limit inferior and infimum (real analysis) I want to prove $$\liminf_{n \to \infty} \left( \inf_{x \in X} f_n(x) \right) \leq \inf_{x \in X} \left( \limsup_{n \to \infty} f_n(x)\right)$$
when $X \subset \mathbb{R}$ and $f_n\colon X \rightarrow \mathbb{R}$ so that $\{f_n(x)\>\colon \> n \in \mathbb{N}, x \in X\}$ is bounded.
I know that should be able to use the property that if $x_n \leq y_n$ then $\limsup_{n} x_n \leq \limsup_{n} y_n$
and that $\inf_{n} x_n \leq \liminf_{n} x_n$.
But how am I supposed finish this proof?
 A: You have that
$$
\inf_{x\in X}f_n(x)\le f_n(x)\:\:\:\: \forall n\in \Bbb N\iff \liminf_{n\to\infty}\left( \inf_{x \in X} f_n(x) \right)\le\liminf_{n\to\infty}f_n(x)\:\:\:\:\forall x\in X.
$$
But we have also that
$$
\liminf_{n\to\infty}f_n(x)\le\limsup_{n\to\infty}f_n(x)\quad \forall x\in X.
$$
Now, since the infimum of a bounded from below set in $\Bbb R$ is the largest of all lower bounds, we get
$$
\liminf_{n \to \infty} \left( \inf_{x \in X} f_n(x) \right) \leq \inf_{x \in X} \left( \limsup_{n \to \infty} f_n(x)\right),
$$
because $\liminf_{n\to\infty}\big( \inf_{x \in X} f_n(x) \big)$ is a lower bound for the set 
$$
\left\{\limsup_{n\to\infty}f_n(x)\:\big|\:x\in X\right\},
$$
which in turn is bounded and thus bounded from below since $\{f_n(x)\>\colon \> n \in \mathbb{N}, x \in X\}$ is.
A: By properties of the infimum, it suffices to prove that for every $x\in X$, we have 
$$\liminf_n(\inf_{z\in X} f_n(z)) \,\le\, \limsup_n f_n(x)$$
But this easily follows, since $\inf_z f_n(z) \le f_n(x)$, hence
$$\liminf_n(\inf_z f_n(z)) \,\le\, \liminf_n f_n(x)\,\le\, \limsup_n f_n(x)$$
