If $A = \{x : x\ \text{is the first three odd numbers}\}$
$B = \{2x + 3 : 0 \leq x < 5, x \in \mathbb{N}\}$, then which of the following is true

The first three odd natural numbers are $1, 3$, and $5$.

$A = \{1, 3, 5\}$

For set $B$, the condition is $0 \leq x < 5$ and $x \in \mathbb{N}$. The natural numbers satisfying this condition are $x = 1, 2, 3, 4$.

Substituting these values into $2x + 3$:
For $x = 1$, $2(1) + 3 = 5$
For $x = 2$, $2(2) + 3 = 7$
For $x = 3$, $2(3) + 3 = 9$
For $x = 4$, $2(4) + 3 = 11$

$B = \{5, 7, 9, 11\}$

Finding the intersection of sets $A$ and $B$:

$A \cap B = \{1, 3, 5\} \cap \{5, 7, 9, 11\} = \{5\}$

Since $A \cap B$ contains exactly one element, it is a singleton set. Also, $n(B) = 4$.

Answer: $A \cap B$ is a singleton set