1. ## Open sets

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.

2. ## Re: Open sets

Originally Posted by Sheld
Open balls with radius less than one are sets with a single element,
Right

i.e. not open.
Why?. If $\displaystyle G=\{a\}$ then, $\displaystyle B(a,1/2)=\{a\}\subset G$ so, $\displaystyle G$ is open.

3. ## Re: Open sets

Oh ok. The single sets are open and are in F.

All unions of the single sets are open under the discrete metric as well.

So I can build any open set in E using unions.

But (x_1,x_2,...,x_n) union (y_1,y_2,...,y_n) could be an open set if F

but it is certainly not in E

4. ## Re: Open sets

Under the discrete metric on a set $\displaystyle X$ every singleton set is open, so every $\displaystyle G\subset X$ has the form $\displaystyle G=\bigcup_{x\in G}\{x\}$ that is, $\displaystyle G$ is open. If $\displaystyle X=\mathbb{R}^n$ a singleton set is not open, hence, $\displaystyle E\subset F$ and $\displaystyle E\neq F$ .