Limit of a sequence of intervals Let $\{a_n\}_{n\geqslant 1} $ be a sequence such that $a_n>0\ \forall n$ and $\lim_{n\to\infty} a_n=0$ . What will be 

$$\bigcup_{n=1}^\infty [a_n,1) \ \ =?$$

Will it be $(0,1)$ or $[0,1)$ ?
Edit:
This was originally a part of a larger question.

Consider $2$ sequences $\{a_n\}\ \&\ \{b_n\}$ with $a_n>0$ and $b_n>1$ for all $n$ and $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}b_n=1$. Define $A_n=[a_n,b_n)$. Find $\limsup_{n\to\infty} A_n$ and $\liminf_{n\to\infty} A_n$

Can someone confirm me whether both the answers are $(0,1)$ or not?
 A: I will show that
$$
\liminf_{n\rightarrow\infty}[a_n,b_n) = \limsup_{n\rightarrow\infty}[a_n,b_n) = (0,1].
$$
Let $x \in \mathbb{R}$.
By definition of $\liminf$, $x \in \liminf_{n\rightarrow\infty}[a_n,b_n)$ iff $x$ belongs to every interval $[a_n,b_n)$ for sufficiently big $n \in \{1, 2, \dots\}$. I claim that this is the case iff $x \in (0,1]$. To see this, suppose firstly that $x \in (0,1]$. Then, since $b_n > 1$ for all $n \in \{1, 2, \dots\}$, we have for every $n \in \{1, 2, \dots\}$ that $x < b_n$. At the same time, since $\lim_{n\rightarrow\infty} a_n = 0$, then, for sufficiently big $n \in \{1, 2, \dots\}$, we have that $a_n \leq x$. Therefore, for sufficiently big $n \in \{1, 2, \dots\}$, $x \in [a_n, b_n)$. Conversely, if $x \in [a_n, b_n)$ for all sufficiently big $n \in \{1, 2, \dots\}$, then, on the one hand, $x$ must be greater than $0$, since $a_n > 0$ for all $n \in \{1, 2, \dots\}$, while, on the other hand, $x$ cannot be greater than $1$, because otherwise, since $\lim_{n\rightarrow\infty}b_n = 1$, then, for sufficiently big $n \in \{1, 2, \dots\}$, we would have $b_n < x$, implying that $x \notin [a_n, b_n)$.
The previous paragraph established that $\liminf_{n\rightarrow\infty}[a_n,b_n) = (0,1]$. Since it is always the case that $\liminf \subseteq \limsup$, we automatically have $(0,1] \subseteq \limsup_{n\rightarrow\infty}[a_n,b_n)$. It remains to show that $(0,1] \supseteq \limsup_{n\rightarrow\infty}[a_n,b_n)$. Indeed, suppose that $x \in \limsup_{n\rightarrow\infty}[a_n,b_n)$. Then, by definition of $\limsup$, $x$ belongs to infinitely many intervals $[a_n,b_n)$. Therefore, since $a_n > 0$ for all $n \in \{1, 2, \dots\}$, we must have $x > 0$. To see that $x \leq 1$, assume the contrary. Then, since $\lim_{n\rightarrow\infty}b_n = 1$, we have for all sufficiently big $n \in \{1, 2, \dots\}$, that $b_n < x$. But this contradicts the fact that for infinitely many $n \in \{1, 2, \dots\}$ we have $x \in [a_n, b_n)$.
A: $(0,1)$ obviously because $\forall n,\, 0\notin [a_n,1)$ is a consequence of $\forall n,\, a_n\gt 0$.
