Convergence in $C(X)$ is uniform convergence.

Question

I read this the convergence in $C(X)$ is uniform convergence.

Where $X$ is compact hausdorff topological space and

$$C(X)=\{f:X\to\mathbb{C}\;\mid \; f\ \text{is continuous}\}$$

And $$\|f\|=\sup\{|f(x)|:x\in X\}$$

What I have done: I suppose $f_n\to f $ pointwise, then for every $\varepsilon>0$ and for every $x\in X$ there exists a positive integer $N$ such that $\forall n\geq N$ we have

$$d(f_n(x),f(x))=\|f_n(x)-f(x)\|<\varepsilon$$

then

$$\sup_{x\in X}\|f_n(x)-f(x)\|<\varepsilon$$

Then further what I should do?

Edited:My question is how to prove that convergence in C(X) is uniform convergence and also is this because of compactness of $X$ or because of supremum norm?

Answer 1

Let $f_n \rightarrow f$ in $C(X)$. This means $\|f_n-f\| \rightarrow 0$. In turn, this means that $\forall \epsilon >0 \space \exists n_\epsilon \space \| f_n-f \|< \epsilon$. Detailing what the norm means, we get $\mathbb{sup}_{x \in X}|f_n(x) - f(x)|< \epsilon$, which in particular implies $|f_n(x) - f(x)|< \epsilon \space \forall x \in X$ (and $n>n_\epsilon$ as previously found, of course). And this is just the definition of the uniform convergence.