1
$\begingroup$

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 .

$\endgroup$
1
$\begingroup$

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 ${\bf R}^{\infty}$.

$\endgroup$
  • $\begingroup$ Product space . $\endgroup$ – Planche Sep 9 '16 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.