Let $X=\{x_1,...,x_n\}$ be a finite set and let $\mathcal{T}=\mathcal{P}(X)$ be the discrete topology on $X$. Then is it true that $C^0(X;\mathbb{R})$ i.e. the set of continuous maps from $X$ to $\mathbb{R}$ together with the uniform metric $d(f,g)=\max_{x\in X}|f(x)-g(x)|$ is isometric to $\mathbb{R}^n$ with the maximum norm $\|\cdot\|_\infty$? I think yes, because :
Define $\varphi:C^0(X;\mathbb{R})\to\mathbb{R}^n,f\mapsto (f(x_1),...,f(x_n))$. Then clearly $\varphi$ is injective. Furthermore, it is also surjective, because if $\vec y=(y_1,...,y_n)\in\mathbb{R}^n$ then we define $f:X\to\mathbb{R}$ by $f(x_i)=y_i$ for all $i\in\{1,...,n\}$. Because we have the discrete topology on $X$, every function from $X$ to another topological space is continuous, so $f\in C^0(X;\mathbb{R})$. Then we have $\varphi(f)=\vec y$. Hence $\varphi$ is bijective. Furthermore, it is an isometry, because $\|\varphi(f)-\varphi(g)\|_\infty=\max_{1≤i≤n}|f(x_i)-g(x_i)|=d(f,g)$. Thus $\varphi$ is a bijective isometry (and therefore a homeomorphism) and thus $(C^0(X;\mathbb{R}),d)$ and $(\mathbb{R}^n,\|\cdot\|_\infty)$ are isometric. Is this correct?
If yes, this is a bit strange, because in todays exam we were asked to prove that $A\subset C^0(X;\mathbb{R})$ is compact if and only if it is bounded. But $A$ is compact iff $\varphi(A)$ is compact iff $\varphi(A)$ is bounded and closed iff $A$ is bounded and closed. Where am I mistaking?
You are entirely correct. It is not true that any bounded subset of $C^0(X;\mathbb{R})$ is compact; you must also require the subset to be closed. You can see this beyond all doubt in the case that $X$ has just one point, and so $C^0(X;\mathbb{R})$ is very obviously isometric to $\mathbb{R}$.