topology in R infinity What does the following sentence mean and why is that true:
"The nonnegative orthant in $R^{\infty}$ has empty interior in product topology"
Thank you!
 A: Likely the following:

The set $A= \{ x = ( x_i )_{i=1}^\infty \in \mathbb{R}^\infty : ( \forall i ) ( x_i \geq 0 ) \}$ has empty interior in the product topology on $\mathbb{R}^\infty$.

It is true because the basic open sets are of the form $U = \prod_{i=1}^\infty U_i$ where each $U_i$ is an open subset of $\mathbb{R}$ and $U_i = \mathbb{R}$ for all but finitely many $i$.  It follows that given any nonempty open $U \subseteq \mathbb{R}^\infty$ there is an $x = (x_i)_{i=1}^\infty \in U$ such that $x_i < 0$ for all but finitely many $i$, and so no nonempty open set is a subset of $A$.

Edit: The above remarks were made under the (likely incorrect) thinking that $\mathbb{R}^\infty$ is what might better be called  $\mathbb{R}^{\mathbb{N}}$, a product of countably many copies of $\mathbb{R}$.  Some (most?) mathematical circles would use $\mathbb{R}^\infty$ to denote the family $$\{ x = (x_i)_i \in \mathbb{R}^{\mathbb{N}} : x_i = 0\text{ for all but finitely many }i \}.$$  The analysis above mostly works for this set as well, but the conclusion 

It follows that given any nonempty open $U \subseteq \mathbb{R}^\infty$ there is an $x = (x_i)_{i=1}^\infty \in U$ such that $x_i < 0$ for all but finitely many $i$ ...

would need to be changed to 

It follows that given any nonempty open $U \subseteq \mathbb{R}^\infty$ there is an $x = (x_i)_{i=1}^\infty \in U$ such that $x_i < 0$ for some $i$ ...

(Which actually suffices for my original assumption as well!)
