Let $(f_n)$ be a sequence of functions from $X$ to $\mathbb N$. We define $f:X \to \mathbb N$ by $f(x) = \limsup_{n \to \infty} f_n (x)$. Then we write $f = \limsup_{n \to \infty} f_n$. Assume that $f(x)$ is finite for all $x \in X$.
In this comment, @TheBridge said that $$\forall \alpha \in \mathbb N:f^{-1} (\alpha) = \bigcap_{m=0}^\infty \bigcup_{n=m+1}^\infty f_n^{-1} (\alpha)$$
Could you please explain the reasoning/intuition to get this formula? Thank you so much!
This is false. Let $f_n(x)=1$ for all $x$ when $n$ is even and $f_n(x)=2$ for all $x$ when $n$ is odd. Then $f(x)=2$ for all $x$. So $x \in \cap_{m=0}^{\infty}\cup_{n=m+1}^{\infty}f_n^{-1}\{1\}$ for any $x$ but $x \notin f^{-1} \{1\}$ for any $x$.