let $(X,d_X)$ and $(\mathbb{R}^n,d_2)$ be metric spaces with $X$ compact. Then the set of continuous functions is defined by $$ C_n(X):=\{f:X\rightarrow \mathbb{R}^n \;|\;f \text{ continuous }\} $$
$\left(C_n(X),d_{\sup}\right)$ is a complete and separable metric space
First question (i): Is it necessary, that $X$ must be a metric space or is it also fine if $X$ is a compact topological space?
(ii) We can write $C_n(X)$ as the cartesian product $C_n(X)=C_1(X)\times C_{n-1}(X)$. Is this really obvious or do we have to proof this? I know, if we have $C_1$ and $C_{n-1}$ than it is not obvious that we can write $C_1(X)\times C_{n-1}(X)=C_n(X)$ because it depends on the properties of $C_1$ and $C_{n-1}$.
If it is clear, that we can write $C_n(X)$ as cartesian product, will $C_1$ and $C_{n-1}(X)$ be also complete and separable and why?
thanks
(i) For $C_n(X)$ to be separable metric, $X$ has to be compact metrisable (e.g. see the references in this post. The result is true for both functions to the reals and finite-dimensional Euclidean space.
(ii) A function $f: X \to \mathbb{R}^n$ is continuous iff all $\pi_i \circ f: X \to \mathbb{R}$ are continuous. So mapping $f \in C_n(X)$ to $(\pi_1\circ f,\ldots, \pi_n \circ f) \in C_1(X)^n$ is a linear isomomorphism between the spaces, and an isometry (I think) if we give $\mathbb{R}^n$ and the product $C_1(X)^n$ the max-metric (and base the $d_{\text{sup}}$ on the max-metric, which is topologically the same).
It's not a long fact to prove, but it does require a small argument such as I gave above, IMHO.
A finite product of completely metrisable separable spaces is still of that type.