Uniform convergence on $X \times X$. I'm studying Rudin's Functional Analysis, and there is a result that just works for real valued functions, so I'm trying to generalized it.
I already verify almost all the proof in that case, I'm just missing to prove that if $\{f_n\}_{n=1}^{\infty}$ is a sequence of continuous functions, such that $|f_n(x)|\leq 1, \forall x \in X, n \in \mathbb{N}$, defined as $f_n: (X, \tau) \longrightarrow (\mathbb{C}, \tau_{\text{usual}})$, where $(X, \tau)$ is  compact topological space, then de metric defined by
$$d(x, y)=\sum_{n=1}^{\infty}2^{-n}|f_n(x)-f_n(y)|,$$
is continous on $X\times X$ with the product topology, so in the proof they use the fact that $d(x, y)$ is uniformly convergent on $X\times X$, but I'm stuck trying to prove it, the result appears on Rudin's Functional Analysis, Chapter 3, result 3.8 (c), page 63.
Any kind of help it will be really appreciate, thank you so much.
 A: What you're not mentioning in your post which is mentioned in Rudin's book is that it is also assumed that $|f_n(x)|\le1$ for all $x\in X$ and $n\in\mathbb{N}$, i.e. the sequence $(f_n)$ is uniformly bounded. This is necessary, please edit your post and add this detail.
Let $\{(x_i,y_i)\}_{i\in I}\subset X\times X$ be a net with $(x_i,y_i)\to (x,y)$, i.e. $x_i\to x$ and $y_i\to y$. We have
$$|d(x,y)-d(x_i,y_i)|=\bigg|\sum_{n=1}^\infty\frac{1}{2^n}\big(|f_n(x)-f_n(y)|-|f_n(x_i)-f_n(y_i)|\big)\bigg|\le$$ $$\le\sum_{n=1}^\infty\frac{1}{2^n}|f_n(x)-f_n(x_i)-f_n(y)+f_n(y_i)|\le\sum_{n=1}^\infty\frac{1}{2^n}|f_n(x)-f_n(x_i)|+\sum_{n=1}^\infty\frac{1}{2^n}|f_n(y)-f_n(y_i)|=$$
$$=d(x_i,x)+d(y_i,y)$$
so it suffices to show that, for any net $(x_i)\subset X$, if $x_i\to x$ in $(X,\tau)$, then $d(x_i,x)\to0$.
Let $\varepsilon>0$. Find $n_0\in\mathbb{N}$ such that $\sum_{n>n_0}\frac{2}{2^n}<\varepsilon/2$. Now since each $f_j$ is continuous, find $U_j\in\text{Nbd}(x)$ such that $|f_j(x)-f_j(x')|<\varepsilon/{2n_0}$ and do this for all $j=1,\dots, n_0$. Set $U:=\bigcap_{j=1}^{n_0}U_j$. As this is a finite intersection, $U$ is an open neighborhood of $x$. Now if $x'\in U$, we have
$$d(x,x')=\sum_n\frac{1}{2^n}|f_n(x)-f_n(x')|=\sum_{n=1}^{n_0}\frac{1}{2^n}\cdot\frac{\varepsilon}{2n_0}+\sum_{n>n_0}\frac{1}{2^n}\cdot 2<\varepsilon/2+\varepsilon/2=\varepsilon $$
Now since $x_i\to x$, find $i_0\in I$ such that $x_i\in U$ for all $i\ge i_0$. Then $d(x_i,x)<\varepsilon$ for all $i\ge i_0$, proving that $d(x_i,x)\to0$, as $\varepsilon>0$ was arbitrary.
A: $d(x, y)=\sum_{n=1}^{\infty}2^{-n}|f_n(x)-f_n(y)|$
Hint : use Weierstrass M-Test
