Interpretation of {Infinitely Often} = {Almost Always}

Question

I am trying to better understand what it means for a sequence $A_n$ of subsets of a set $S$
to be such that $\bigcap^\infty_{n = 1} \bigcup^\infty_{m = n} A_n = \limsup A_n = \liminf A_n = \bigcup^\infty_{n = 1} \bigcap^\infty_{m = n} A_n$

I find the interpretation infinitely often and eventually always as in

\begin{equation} \bigcap^\infty_{n = 1} \bigcup^\infty_{m = n} A_n = \limsup A_n = \{ w \, \colon w \in A_n \quad \text{infinitely often} \} \end{equation} and \begin{equation}
\bigcup^\infty_{n = 1} \bigcap^\infty_{m = n} A_n = \liminf A_n = \{ w \, \colon w \in A_n \quad\text{eventually always} \} \end{equation}

very helpful and I am looking for an analogous interpretation what it means for the two to be equal.

Answer 1

Suppose that $\limsup A_n=\liminf A_n$. Then if $w$ is in $A_n$ infinitely often, it is eventually always in $A_n$. If $w$ is not in $A_n$ infinitely often, then it’s eventually not in any $A_n$. Thus, for each $w$ there is an $n_w$ such that either $w\in A_n$ whenever $n\ge n_w$, or $w\notin A_n$ whenever $n\ge n_w$. There are no points $w$ that are in infinitely many $A_n$’s and out of infinitely many $A_n$’s.

Of course the points that are in every $A_n$ from some point on are the ones that in the limit of the $A_n$’s, and the ones that are not in any $A_n$ from some point on are the ones that are not in the limit of the $A_n$’s.

Answer 2

Notice that your interpretation tells us easily that the latter is a subset of the former. I assume you're looking for a 'plain English' description of the case where the two are equal. I would word it as follows: membership in $A_n$ stabilizes over $n$. By this I mean that for any $w$, either $w$ is in finitely many $A$'s (so that the situation "stabilizes" to $w\not\in A$) or there is an $N$ such that $w\in A_n$ for every $n\ge N$ (so that the situation "stabilizes" to $w\in A$).