Hint for proof of equivalence of lim sup

Question

I am major in physics, and real analysis is my first pure math course. Thus, I am not sure how to proof a thing mathematically, that is in what form, in what kind of format. I have read some analysis books, like Principle of mathematical analysis (still reading), real and complex analysis & Functional analysis (all series by Rutin). Not all books, but parts of these. One can always know that understand a proof is much easier than complete a proof by yourself. No more pre-words, let us see the problem:

Prove that$$\lim \sup_{n\to \infty} x_n=inf\{\gamma\in\mathbb R:\text{there exists $N>0$ such that $x_n\lt\gamma$ for all $n\ge N$ }\}$$where $\lim \sup_{n\to\infty} x_n=\lim_{n\to\infty}\sup\{x_n,x_{n+1},{...}\}$

I tried to proof LHS to have three conditions, i.e. =$+\infty$, =some constant, =$-\infty$, coorespond to RHS with same n. Could this approach be valid in math? Could someone figure out a proof by letting LHS$\ge$RHS and RHS$\le$LHS?

Thanks.

Answer 1

My attenpt to the proof:
Monotonic decreasing sequence $\sup \{x_n\}$:
Denote $z_k=\sup \{x_n\}_{n=k}^{\infty}=\sup\{x_k,x_{k+1},...\}$.
Notice that
$z_k=\sup\{x_k,x_{k+1},x_{k+2},...\}=\sup\{x_k,\sup\{x_{k+1},x_{k+2},...\}\}=\sup\{x_k,z_{k+1}\}\ge z_{k+1},k\in\{1,2,3,...\}$.
Since $\{z_k\}_{k=1}^{\infty}$ is a monotonic decreasing sequence,$$\lim_{n\to\infty}z_n=\inf\{z_k\}_{k=1}^{\infty}$$ We now have a look at the RHS set, denote it as A: $$A=\{\gamma\in\Bbb R|\text{there exists N>0 such that $x_n<\gamma$, $\forall\ n\ge N$ }\}$$ In other words, $$A=\{\gamma\in\Bbb R|\text{$\exists$ N>0, such that $\gamma\gt\sup \{x_n\}_{n=N}^{\infty}=z_N$}\}$$ To make sure N always exists, we need $\gamma$ to be at least greater than $\inf \{z_k\}_{k=1}^{\infty}$, otherwise there will be no such a N. Hence, one can rewrite A as: $$A=\{\gamma\in\Bbb R|\gamma=\inf\{z_k\}_{k=1}^{\infty}+\epsilon,\forall\ \epsilon\gt0\}$$ One can notice that $\inf A \in \Bbb R$ since $\Bbb R$ has least upper bound property. Moreover, $\inf A=\inf\{z_k\}_{k=1}^{\infty}=\lim_{n\to\infty}z_n$. That is $$\lim \sup_{n\to\infty} x_n=\inf\{\gamma \in \Bbb R|\exists\text{ N>0, such that $\gamma>x_n,\forall n>$N}\}$$ $\square$