The cardinal number of a set from Cylindrical Sigma-algebra.

Question

Let $\mathbb{R}^{[0,1]}$ be the set of all function on $[0,1]$, and $\mathcal{B}(\mathbb{R}^{[0,1]})$ be the sigma-algebra generated by all cylinder sets: $$\{ x=x(t):(x(t_1),\ldots,x(t_n))\in B \}$$ for some $B\in \mathcal{B}(\mathbb{R}^n)$. We know that every set $A\in\mathcal{B}(\mathbb{R}^{[0,1]})$ can be presented as $A=\left(x=x(t):(x(t_1),x(t_2),\ldots)\in\mathcal{B}(\mathbb{R}^\infty)\right)$ for some ${t_j}\in [0,1],j=1,2,\ldots$. Since we only have conditions on a countable set, is it true that every set in $\mathcal{B}(\mathbb{R}^{[0,1]})$ has a cardinal number $2^c$? Therefore $C[0,1]$ can't belong to $\mathcal{B}(\mathbb{R}^{[0,1]})$ since its cardinal number is $c$.