I have been trying to understand the basis elements of the uniform topology on $\mathbb{R}^{\omega}$. For some time, I thought they would be:
$B_\bar{p} (x,\epsilon) = \prod (x_i - \epsilon, x_i + \epsilon)$ if $\epsilon < 1$
However, after reading a bit online, I learned that this set is not even open in the uniform topology. The actual basis elements are:
$B_\bar{p} (x,\epsilon) = \bigcup_{\delta < \epsilon} \prod (x_i - \delta, x_i + \delta)$ if $\epsilon < 1$
Why is this the case?
Note that under the uniform topology the corresponding metric is $d(x,y) = \min(1, \sup_i (|x_i-y_i|))$. In a finite product of intervals $(x_i-\epsilon, x_i + \epsilon)$ any particular element will thus be strictly less than $\epsilon$ distance away from $x$, but in an infinite product you have elements like $y = (y_i) = (x_i + (1-1/n)\epsilon)$, where the supremum is equal to $\epsilon$. Note that any open neighborhood of this particular point $y$ will contain elements not in the first given product, demonstrating that the product is not open.
Such elements must be excluded from the basis elements, which should instead be the set of elements of distance strictly less than $\epsilon$ from $x$. This set can be constructed as the given union over $\delta <\epsilon$.
