# Borel Sigma Algebra on $\mathbb{R^{\infty}}$

I read some book which contains

Theorem Borel Sigma Algebra on $\mathbb{R^{\infty}}$ is generated by $A=\{ \Pi_1^\infty X_i$ : For finite indices $X_i=\{x_i\}$ and for remaining indices $X_i=\mathbb{R} \}.$

I know Borel sigma Algebra is sigma algebra generated by open sets.

If this is true $(0,1) × \mathbb{R} × \mathbb{R} ...$ is generated by sets in A. But I don't understand this situation.

Could you explain?

$R^\infty$ is product space .

## 1 Answer

The source of your confusion is the absence of a choice (at least in your question) of a specific topology on the set in question. In detail:

To call a set open, we first need to choose a topology on the given space. On the real line ${\bf R}$, the most common assumed topology is the Euclidean one. But, on ${\bf R}^{\infty}$, there are several choices, and none is obvious. And so, before answering your question, there has to be a clear definition of which topology is intended on ${\bf R}^{\infty}$.

• Product space . – Planche Sep 9 '16 at 18:49