# Generating Borel $\sigma$ -algebra on metric spaces

I have troubles fully grasping the following statement:

'Let $(U, \rho)$ be a metric space. We equip U with the Borel $\sigma$ -algebra generated by the open sets in U (in the metric topology).'

Even though my knowledge of topology is very limited I am able to get the basic intuition: since every metric space is also topological space, we can use naturally defined topology (generated by the open balls) to generate Borel sigma algebra (hopefully I am getting this right). In the case of, let's say, $U=\Bbb {R}^d$ things are pretty clear to me (what I find helpful here is the relation $\tau_{d} \subset B_{d}\subset \Bbb{P}(\Bbb{R}^d)$ where $\tau$, $B$ and $\Bbb{P}$ denote topology, Borel sigma algebra and power set, respectively).

However what happens in case of $U=C([a,b])$ (space of continuous functions, for $\rho$ we can take $||.||_{\infty})$ ? It is not so straightforward to me what do we take for topology here and how does generated Borel algebra look like (I understand there is not precise description, I just need the intuition). Are there any additional problems here since C([a,b]) is infinite dimensional?

• What's Pot(R^d)?? – mathworker21 Jan 8 '18 at 23:36
• Power set, I will change the notation. – Metod Jazbec Jan 9 '18 at 6:51

The topology is still generated by open balls. An open ball in $C([a,b])$ looks like $$B_{r}(f) := \{g \in C([a,b]) \mid ||f-g||_{\infty} < r\}.$$ Since $f$ and $g$ are continuous on the compact set $[a,b]$, they are bounded, so you can think of $B_{r}(f)$ as the set of functions $g$ such that $g(x)$ is always within distance $r$ of $f(x)$ for $x \in [a,b]$. The topology on $C([a,b])$ is generated by the basis $$\mathcal{B} = \{B_{r}(f) \mid f \in C([a,b]), r > 0\}.$$
As for what the Borel $\sigma$-algebra looks like... Well, you still have that the topology is contained in the $\sigma$-algebra, which in turn is contained in the power set of $C([a,b])$. Beyond that, it's hard to say what it "looks like."