Is this function with $\sup$ continuous?

Question

Prove or disprove:

Let $f_n=\mathbb{R}\rightarrow[0,1]$ be a continuous function for all $n\in\mathbb{N}$. Then, $h(x):=\sup_{n\in\mathbb{N}}f_n(x)$ is continuous.

All right, I would say this statement is wrong, but I don't really know how to show it. It's kinda hard to explain, but I thought about maybe using a sequence, that converges to a irrational number, and doing some clever stuff with the sup... I will try a little a bit, but can someone give me some advice? Thanks in advance!

Answer 1

Take$$f_n(x)=\begin{cases}nx&\text{ if }x<\frac1n\\1&\text{ otherwise.}\end{cases}$$Each $f_n$ is continuous, but$$\sup_{n\in\mathbb N}f_n(x)=\begin{cases}0&\text{ if }x=0\\1&\text{ otherwise.}\end{cases}$$

Answer 2

See what happens with $f_n(x)=1-\min\{\lvert x\rvert^n, \lvert x\rvert^{-n}\}$ (with the notational assumption $\min\{\lvert 0\rvert^n,\lvert 0\rvert^{-n}\}=0$ for all $n>0$).