# Swapping a limit and a $\sup$

In this proof of the completeness of $(C(K), \| \cdot \|_\infty)$ they use the following inequality:

$$\sup_{x \in K} \lim_{m \to \infty} | f_n(x) - f_m(x) | \leq \liminf_{m \to \infty}\, \sup_{x \in K} | f_n(x) - f_m(x) |.$$

Can someone explain to me why this is true? Presumably, they first write $$\sup_{x \in K} \lim_{m \to \infty} | f_n(x) - f_m(x) | = \sup_{x \in K} \,\liminf_{m \to \infty} | f_n(x) - f_m(x) |$$ and then swap but the exact argument why this gives the inequality is unclear to me.

We have for all $x\in K$, $m,n$ integers: $|f_n(x)-f_m(x)|\leq\sup_{x\in K}|f_n(x)-f_m(x)|$ and taking the $\liminf$: $$\forall n\in\mathbb N,\forall x\in K\quad\liminf_m|f_n(x)-f_m(x)|\leq\liminf_m\,\sup_{z\in K}|f_n(z)-f_m(z)|,$$ and now taking the supremum over $x\in K$ we get $$\forall n\in\mathbb N: \quad\sup_{x\in K}\,\liminf_m|f_n(x)-f_m(x)|\leq\liminf_m\,\sup_{z\in K}|f_n(z)-f_m(z)|,$$ hence $$\forall n\in\mathbb N: \quad\sup_{x\in K}\,\lim_m|f_n(x)-f_m(x)|\leq\liminf_m\,\sup_{z\in K}|f_n(z)-f_m(z)|,$$ since we got the convergence of $\{f_n(x)-f_n(x)\}_m$.