Originally Posted by

**Deveno** there are a couple of things to note, here:

first of all, if the index set J is finite, you get the same topology in either case. I understand this

for reasons that are often not made clear from the definition itself, the product topology is the smallest (coarsest) topology that makes the projection functions continuous.

so what can go wrong in the box topology? suppose you have $\displaystyle \mathbb{R}^\omega$.

clearly, the set $\displaystyle U = \prod_{n \in \mathbb{N}}(-1/n,1/n)$ is open in the box topology.