A problem related to limsup and liminf. Suppose $a_n &gt;0$ and $b_n &lt;1$ and lim $a_n=0$, lim $b_n =1$ and $A_n =\{x: a_n \leq x  &lt;b_n\}$   then find limsup$ (A_n)$ and liminf$(A_n) $.

As $a_n$ goes to 0 and $b_n$ goes to 1, How can I use this condition to find  limsupA_n and liminf A_n, I got stuck.
It is generally difficult to express $\limsup_n A_n$ and $\liminf_n A_n$ in terms of $\lim_n a_n$ and $\lim_n b_n$. The reason is that $\limsup_n A_n$ and $\liminf_n A_n$ are not topological notions, that is, they do not require the existence of a topology on $[0,1]$, whereas $\lim_n a_n$ and $\lim_n b_n$ are topological notions.

Since $a_n\to 0^+$ and $b_n\to 1^-$ (we know that $0 &lt; a_n \leq b_n &lt; 1$), we have that for every $\epsilon \in (0,1)$, there is an $n_0=n_0(\epsilon)\in\mathbb{N}$ so that $a_n &lt; \epsilon$ and $1-b_n&lt;\epsilon$ for all $n\geq n_0$, therefore $[a_n, b_n] \subseteq [a_{n_0}, b_{n_0}]$.

Then,

$$
\liminf_n A_n = \bigcup_{n\geq 1}\bigcap_{j\geq n}[a_j, b_j]
\supseteq \bigcup_{n\geq n_0(\epsilon)}\bigcap_{j\geq n}[a_j, b_j]
\supseteq \bigcup_{n\geq n_0(\epsilon)}[a_{n_0}, b_{n_0}] = [a_{n_0(\epsilon)}, b_{n_0(\epsilon)}]
$$ 

Since this holds for all $\epsilon &gt; 0$, we can take $\epsilon \to 0^+$ to conclude that 

$$
\liminf_n A_n \supseteq (0, 1).
$$

In fact, $(0,1)$ is the largest possible such set because we can see by the definition that $0,1\notin \liminf_n A_n$. Therefore, $\liminf_n A_n = (0,1)$.

Similarly, we may show that $\limsup_n A_n = (0,1)$.
$(A_n)_n$ is an increasing sequence of sets, thus it converges to $\cup_n A_n=\liminf A_n=\limsup A_n$ and $\cup_nA_n = [0,1).$