∖ Y = Y^{'} ∖ X^{'})$ .
Recall the definition of set difference: $A \setminus B = A \cap B'$ $A ∖ B = A \cap B^{'}$ .
- Left-hand side: $X \setminus Y = X \cap Y'$
- Right-hand side: $Y' \setminus X' = Y' \cap (X')' = Y' \cap X = X \cap Y'$
  Since set intersection is commutative, $X \setminus Y = Y' \setminus X'$ holds identically for all sets $X, Y$ . True!

Hence, Option D is true.

Final answer

Common trap

Overlooking the set identity $A \setminus B = A \cap B'$ and that $(B')' = B$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 3
Question: 33 of 65, printed as CS Q.23
Section: Computer Science and Information Technology

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