If $\displaystyle E$ denotes the collection of open sets in $\displaystyle \mathbb{R}^{n}$ under the Euclidean metric

and $\displaystyle F$ denotes the collection of open sets in $\displaystyle \mathbb{R}^{n}$ under the discrete metric

Show $\displaystyle E$ $\displaystyle \subset$ $\displaystyle F$ , but $\displaystyle E \neq F$

I am pretty sure the only open set in $\displaystyle \mathbb{R}^{n}$ under the discrete metric is $\displaystyle \mathbb{R}^{n}$. Open balls with radius less than one are sets with a single element, i.e. not open. Open balls with radius greater than 1 are the whole set.

$\displaystyle \mathbb{R}^{n}$ can be expressed as a union of open balls. None of which would be in $\displaystyle F$.

But would these sets be in $\displaystyle E$? If they are, how is $\displaystyle E$ $\displaystyle \subset$ $\displaystyle F$? Are $\displaystyle E$ and $\displaystyle F$ sets of sets? Is $\displaystyle F$ = $\displaystyle \mathbb{R}^{n}$ or is $\displaystyle F$ = {$\displaystyle \mathbb{R}^{n}$ } (that is the set containing the set $\displaystyle \mathbb{R}^{n}$ as its only element)

Could someone clear up this confusion for me? Thank you.