Let $A=[0,1]\cap \mathbb{Q}$ and suppose $B_n=\{x \in A | x=\frac{p}{q}, q \leq n, q \in \mathbb{Z}^+\}$ for all $n \in \mathbb{N}$, then choose the correct options-
(a) $B_n$ is countable for all $n\in \mathbb{N}.$
(b) $B_n$ is countable for finite $n \in \mathbb{N}.$
(c) $\cup_{n=1}^{\infty}B_n$ is countable.
(d) Cardinality of $B_{100}$ is 40.
Explanation: (a) and (c) are correct.
(a) and (b): As $B_n$ is set of some rationals in $[0,1]$ so it is subset of $\mathbb{Q}$. Hence it is countable for all $n \in \mathbb{N}.$
(c) It is countable being countable union of countable sets.
(d) Here $B_1=\{0,1\}$, $B_2=\{\frac{1}{2}\}$, $B_3=\{\frac{1}{3}, \frac{2}{3}\} \ldots$. In general, for each denominator $q(\neq 1)$ we have $\phi(q)$ choices for $p$, thus cardinality of $B_{100}$ will be $2+\phi(2)+\phi(3)+\ldots +\phi(100).$