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Isomorphism between R^inf and a proper subset of R^inf

  1. Jan 8, 2009 #1
    I'm fairly certain the following is a vector space isomorphism [tex]\phi :\mathbb{R}^\infty\rightarrow\mathbb{R}^\infty[/tex] where the vector space is the space of infinite sequences of real numbers and phi is defined by [tex] \phi(a_1,a_2,...)=(0,a_1,a_2,...) [/tex]. The mapping is linear and the inverse seems to be well defined. Is my logic flawed?
     
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  3. Jan 8, 2009 #2

    Hurkyl

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    What's the inverse of (1, 0, 0, 0, ...)?

    (Why are you asking about the inverse anyways?)
     
  4. Jan 8, 2009 #3
    You're right. As I wrote it, phi isn't an isomorphism. I'll try again. Define [tex]\phi :\mathbb{R}^\infty\rightarrow S[/tex] ,where [tex]S\subset \mathbb{R}^\infty[/tex] is the subspace containing all infinite sequences of reals of the form [tex](0,a_1,a_2,...)[/tex], as: [tex]\phi(a_1,a_2,...)=(0,a_1,a_2,...)[/tex]. Define the inverse function [tex]\phi^{-1}:S\rightarrow\mathbb{R}^\infty[/tex] as: [tex]\phi^{-1}(0,a_1,a_2,...)=(a_1,a_2,...)[/tex]. Does this establish an isomorphism between [tex]\mathbb{R}^\infty[/tex] and [tex]S[/tex]?
     
  5. Jan 8, 2009 #4

    Hurkyl

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    Let's see... the things you need are
    1. phi is well-defined
    2. phi is linear
    3. phi^-1 is well-defined
    4. phi^-1 is linear
    5. The composition of phi with phi^-1 is the identity
    6. The composition of phi^-1 with phi is the identity

    You've checked all of those, right? Then congratulations, you have an isomorphism!


    Incidentally, you can prove a more general theorem: if phi is injective can you show it's an isomorphism onto its image? (And thus, all you need for your problem is an injection that is not surjective)
     
  6. Jan 8, 2009 #5
    yes, it must be because it's surjective onto its image. Also, I think that 1,2,3,5,6 combined imply 4. Thanks a lot!
     
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