Borel $\sigma$-algebra of continuous functions Let $B(C(T,\mathbb{R}))$ be the Borel sigma algebra of continuous functions mapping from the compact metric space $T$ to $S$ defined by the canonical metric $\|\cdot\|_\infty.$ 
Now I was wondering whether we can write this sigma algebra also as a sigma-algebra generated by some cylinder sets? 
If anything is unclear, please let me know.
 A: If we add the assumption that $T$ is separable then the answer is yes. Let's prove it.
Let $\cal P$ denote the $\sigma$-algebra generated by the cylinder sets, and let $\cal L$ denote the $\sigma$-algebra generated by the open sets in the uniform metric.
Note that $\cal P \subset L$, since the product topology is weaker than the uniform topology on $C(T, \Bbb R)$. This is true on any topological space, since uniform convergence always implies pointwise convergence.
The reverse direction (i.e, proving that $\cal L \subset P$) is a bit trickier.
First I claim that $C(T, \Bbb R)$ is separable (in the uniform metric). To prove this, let $\mathcal D \subset T$ be a countable dense subset. For $x \in \cal D$ and $r \in \Bbb Q_+$, define a continuous map $f_{x,r}: T \to \Bbb R$ as $f_{x,r}(y)=\max \{ 0, 1-r\cdot d(x,y)\}$. By the Stone-Weierstrass theorem, the set of all rational linear combinations of finite products of the $f_{x,r}$ forms a countable dense subset of $C(T,\Bbb R)$. Hence $C(T,\Bbb R)$ is separable.
Now, in order to show that $\cal L \subset P$, it suffices to show that any closed ball in $C(T,\Bbb R)$ lies in $\cal P$. Indeed, this would imply that any open ball lies in $\cal P$, since open balls are countable unions of closed balls. From open balls, you can conclude by separability of $C(T, \Bbb R)$ that any open set in $C(T, \Bbb R)$ lies in $\cal P$ (because open sets  in separable metric spaces are just countable unions of open balls). And from open sets, you can conclude that any Borel set in $C(T,\Bbb R)$ (i.e, any set in $\cal L$), lies in $\cal P$.
So to finish off the proof we just need to show that any closed ball $B(f, \epsilon) \subset C(T, \Bbb R)$ lies in $\cal P$. But this is easy: just write $$B(f,\epsilon) = \bigcap_{x \in \cal D} \pi_x^{-1}\big( [f(x)-\epsilon\;,\; f(x)+\epsilon] \big)$$
where the $\pi_x: C(T,\Bbb R) \to \Bbb R$ are the canonical projections $g \mapsto g(x)$, and $\cal D$ is a countable dense subset of $T$. Since $\cal P$ is generated by the $\pi_x$, and since $\cal D$ is countable, it follows that the right-hand side of the above expression is in $\cal P$, and thus $B(f, \epsilon) \in \cal P$, as desired.
