Prove that $\limsup_{n\ge 1}x_n$ can also be expressed as the $\inf$ of a certain set. I've been asked to show that
$$
\limsup_{n\ge 1}x_n = \inf\{y\in\mathbb{R}\ |\ \exists N>0\ \text{s.t.}\ x_n< y\ \forall n\ge N\}
$$
for bounded sequences $x_n$.
Last night I came up with a rather uninspired argument to prove this that just hinged on showing that the LHS above is the greatest lower bound of the set on the right. I don't think it's a very interesting or clever proof, and I haven't even checked it in a clear state of mind so it might not be completely correct, but I was wondering if someone could verify if I'm in the right direction.
Additionally, if the argument is correct, but overcomplicated or contrived, I would appreciate a hint pointing to a nicer solution!
Proof
Let $z_n = \sup_{k\ge n}x_n$, and $\Gamma = \{\gamma\in\mathbb{R}\ |\ \exists N>0\ \text{s.t.}\ \forall n\ge N,\ x_n<\gamma\}$. We will show that for all $\gamma\in\Gamma$, $\lim_{n\to\infty}z_n \le \gamma$, and that if $z\in\mathbb{R}$ is such that $z\le \gamma$ for all $\gamma\in\Gamma$, then $z\le\lim_{n\to\infty}z_n$.
Take $\gamma\in\Gamma$. To show that $\lim_{n\to\infty}z_n\le \gamma$, we need to show that $\exists N\ \text{s.t.}\ \forall n\ge N,\ z_n\le\gamma$. However, $\gamma\in\Gamma$ implies $\exists N = N_0\ \text{s.t.}\ \forall n\ge N,\ x_n\le \gamma$. Then for all $n\ge N_0$, $x_n\le\gamma$, implying $\sup_{k\ge n}x_n \le \gamma$, and so $z_n\le\gamma$ for all $n\ge N_0$. Thus, $\lim_{n\to\infty} z_n\le \gamma$, so $\limsup_{n\ge 1}x_n$ is a lower bound for $\Gamma$.
Now, let $z\in\mathbb{R}\backslash\Gamma$ be an arbitrary lower bound for $\Gamma$. Suppose $\limsup_{n\ge 1}x_n < z$. Then $\exists N$ such that for all $n\ge N$, $\sup_{k\ge n}x_n < z$. For this same $N$, for all $n\ge N$, $x_n \le \sup_{k\ge n}x_n < z$. This, however, implies $z\in\Gamma$, a contradiction. Thus $\limsup_{n\ge 1}x_n \ge z$ for all lower bounds $z\not\in\Gamma$.
If $\gamma = \inf(\Gamma)\in\Gamma$, then $\exists z_n\in B_{1/n}(\gamma)\backslash\Gamma$ for all $n\ge N$ which is a lower bound for $\Gamma$ not in $\Gamma$, and $\limsup_{n\ge 1}x_n\ge z_n$ for all $n$. Since this is true for arbitrary $n$, we can only conclude that $\limsup_{n\ge 1}x_n = \gamma$.
 A: Some minor structural suggestions that may simplify things.  
[Your posted proof has 4 paragraphs, I refer to the second and third.]
1) Preliminary:
Note that boundedness of $\{x_n\}_{n=1}^{\infty}$ implies the set $\Gamma$ is non-empty and lower-bounded.  Define $\gamma^*=\inf \Gamma$ and note that $\gamma^*$ is finite. 
2) Show $\limsup_{n\rightarrow\infty} x_n\leq \gamma^*$:
Your paragraph 2 already shows
$$ \limsup_{n\rightarrow\infty} x_n \leq \gamma \quad \forall \gamma \in \Gamma$$
Taking the "infimum of both sides" concludes
$$  \limsup_{n\rightarrow\infty} x_n \leq \gamma^* $$
3) Show $\limsup_{n\rightarrow\infty} x_n \geq \gamma^*$:  
Suppose not (we reach a contradiction). Then the $\limsup$ is strictly less than $\gamma^*$ and there is a number $z$ in between:
$$\limsup_{n\rightarrow\infty} x_n < z< \gamma^*$$
Your paragraph 3 already shows this yields a contradiction. 
A: Defining $s_{n}=\sup_{k\geq n}x_{k}$ we have $s_{1}\geq s_{2}\geq s_{3}\geq\cdots$
so it is direct that: $$\limsup x_{n}=\lim_{n\to\infty}s_{n}=\inf\left\{ s_{n}\mid n\in\mathbb{N}\right\} $$
It remains to be shown that the sets


*

*$A:=\left\{ s_{n}\mid n\in\mathbb{N}\right\}$

*$B:=\left\{ y\in\mathbb{R}\mid\exists N\in\mathbb{N}\forall n[n>N\implies x_{n}<y\right\}$
have the same infimum or equivalently that the set of lower bounds
of $A$ coincides with the set of lower bounds of $B$.

If $z$ is not a lower bound of $A$ then some $n$ exists with $s_{n}<z$.
Then we can choose some $y$ satisfying $s_{n}<y<z$. 
We have $x_{k}<y$
for every $k\geq n$ showing that $y\in B\wedge y<z$ and conclude
that $z$ is not a lower bound of $B$.

If $z$ is not a lower bound of $B$ then some $y\in B$ exists with
$y<z$ and from $y\in B$ we conclude that some $n$ exists with $s_{n}\leq y$.
Then $s_{n}<z$ and we conclude that $z$ is not a lower bound of
$A$.

This completes the proof.
