1. ## Open sets

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

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

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

I am pretty sure the only open set in $\mathbb{R}^{n}$ under the discrete metric is $\mathbb{R}^{n}$. 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.

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

But would these sets be in $E$? If they are, how is $E$ $\subset$ $F$? Are $E$ and $F$ sets of sets? Is $F$ = $\mathbb{R}^{n}$ or is $F$ = { $\mathbb{R}^{n}$ } (that is the set containing the set $\mathbb{R}^{n}$ 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 $G=\{a\}$ then, $B(a,1/2)=\{a\}\subset G$ so, $G$ 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 $X$ every singleton set is open, so every $G\subset X$ has the form $G=\bigcup_{x\in G}\{x\}$ that is, $G$ is open. If $X=\mathbb{R}^n$ a singleton set is not open, hence, $E\subset F$ and $E\neq F$ .