Let $C(\mathbb{R}_{\geq 0}, \mathbb{R})$ be a space of continuous functions. For some $n\in\mathbb{N}$ and $0< t_1 < t_1 < ... < t_n$, let us call a set $\{f \in C(\mathbb{R}_{\geq 0}, \mathbb{R}) ; f(t_1)<a_1,f(t_2)<a_2,...,f(t_n)<a_n\}$ a cylinder set. When I read the book , "Diffusion Processes and Stochastic Calculus, Fabrice Baudoin", I found the argument (Proposition 1.3) that this cylinder is open with respect to the topology induced by the following distance. For $f,g \in C(\mathbb{R}_{\geq 0}, \mathbb{R})$ $$d(f,g) = \sum_{n=1}^{\infty}\frac{1}{2^n} \min(\sup_{0\leq t\leq n} |f(t)-g(t)|, 1)$$
But I don't see why. Any ideas would be highly appreciated!
Hints: In the metric you have defined we have $f_k \to f$ iff $f_k(x) \to f(x)$ uniformly on $[0,n]$ for each $n$. Take a sequence $(f_k)$ in the complement of the given set converging to $f$. Then $f_k (t_i) \to f(t_i)$ for each $i$. There exist $i$ such that $f_k(t_i) \geq a_i$ for infinitely many $k$ and this gives $f(t_i) \geq a_i$ proving that the complement is closed.