Suppose we have a set $X_b = \{\frac{a}{b}:a \in \mathbb{Z^{+}}\} $ where $b \in \mathbb{Z^{+}}$. We want to find $\lim_{b \to +\infty} \inf{X_b}$ and also find find $\lim_{b \to +\infty} \sup{X_b}$.
Attempt: Clearly as $b \rightarrow \infty$, $\frac{a}{b} \rightarrow 0$ therefore the sequence inside $X_b$ will tend to $0$ faster as b gets larger. We also see this sequence is not monotone since its cannot be a sequence of subsets. By definition we know that:
$\lim_{b \to +\infty} \inf{X_b} = \bigcup _{b=1}^{\infty} \bigcap_{i=b}^{\infty}X_i$
$\lim_{b \to +\infty} \sup{X_b} = \bigcap _{b=1}^{\infty} \bigcup_{i=b}^{\infty}X_i$
To get a better picture we can see that: $X_1$ = $\{1,2,3,...\} = \mathbb{Z^{+}}$ , $X_2$ = $\{1/2,1,3/2,2,...\}$ so we can see that $X_1 \subset X_2$. Lets look at $X_3$ = $\{1/3,2/3,1,4/3,...\}$ but here $X_2$ not subset of $X_3$. Noticing this pattern we see that $\forall b \in \mathbb{Z^{+}}$, {b/b,2b/b,3b/b,...} $\subset$ $X_b$. From this I want to conclude $\bigcap_{i=b}^{\infty}X_i = \mathbb{Z^{+}}, b \in \mathbb{Z^{+}}$. If we can prove this, then from that could we conclude $\lim_{b \to +\infty} \inf{X_b} = \bigcup _{b=1}^{\infty} \bigcap_{i=b}^{\infty}X_i = \mathbb{Z^{+}}?$
Note that $m|n \Rightarrow X_m \subseteq X_n$. Hence we see $X_1 \subseteq X_n$ for all $n$ and so $S_b =\bigcap_{i=b}^{\infty}X_i \supseteq X_1 = \Bbb Z^+$. For the reverse inclusion suppose ${m \over n} \in S_b$, $(m,n)=1$, $n \gt 1$. Choose $p$ prime with $p \gt n,b$ then ${m \over n} \notin X_p \supseteq S_b$, contradiction.
For the other part we use $\bigcup_{i=1}^{\infty}X_i= \Bbb Q^+ $ (this is easy to prove) and $X_n \subseteq X_{nb}$ to find for all $b$: $$\bigcup_{i=b}^{\infty}X_i= \bigcup_{i=1}^{\infty}X_i= \Bbb Q^+$$