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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?

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    Are you assuming the discrete topology on $X$? If not, then the functions that are in $C^0(X;\mathbb R)$ will depend on what topology you put on $X$, so it seems likely the spaces will not be isomorphic. **EDIT:** Ah, I just saw you do indeed refer to the discrete topology on $X$. Still, this poses an interesting variation, I think. – 2017-01-26
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    I think the exam question was wrong. Take $X=\{1,...,n\}$ and $A$ to be the set of those functions that send $1$ to $2^{-k}$ for some $k$ and $i$ to $0$ for $i\neq 1$. Then obviously $A$ is bounded, as for each $f\in A$, $d(0,f) \leq 1$ (where $0$ denotes the function that sends everyone to $0$). However it isn't compact, as the sequence $(f_k)$ in $A$ of functions that send $1$ to $2^{-k}$ has no limit point in $A$. – 2017-01-26
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    I think that perhaps they wanted to write 'pre-compact' instead of 'compact'; in this case the equivalence would hold. And such a tiny syllable can easily be forgotten... In this case, one could even apply Arzela-Ascoli to directly obtain the equivalence (although this would be a bit of an overkill ^^). – 2017-01-26

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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 $\mathbb{R}$.

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    Thanks for the confirmation! – 2017-01-27