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Trying to prove F^infinity is infinite dimensional

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Theorem: Prove that [tex]\mathbb{F}^{\infty}[/tex] infinite dimensional


    2. Relevant equations

    Definition of Infinite Dimensional Vector Space: A vector space that
    is not finite dimension

    Definition of Finite Dimensional Vector Space: [tex]\exists[/tex] list of vectors
    in it that spans the space

    Definition of List of Length n: An ordered collection of [tex]n[/tex] objects.
    It has finite length.


    3. The attempt at a solution


    1) [tex]\mathbb{F}^{n}[/tex] is the set of all list of length n

    2) [tex]\mathbb{F}^{\infty}[/tex] is the set of all list of length [tex]\infty[/tex]

    2.1) This contradicts the definition of list [tex]\therefore[/tex] there is
    no list of infinite length [tex]\therefore[/tex] it cannot be a finite dimensional
    vector space because it doesn't contain a list [tex]\therefore[/tex] it is
    an infinite dimensional vector space.

    My solution seems weak in that it doesn't use any math. But does
    it actually prove the theorem? If not can someone point me in the right direction?
     
    Last edited: Sep 7, 2009
  2. jcsd
  3. Sep 7, 2009 #2
    That's a good outline for a proof. Suppose that [tex] F^{\infty} [/tex] isn't of infinite dimension. Then it must be of dimension n. So the set (consisting of n vectors) [tex] \beta = (1, 0, 0, ... 0), (0, 1, 0, ... 0), ... , (0, 0, ... , 1) [/tex] must span [tex] F^{\infty} [/tex].

    Obviously [tex] (1, 2, 3, ... , n+1) \in F^{\infty} [/tex]. Is that vector in [tex] \beta [/tex] ?
     
  4. Sep 7, 2009 #3
    It's not in the list [tex]\beta[/tex], it's in the span of [tex]\beta[/tex]. Right?
     
  5. Sep 7, 2009 #4
    Sorry, my question should have been, "is that vector in the span of [tex] \beta [/tex]?"
     
  6. Sep 8, 2009 #5
    I don't understand where you are trying to go with this. There is no contradiction if you suppose that it is not infinite dimensional.
     
  7. Sep 11, 2009 #6
    Never mind this was due last week and my proof was correct according to the solutions.
     
  8. Sep 12, 2009 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since you are trying to prove it is infinite dimensional there had better be!
     
  9. Sep 12, 2009 #8
    This reasoning is not clear. A finite dimensional vector space does not need to contain a list; it need only satisfy the axioms defining a vector space and be spanned by a finite amount of elements of that space under scalar multiplication and vector addition. Ie., the set of sequences of the form (n, 0, 0, ...) for all real n forms a 1-dimensional vector space under pointwise addition, but it does not consist of any lists.
    J89's outline has a similar missing link: ie., it refers to a list of vectors that are not themselves elements of Foo as potential spanning vectors for Foo, which is unfounded.
     
    Last edited: Sep 12, 2009
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