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Math Help - Open sets

  1. #1
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    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.
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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 G=\{a\} then, B(a,1/2)=\{a\}\subset G so, 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 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 .
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