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.


2 Answers 2


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.

This result can be found in: Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.

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.

  • $\begingroup$ +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 $\endgroup$ Mar 20, 2019 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:

In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.


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