Q.Let $A = \{p, q, r\}$. Verify that $\emptyset \subseteq A$, then write the power set $P(A)$ and state $n(P(A))$.

The empty set is a subset of every set, so it must appear in the power set; list every possible subset of the 3-element set systematically by size, then count them.

Listing subsets of every size from 0 to 3 elements gives all 8 members of $P(A)$.

Step 1 — Confirm $\emptyset \subseteq A$. By the theorem from this chapter's opening section, the empty set is a subset of every set, since it has no elements that could ever fail to belong to $A$. So $\emptyset \subseteq A$ holds, and $\emptyset$ must be included as a member of $P(A)$.

Step 2 — List subsets by size.

• Size 0 (the empty set): $\emptyset$
• Size 1: $\{p\}, \{q\}, \{r\}$
• Size 2: $\{p,q\}, \{p,r\}, \{q,r\}$
• Size 3 (the whole set): $\{p,q,r\}$

Step 3 — Collect and count. $P(A) = \{\emptyset, \{p\}, \{q\}, \{r\}, \{p,q\}, \{p,r\}, \{q,r\}, \{p,q,r\}\}$ — that's $1+3+3+1 = 8$ subsets.

Step 4 — Verify using the formula. $n(A) = 3$, so $n(P(A)) = 2^3 = 8$, matching the direct count.

Forgetting to include $\emptyset$ or the full set $A$ itself as members of $P(A)$ — both are always subsets of $A$ and must be listed. Also, listing a subset more than once (e.g. writing $\{p,q\}$ and $\{q,p\}$ as if they were different) — since order doesn't matter inside a set, these are the same subset and should be counted only once.