Infinitely often and almost always

Let $A_{1},A_{2},A_{3},\ldots$ be a sequence of events in some sample space $\Omega$. Let $\omega \in \Omega$ be an outcome.

In the following table, all statements in the "infinitely often" column are logically equivalent. Similarly, all statements in the "almost always" column are logically equivalent.

perspective	infinitely often	almost always
unions and intersections	$\omega \in \bigcap _{N=1}^{\infty }\bigcup _{n=N}^{\infty }A_{n}$	$\omega \in \bigcup _{N=1}^{\infty }\bigcap _{n=N}^{\infty }A_{n}$
first-order quantifiers	$\forall N\geq 1\ \exists n\geq N\colon \omega \in A_{n}$	$\exists N\geq 1\ \forall n\geq N\colon \omega \in A_{n}$
verbal expression	$\omega \in A_{n}$ for infinitely many $n\geq 1$	$\omega \in A_{n}$ for almost all $n\geq 1$, i.e. $\omega \in A_{n}$ for all but finitely many $n\geq 1$, i.e. $\omega \notin A_{n}$ for finitely many $n\geq 1$
lim sup/lim inf	$\omega \in \limsup _{n\to \infty }A_{n}$	$\omega \in \liminf _{n\to \infty }A_{n}$
limit of sup/inf	$\omega \in \lim _{N\to \infty }\bigcup _{n=N}^{\infty }A_{n}$	$\omega \in \lim _{N\to \infty }\bigcap _{n=N}^{\infty }A_{n}$

Analogy with sequences of real numbers

Let $(a_{n})_{n=1}^{\infty }$ be a sequence of real numbers, let $\epsilon >0$ be a real number, and let $x$ be a real number.

We say $(a_{n})_{n=1}^{\infty }$ is eventually $\epsilon$-close to $x$ iff there exists some $N\geq 1$ such that for all $n\geq N$ we have $|a_{n}-x|\leq \epsilon$.

We say that $(a_{n})_{n=1}^{\infty }$ is continually $\epsilon$-adherent iff for every $N\geq 1$ there exists some $n\geq N$ such that $|a_{n}-x|\leq \epsilon$.

I think we can even define $A_{n}:=\{x\in \mathbf {R} :|a_{n}-x|\leq \epsilon \}$.

$\limsup _{n\to \infty }|a_{n}-x|=\inf _{N\geq 1}\sup _{n\geq N}|a_{n}-x|\leq \epsilon$

$\liminf _{n\to \infty }|a_{n}-x|=\sup _{N\geq 1}\inf _{n\geq N}|a_{n}-x|\leq \epsilon$ -- I think this one is equivalent to infinitely often, which is confusing since now the quantifier order has seemingly switched.

but this makes sense in terms of strength of "infinitely often" vs "almost always". We have $\liminf _{n\to \infty }|a_{n}-x|\leq \limsup _{n\to \infty }|a_{n}-x|$, so if $\limsup _{n\to \infty }|a_{n}-x|\leq \epsilon$ (i.e. $a_{n}$ is $\epsilon$-close to $x$ almost always) then $\liminf _{n\to \infty }|a_{n}-x|\leq \epsilon$ (i.e. $a_{n}$ is $\epsilon$-close to $x$ infinitely often).

perspective	infinitely often	almost always
Tao's terminology (see his Analysis)	$(a_{n})_{n=1}^{\infty }$ is continually $\epsilon$-adherent	$(a_{n})_{n=1}^{\infty }$ is eventually $\epsilon$-close to $x$
first-order quantifier	for every $N\geq 1$ there exists some $n\geq N$ such that $|a_{n}-x|\leq \epsilon$	there exists some $N\geq 1$ such that for all $n\geq N$ we have $|a_{n}-x|\leq \epsilon$
	$|a_{n}-x|\leq \epsilon$ for infinitely many $n$	$|a_{n}-x|\leq \epsilon$ for all but finitely many $n$
	$\liminf _{n\to \infty }|a_{n}-x|\leq \epsilon$	$\limsup _{n\to \infty }|a_{n}-x|\leq \epsilon$
	$x\in \limsup _{n\to \infty }A_{n}$	$x\in \liminf _{n\to \infty }A_{n}$