prove a set is closed in $C^0$

Question

Let $C^0([a,b), \mathbb{R})$ be the vector space of continuous functions $f:[a,b]\to\mathbb{R}$ equipped with the sup norm and sup metric. Prove that the set $A=\{f \in C^0:f(x)\geq 0 \textrm{ for all }x\in[a,b]\}$ is closed in $C^0$.

I have proved that any sequence of functions $f_n\in A$ converges uniformly to a function $f\in C^0$. This is because $C^0$ is a complete metric space. But now I'm stuck. Is there something I'm missing?

Answer 1

You have proved that any sequence of functions $f_n \in A$ converges uniformly to a function $ f \in C^0$ ???? This is not true: take $f_n(x)=x^n$ , $[a,b]=[0,1]$

Let $ (f_n)$ be a uniformly convergent sequence in $A$ and denote by $f$ the limit function of $(f_n)$. It is well known that $f \in C^0([a,b], \mathbb{R})$

You have to show that $f(x) \ge 0 $ for all $x \in [a,b]$. But this is easy to see, since each $f_n \in A$ and $(f_n)$ converges pointwise to $f$.

Answer 2

Choose a convergent (for the sup metric) sequence $(f_n)_{n\in \mathbb{N}}$ in $A$ and suppose that $f_n\rightarrow f$ with $f\notin A$. Then there exists and $x\in [a,b]$ such that $f(x)<0$. Let $\varepsilon=\frac{|f(x)|}{2}$. Then there exists an $N\in \mathbb{N}$ such that for all $n\geq N$ we have that $d(f_n,f)=\left\|f_n-f\right\|<\varepsilon$. Hence for all $y\in [a,b]$ and all $n\geq N$ we have that $|f_n(y)-f(y)|<\varepsilon$. Hence $f_n(x)\in (f(x)-\varepsilon, f(x)+\varepsilon)$ and thus $f_n(x)<f(x)+\varepsilon<0$. But this is impossible since $f_n\in A$.