Currently, I am stuck with following the final conclusion in the proof of corollary V.6.1 in Schaeffer & Wolff's Topological Vector Spaces, 2nd edition, pp. 230-231. Probably it is trivial and I would appreciate a bit of help here (all vector spaces are over $\mathbb{R}$, $L_i$ are equipped with their order topology):
I understand why $p_i$ are continuous when $\prod_{i=1}^n L_i$ is equipped with the canonical order topology and why $p_i$ are open maps then. However, I fail to see why the product topology and the order topology have to coincide. It follows that every neighbourhood of $0$ in $\prod_{i=1}^n L_i$ is a neighbourhood in the order topology. If, on the other hand, I let $W$ be a convex neighbourhood of $0$ in the order topology, then we have to show that it contains a ¨rectangle¨, i.e. $$ \bigcap_{i=1}^n p_i^{-1}(U_i) \subset W $$ for $U_i \subset L_i$ neighbourhoods of $0$ in $L_i$.
Setting $U_i = p_i(W)$ does not work. What other choice could work? Thanks!