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Thread: Open sets

  1. #1
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    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.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Open sets

    Quote Originally Posted by Sheld View Post
    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.
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  3. #3
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    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
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    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 $\displaystyle E\neq F$ .
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