Non-Borel set in arbitrary metric space

Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $$C[0,1]$$ all continuous functions on $$[0,1]$$ where metrics is supremum.

Yes, there is indeed examples of non-Borel sets in $$C[0,1]$$ of all continuous functions from $$[0,1]$$ to $$\mathbb{R}$$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.
In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $$(\{x,y\},d)$$ equipped with the discrete metric $$d:\{x,y\}\times \{x,y\} \to \{0,1\}$$ given by $$d(x,y)=1, \quad d(x,x)=d(y,y)=0.$$ The Borel sigma algebra on this metric space is given by $$\{\{x\},\{y\},\{x,y\},\emptyset\} = \mathcal{P}(\{x,y\})$$ where $$\mathcal{P}(\{x,y\})$$ is the powerset of $$\{x,y\}$$, so all subsets are Borel measurable sets.
• +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Y\subset X$ is equal to $\cup \{\{y\}:y\in Y\},$ which is a countable union of closed sets – DanielWainfleet Mar 20 at 4:18
Martin gave a specific example in $$C[0,1]$$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer: