For any $x \in X$, $f_n(x) \to \infty$ as $n \to \infty$.
Is this equivalent to saying $\cap_{K=1}^{\infty} \cup_{N=1}^{\infty} \cap_{n=N}^{\infty} \{x: f_n(x)\geq K\}$? Why?
My understanding: for any a positive integer K, we can find N, such that when n>N, then $f_n(x)>K$?