Denote $\{x_n\}$ as a sequence of real numbers, and there exist a $x\in\mathbb{R}$ such that
$\limsup_{n\to\infty}{x_n}=x$.
We can write "$\lim_{n\to\infty}x_n=x$"
as "$\forall \epsilon>0,\exists N\in\mathbb{N} ,s.t\quad n\geq N\Rightarrow|x_n-x|<\epsilon$".
So I wonder if it's possible for the "$\limsup$".
Is there a way to state $\limsup_{n\to\infty}{x_n}=x$ by "$\forall$" and "$\exists$"?
Thanks in advance.
Best Answer
Yes, there is. Here is one way. The definition of $\limsup_{n\to\infty} x_n = x$ consists of two parts:
So, with that in mind, let's see if we can write something down. For part 1: $$ \forall N\in \Bbb N, \varepsilon >0\,\exists n>N s.t. |x_n-x|<\varepsilon $$ And then the second part. It's going to be a bit longer, but the idea is that $x$ is the largest number satisfying the above relation. Thus any $y$ which satisfies it must be less than or equal to $x$. If you keep that in mind, it ought to be readable: $$ \forall y\in \Bbb R\Big(\forall N\in \Bbb N, \varepsilon >0\,\exists n>N s.t. |x_n-y|<\varepsilon\implies y\leq x\Big) $$ Sticking these two statements together with a $\land$ should get you a nice symbolic definition of $$ \limsup_{n\to\infty}x_n = x $$
From this it is also easy to define $\liminf$ by simply switching the inequality in the second statement.