$a_n\leq b_n$ for almost all $n\in\mathbb{N}\implies \limsup_n a_n\leq\limsup_n b_n$?

Question

Let $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ be two sequences such that $a_n\le b_n$ for almost all $n$. Does this imply that $$ \limsup_{n\to\infty}a_n\le\limsup_{n\to\infty}b_n? $$

To my opinion, yes! Since set $$ a:=\limsup_{n\to\infty}a_n,~~~~~b:=\limsup_{n\to\infty}b_n $$ and suppose that $a>b$. Then, choosing a neighborhood around $a$ that does not contain $b$, there are infinitely many $a_n$ in this neighborhood, i.e. almost all $a_n$ are bigger than $b$ and almost all $a_n>b_n$.

But I do not know if this makes sense.

Edit

Maybe it would be better to choose $\varepsilon:=\lvert a-b\rvert/3$ and to consider $B_{\varepsilon}(a)$ and $B_{\varepsilon}(b)$, the open balls around a and b with radius $\varepsilon$. Then there are infinitely many $a_n$ in $B_{\varepsilon}(a)$ and infinitely many $b_n$ in $B_{\varepsilon}(b)$, hence $a_n>b_n$ for almost all n.

Answer 1

I assume that by "almost all" you mean "an infinite number" (if instead you mean "all but a finite number" then the result is truly obvious).

In this case the thesis is false. Take $a_n=(-1)^n$ and $b_n=0$. Then $a_n<b_n$ infinitely often but $\limsup a_n=1>0=\limsup b_n$.

The problem is that in fact in this situation, "almost always" $a_n>b_n$ and "almost always" $a_n<b_n$, so that your language is not really justified. To formalize it, we use ultrafilters.