A member of
$$
\bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n
$$
is a member of at least one of the sets
$$
\bigcap_{n\ge N} A_n,
$$
meaning it's a member of either $A_1\cap A_2 \cap A_3 \cap \cdots$ or $A_2\cap A_3 \cap A_4 \cap \cdots$ or $A_3\cap A_4 \cap A_5 \cap \cdots$ or $A_4\cap A_5 \cap A_6 \cap \cdots$ or $\ldots$ etc. That means it's a member of all except finitely many of the $A$.
A member of
$$
\bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n
$$
is a member of all of the sets
$$
\bigcup_{n\ge N} A_n,
$$
so it's a member of $A_1\cup A_2 \cup A_3 \cup \cdots$ and of $A_2\cup A_3 \cup A_4 \cup \cdots$ and of $A_3\cup A_4 \cup A_5 \cup \cdots$ and of $A_4\cup A_5 \cup A_6 \cup \cdots$ and of $\ldots$ etc. That means no matter how far down the sequence you go, it's a member of at least one of the sets that come later. That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to.
You cannot guarantee that $\langle a_n:n\in\Bbb N\rangle$ is a decreasing sequence: it might start out $$\left\langle 1,2,\frac1{10},\pi,\frac12,\frac13,\frac1{2^{100}},4,\frac1{100},\dots\right\rangle\;,$$ for instance. All you know is that all of its terms are positive, and for each $\epsilon>0$ there is some $n_a(\epsilon)\in\Bbb N$ such that $0<a_n<\epsilon$ whenever $n\ge n_a(\epsilon)$.
Similarly, all you can say for sure about $\langle b_n:n\in\Bbb N\rangle$ is that for every $\epsilon>0$ there is some $n_b(\epsilon)$ such that $1<b_n<1+\epsilon$ whenever $n\ge n_b(\epsilon)$. And above all you cannot assign specific values to the numbers $a_n$ and $b_n$: that’s changing the problem. (Of course, you can do so to look at an example or two in order to get a better idea of what’s going on, but that’s a different matter altogether.)
Now let’s take a look at $\liminf_n A_n$, where $A_n=[a_n,b_n)$: we want to determine which real numbers are eventually in the sets $A_n$, i.e., which are in all $A_n$’s from some point on. Here’s where those numbers $n_a(\epsilon)$ and $n_b(\epsilon)$ come in handy. For $\epsilon>0$ let $n(\epsilon)=\max\{n_a(\epsilon),n_b(\epsilon)\}$; then $0<a_n<\epsilon$ and $1<b_n<1+\epsilon$, and hence $$[\epsilon,1]\subseteq[a_n,b_n)\subseteq(0,1+\epsilon)\;.$$ for all $n\ge n(\epsilon)$. In other words, $[\epsilon,1]\subseteq A_n$ for all $n\ge n(\epsilon)$, and we conclude that for each $\epsilon>0$, $[\epsilon,1]\subseteq\liminf_n A_n$. It follows that $$\liminf_n A_n\supseteq\bigcup_{\epsilon>0}[\epsilon,1]=(0,1]\;.$$ On the other hand, if $x>1$, let $\epsilon=x-1$: for every $n\ge n(\epsilon)$ we have $1<b_n<1+\epsilon=x$, so $x\notin A_n$ whenever $n\ge n(\epsilon)$. This shows that $x$ isn’t even in infinitely many of the $A_n$’s, let alone in a tail of them, so $x\notin\limsup_n A_n$, and therefore certainly $x\notin\liminf_n A_n$. It’s also clear that no $x\le 0$ belongs to any of the $A_n$’s, so we’ve established that $\liminf_n A_n=(0,1]$, as you thought.
Along the way we’ve also seen that $\limsup_n A_n\subseteq(0,1]$, so $$\liminf_n A_n\subseteq\limsup_n A_n\subseteq(0,1]=\limsup_n A_n\;,$$ and it follows that $\limsup_n A_n=(0,1]$ as well, also as you thought.
It appears that you’re getting the concepts but might have a bit of difficulty actually writing down an argument to justify your reckoning of $\liminf$ or $\limsup$ of a sequence of sets.
Best Answer
Let $A_n=\emptyset $ for $n$ even and $X$ for $n$ odd. Then $\lim \inf A_n=\emptyset$ and $\lim \sup A_n=X$.
Let $A_n=\{n,2n,2n+1,2n+2,...\}$ with $X=\mathbb N$. Then $\lim \inf A_n=\emptyset=\lim \sup A_n$ but $(A_n)$ is not monotone.
Note that $n \in A_n$ but $n \notin A_{n+1}$. Also $n+1 \in A_{n+1}$ but $n \notin A_{n}$ if $ n>1$.