Trying to get a better understanding of limsup/liminf in terms of probability.
An intuitive explanation of the limsup of a sequence of sets $(A_n)_{n\geq 1}$, in the context of probability, is that $\limsup_{\substack{n \to \infty}}A_n$ is "the event that infinitely many events occur".
Events are subsets of the sample space. Take the simple example of rolling a die, so $\Omega = \{1,2,3,4,5,6\}$.
Consider the sequence of events $A_n$ defined by (for $i\in\mathbb{N}$)
$$ \begin{align*} A_{3i-2}&=\{1,2,3\}\\ A_{3i-1}&=\{1\}\\ A_{3i}&=\{1,2,3,6\} \end{align*} $$
Then, $$ \begin{align*} \limsup_{n\rightarrow\infty}A_n&=\bigcap_{n=1}^{\infty}\bigcup_{j=n}^{\infty}A_j\\ &=\{1,2,3,6\}\cap \{1,2,3,6\}\cap\{1,2,3,6\}\cap\dots\\ &=\{1,2,3,6\} \end{align*} $$
Also, $$ \begin{align*} \liminf_{n\rightarrow\infty}A_n&=\bigcup_{n=1}^{\infty}\bigcap_{j=n}^{\infty}A_j\\ &=\{1\}\cup \{1\}\cup\{1\}\cup\dots\\ &=\{1\} \end{align*} $$
Firstly, just want to check my understanding is correct with this trivial example. Limsup contains $\{1,2,3,6\}$ because limsup is "the set of all $x\in\Omega$ that occur infinitely many times". The liminf set is smaller because $1$ is the only element that occurs "all but finitely many times". Sound correct?
Now, when we say that limsup is "the event that infinitely many events occur", does the "the" imply we take all possible values of $x\in\Omega$ contained infinitely many times within the $A_n$-s. I guess my point is that the set $A=\{1,2\}$ is an event where infinitely many of the $A_n$ occur... but we don't say it is "the" event?