Prove $\overline{\lim\limits_{n \to \infty}} \Big(\inf f_n(x) \Big) \leq \inf \Big( \overline{\lim\limits_{n \to \infty}} f_n(x) \Big)$ I have proven this problem, but I am not sure if it is correct, especially the last inequality from the set being bounded. 
Would appreciate tips!
Let $X \subset \mathbb{R}$ and functions $f_n: X \xrightarrow{} \mathbb{R}, n\in \mathbb{N}$, like so the set
$\{f_n(x): n\in \mathbb{N}, x\in X\}$ is bounded.
Firstly we know from the definition of infimum that
$$
\underset{x \in X}{\text{inf}}f_n(x) \leq f_n(x)
$$
From the limit's monotonicity we get
$$
\overline{\lim\limits_{n \to \infty}} \Big( \underset{x \in X}{\text{inf}}f_n(x) \Big) \leq \overline{\lim\limits_{n \to \infty}} f_n(x)
$$
Since we know that the set is bounded, then the set has minima and maxima. Therefore infimum equals minimum, which means, that infimum is a function value.
So if the inequality works for all $x$, then the right side's greatest lower bound must also be larger or equal than the left side. So we have proven that 
$$
\overline{\lim\limits_{n \to \infty}} \Big( \underset{x \in X}{\text{inf}}f_n(x) \Big) \leq \underset{x \in X}{\text{inf}}\Big( \overline{\lim\limits_{n \to \infty}} f_n(x) \Big)
$$
Thank you
 A: Bounded subsets of $\Bbb R$ have least upper bounds and greatest lower bounds, but they need not have a maximum or a minimum. For instance, I don't see what you propose the minimum of $\{f_n(x)\,:\, n\in\Bbb N, x\in X\}$ to be for $f_n(x)=\arctan\frac1x$ and $X=(-1,0)$.
I propose:
\begin{align}\limsup_{n\to\infty}\inf_{x\in X}f_n(x)&=\inf_{n\in\Bbb N}\sup_{k\ge n}\inf_{x\in X}f_k(x)\\\inf_{x\in X}\limsup_{n\to\infty}f_n(x)&=\inf_{x\in X}\inf_{n\in\Bbb N}\sup_{k\ge n}f_k(x)\end{align}
For all $x$ and $k$, $$\inf_{t\in X}f_k(t)\le f_k(x)$$
Therefore, for all $x$ and $n$, $$\sup_{k\ge n}\inf_{t\in X}f_k(t)\le \sup_{k\ge n}f_k(x)$$
Therefore, for all $x$, $$\inf_{n\in\Bbb N}\sup_{k\ge n}\inf_{t\in X}f_k(t)\le \inf_{n\in\Bbb N}\sup_{k\ge n}f_k(x)$$
Therefore, $$\inf_{n\in\Bbb N}\sup_{k\ge n}\inf_{t\in X}f_k(t)\le \inf_{x\in X}\inf_{n\in\Bbb N}\sup_{k\ge n}f_k(x)$$
This holds verbatim under the more general convention that $\sup$, $\inf$, $\limsup$ and $\liminf$ are defined as extended real numbers.
