Sequence of continuous functions $f_n $ with $\lim_{n \rightarrow +\infty} f_n(x)=f(x)$ and property

Question

Let be $(X, d)$ a complete metric space, $f_n : X \rightarrow \mathbb{R}$ a sequence of continuous functions and a function $f: X \rightarrow \mathbb{R}$ such that: $$\lim_{n \rightarrow +\infty} f_n(x)=f(x)$$ Why in the world my textbook says that: $$X=\bigcup_{k\geq1}\bigcap_{n\geq1}\{|f_n|\leq k\}$$

Answer 1

Suppose $x \in X$ and $x \notin \bigcup_{k=1}^{\infty}\bigcap_{n=1}^{\infty}\{ |f_n|\leqslant k\}$. This means for every $k$ we have $x \notin \bigcap_{n=1}^{\infty}\{|f_n|\leqslant k\}$. Therefore, for every $k$ we have $x \in \bigcup_{n=1}^{\infty}\{|f_n|>k\} $.

Since $f_n$ is continuous with compact domain, it is bounded. That is, there exists $M_n\in [ 0,\infty)$ such that $|f_n(x)|\leqslant M_n$ for every $x \in K$. Since $f_n(x) \to f(x)$ as $n \to \infty$, we can choose positive integer $N$ sufficiently large such that $f_n(x) \leqslant f(x) + 1$ for every positive integer $n \geqslant N$. Therefore, $|f_n(x)|\leqslant \max \{M_1,\cdots, M_{N-1},|f(x)|+1\}$ for every positive integer $n$. Now choose $k > \max \{M_1,\cdots, M_{N-1},|f(x)|+1\}$. We find out that no positive integer $n$ satisfy the property $\{|f_n(x)|>k\}$. In other words, $x \notin \bigcup_{n=1}^{\infty}\{|f_n|>k\} $, which is a contradiction.