Excercise of semi-ring and sigma-ring on real line interval 
Exercise: Let $\Omega=(0,1]\cap\mathbb{Q}$ where $\mathbb{Q}$ denotes the set of rational numbers. Let $\mathscr{L}$ be  the class of subsets of $\Omega$ of the form $(a,b]\cap\mathbb{Q}$ where $0\leqslant a\leqslant b\leqslant 1$.
i) Show that any set in $\mathscr{L}$ is either empty or infinite.
ii)Show that the σ-ring generated by $\mathscr{L}$ contains 
  all the subsets of $\Omega$.

i) I thought that since $\mathbb{Q}$ is dense in $\mathbb{R}$, then $(a,b]\cap\mathbb{Q}$must be countably infinite for any $0\leqslant a\leqslant b\leqslant 1$. So the only set that would not be infite would be the empty set.
ii) Since we have the $\sigma$-ring generated by $\mathscr{L}$ I can have $[a_n,b_n)\cap\mathbb{Q}$ for $a_n(x),b_n(x)\in\mathbb{Q}\:\forall n$ so that $\bigcup_\limits{x\in\Omega}^{\infty}(a_n(x),b_n(x)]\cap\mathbb{Q}=(0,1]\cap\mathbb{Q}=\Omega$.
Questions: 
1) Are my proofs right? Regarding i). How can I formalise it?
2) If the proofs are wrong. How should I answer the questions?
Thanks in advance!
 A: $\def\Ω{{\mit Ω}}$For question 1, note that for any $A = (a, b] \cap\mathbb{Q}$ where $0 \leqslant a \leqslant b \leqslant 1$, if $A ≠ \varnothing$, then $a < b$. Because $\mathbb{Q}$ is dense in $\mathbb{R}$, then $(a, b) \cap \mathbb{Q} ≠ \varnothing$. Arbitrarily pick $x_0 \in (a, b) \cap \mathbb{Q} \subseteq A$, again the density of $\mathbb{Q}$ implies there exists $\{x_n\}_{n \geqslant 1} \subseteq \mathbb{Q}$ of distinct elements such that $x_n → x_0\ (n → ∞)$. For $ε = \min(x_0 - a, b - x_0)$, there exists $N \geqslant 1$ such that $|x_n - x_0| < ε$ for $n \geqslant N$, which implies $x_n \in A$ for $n \geqslant N$. Therefore, $\{x_n\}_{n \geqslant N} \subseteq A$, which implies $A$ is infinite.
For question 2, note that $\Ω = (0, 1] \cap \mathbb{Q} \in \mathscr{L}$. For any $A_1, A_2, \cdots \in \mathscr{L}$, since $A_n^c = \Ω \setminus A_n \in \mathscr{L}$ for any $n$, then$$
\bigcup_{n = 1}^∞ A_n^c \in \mathscr{L} \Longrightarrow \bigcap_{n = 1}^∞ A_n = \left( \bigcup_{n = 1}^∞ A_n^c \right)^c \in \mathscr{L}.
$$
For any $r \in \Ω = (0, 1] \cap \mathbb{Q}$, denote $N = \left[ \dfrac{1}{r} \right] + 1$, then taking $A_n = \left( r - \dfrac{1}{n}, r \right] \cap \mathbb{Q} \in \mathscr{L}$ yields$$
\{r\} = \bigcap_{n = N}^∞ A_n \in \mathscr{L}.
$$
Finally, for any $B \subseteq \Ω = (0, 1] \cap \mathbb{Q}$, if $|B| = n < ∞$, suppose $B = \{r_1, \cdots, r_n\}$, then $B = \bigcup\limits_{k = 1}^n \{r_k\} \in \mathscr{L}$. Otherwise, note that $\mathbb{Q}$ is countable and suppose $B = \{r_1, r_2, \cdots\}$, then $B = \bigcup\limits_{n = 1}^∞ \{r_n\} \in \mathscr{L}$.
