What are the open set of $[0,1]^{\mathbb Z}$?

Question

I'm reading an article in probability, and they set $\Omega =[0,1]^{\mathbb Z}$ and they equip $\Omega $ with its Borel $\sigma -$field. What couls look the Borel $\sigma -$field of $\Omega $ ? In other word, what are the open set of $[0,1]^{\mathbb Z}$ ? Are they of the form $]a_1,b_1[\times ]a_2,b_2[\times ...$ ?

Answer 1

The base of topology of $[0,1]^\mathbb Z$ is $$\left\{\prod_{i=1}^k (]a_{n_i},b_{n_i}[\cap [0,1])\times \prod_{i\in \mathbb Z\backslash \{n_1,...,n_k\}}[0,1]\mid (n_i)_{i=1}^k\subset \mathbb Z\right\}.$$

Answer 2

Almost: a basis for this topology is given by taking an open interval on finitely many of the coordinates (finitely many elements of $\mathbb Z$) and taking the full interval $[0, 1]$ elsewhere. In particular, even a general open set can only "impose restrictions" on finitely many of the coordinates.

However, for the purposes of the Borel $\sigma$-algebra the difference doesn't matter, since it includes countable intersections of its elements; in particular, the sets you suggest are the open sets are still Borel sets.