Limsup = sup(inf) for a series of functions

Question

Let $X$ be a set and $(f_k)_{k \in \Bbb N}$ a sequence of functions $f_k : X \to \bar {\Bbb R} $ for all $k$. I have to show that:

I know that for a normal series $(a_n)_{n\in\Bbb N}$ in $\Bbb R$ instead of a series of functions it's a simple theorem but i don't know how to prove it with the functions.

Any ideas or tipps? Thanks in advance!

Answer 1

You are asked to show that these are equal as functions; doing so means showing that they agree when applied to any $x\in X$.

Given $x\in X$, let us define $x_k:=f_k(x)$. Then showing that $$ \liminf_{k\to\infty}f_k(x)=\sup_{k\in\mathbb{N}}\left(\inf_{i\geq k}f_i(x)\right)\qquad\text{and}\qquad\limsup_{k\to\infty}f_k(x)=\inf_{k\in\mathbb{N}}\left(\sup_{i\geq k}f_i(x)\right) $$ holds for that specific $x$ immediately simplifies to showing that $$ \liminf_{k\to\infty}x_k=\sup_{k\in\mathbb{N}}\left(\inf_{i\geq k}x_i\right)\qquad\text{and}\qquad\limsup_{k\to\infty}x_k=\inf_{k\in\mathbb{N}}\left(\sup_{i\geq k}x_i\right). $$

So, if you do already know how to do this for sequences, you're in good shape.

Answer 2

Note that the all the lim sup inf are taken with respect to the sub index, which means the argument x is assumed fixed.

Once you admit( or prove ) the case for (argumented) real number series, you can apply such result pointwisely( at each x) to get the required one.