Question 32

Let $|A| = n$, where $n \ge 1$ since $A$ is a non-empty finite set.

1. Since there is a bijection (one-to-one and onto function) from $A$ to $B$, we have $|B| = |A| = n$.
2. Since there is a bijection from $A \times A$ to $A \cup B$, we have $|A \cup B| = |A \times A| = n^2$.
3. For any two sets $A$ and $B$ of cardinality $n$, the size of their union satisfies:
   $$\max(|A|, |B|) \le |A \cup B| \le |A| + |B| \implies n \le n^2 \le 2n$$

Since $n$ is a positive integer, $n^2 \le 2n \implies n \le 2$, so $n \in \{1, 2\}$:

• For $n = 1$: Taking $A = B = \{1\}$ gives $|A \cup B| = 1 = 1^2$.
• For $n = 2$: Taking disjoint sets $A = \{1, 2\}$ and $B = \{3, 4\}$ gives $|A \cup B| = 4 = 2^2$.

Thus, there are 2 possible values of $|A|$ (namely $1$ and $2$).