Borel $\sigma$-algebras on paths of bounded variation Let $(C, \|\cdot\|)$ be the Banach space of continuous paths $x: [0,1]\rightarrow\mathbb{R}^d$ starting at zero with sup-norm $\|\cdot\|$.
Let further $B\subset C$ be the subspace of $0$-started continuous paths of bounded variation, i.e.
$$B = \big\{x\in C \ \big| \ \|x\|_1:=\sup\nolimits_{(t_\nu)}{\textstyle\sum_\nu}|x_{t_\nu} - x_{t_{\nu-1}}|<\infty \ \text{ and } \ x(0)=0\big\}$$
where the above sup runs over all partitions $(t_\nu)$ of $[0,1]$.
The set $B$ can be endowed with both the (uniform) norm $\|\cdot\|$ and the (1-variation) norm $\|\cdot\|_1$.
Do you know if the Borel $\sigma$-algebras on $(B, \|\cdot\|)$ and $(B, \|\cdot\|_1)$ coincide?
Any references or hints are welcome.
Edit: As pointed out by Gerald Edgar, we narrowed $B$ down to paths started at zero to make the question more meaningful.
Edit 2: As Gerald Edgar pointed out in the comments to his answer below, the answer to this question is no, the Borel $\sigma$-algebras cannot coincide because there are more $\|\cdot\|_1$-open sets in $B$ than there are $\|\cdot\|$-Borel sets in $C$.
 A: I think there is a problem with a mere semi-norm.  The constant functions have variation distance $0$ from each other. Any variation-open set contains either all the constants, or none of them.  Therefore, any variation-Borel set contains either all the constants, or none of them.
Choose $\mathbf u \in \mathbb R^n \setminus \{0\}$.  Take the subset $\{0,\mathbf{u}\} \subseteq B$ of two constant functions.  It is sup-norm closed, so it is a uniform-Borel set.  But it is not a variation-Borel set.

Maybe a better question would restrict to the subset of $B$ with $x(0) = 0$.  Then at least we have a norm.

rmcerafl pointed out a problem with the following.  For now, I will leave it here for reference.

Let $C_0 = \{x \in C : x(0)=0\}$ with the uniform norm.  Let
$B_0 = \{x \in B : x(0)=0\}$ with the variation norm.  Both of these are complete separable metric spaces.  The inclusion map $i : B_0 \to C_0$ is continuous and injective.   Then citing a result or two from descriptive set theory*, we get: $i(B_0)$ is a Borel subset of $C_0$ and $i : B_0 \to i(B_0)$ is a Borel-isomorphic map.  This shows, in particular, the variation-Borel subsets of $B_0$ coincide with the uniform-Borel subsets of $i(B_0)$.


${}^*$ Proposition 8.3.5 and Theorem 8.3.7 in:
Cohn, Donald L., Measure theory, Boston, Basel, Stuttgart: Birkhäuser. IX, 373 p. DM 42.00 (1980). ZBL0436.28001.
