سوال ۹۸ کنکور ارشد علوم کامپیوتر ۱۴۰۰

We test each power-set identity. Recall $X \in P(S) \iff X \subseteq S$.

Option 1 & 2 — the intersection identity. The equality $P(A) \cap P(B) = P(A \cap B)$ is a standard theorem: $X \in P(A) \cap P(B) \iff X \subseteq A \text{ and } X \subseteq B \iff X \subseteq A \cap B \iff X \in P(A \cap B).$ So option 2 (the equality) is always true, and option 1 (the $\subseteq$ half) is therefore also always true.

Option 4 — the union inclusion. If $X \in P(A) \cup P(B)$, then $X \subseteq A$ or $X \subseteq B$; either way $X \subseteq A \cup B$, so $X \in P(A \cup B)$. Thus $P(A) \cup P(B) \subseteq P(A \cup B)$ is always true. (The reverse fails — e.g. $A = \{1\}, B = \{2\}$ gives $\{1,2\} \in P(A \cup B)$ but $\{1,2\} \notin P(A) \cup P(B)$ — but the printed direction holds.)

Option 3 — the difference inclusion. Claim: $P(A) - P(B) \subseteq P(A - B)$. Take $X \in P(A) - P(B)$, i.e. $X \subseteq A$ but $X \not\subseteq B$. We would need $X \subseteq A - B$. This fails.

Counterexample: let $A = \{1, 2\}$ and $B = \{2\}$.

• $X = \{1, 2\} = A$ satisfies $X \subseteq A$ and $X \not\subseteq B$, so $X \in P(A) - P(B)$.
• But $A - B = \{1\}$, and $\{1, 2\} \not\subseteq \{1\}$, so $X \notin P(A - B)$.

Hence the inclusion of option 3 can be violated. The correct answer is option 3.

Pitfall: $P$ commutes with $\cap$ but not with $-$ or $\cup$. The reason $A - B$ misbehaves is that subtracting from $A$ only removes elements, while a subset $X$ of $A$ may still keep elements of $B$ — so $X \subseteq A$ and $X \not\subseteq B$ does not force $X$ to avoid $B$ entirely.