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Basis of a vector space

  1. Dec 5, 2007 #1
    Maximal subspace

    Problem: Prove that every vector space V has maximal subspace, i.e. a proper subspace that is not properly contained in a proper subspace of V.

    I let A be the collection of all proper subspaces of V, but I can't prove that every totally ordered subcollection of A has an upper bound in A. The problem that the union of proper subspaces is not necessarily a proper subspace of V. What do I do now?
     
    Last edited: Dec 5, 2007
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  3. Dec 5, 2007 #2

    morphism

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    But the union of a chain of subspaces is a subspace.
     
  4. Dec 5, 2007 #3

    JasonRox

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    Think basis elements.
     
  5. Dec 5, 2007 #4
    But it has to be a proper subspace of V.

    For example { span{1}, span{1,x}, span{1,x,x^2}, span{1,x,x^2,x^3}, ..... } is a chain of proper subspaces of R[x], but its union is all of R[x], which is not a proper subspace of R[x].
     
    Last edited: Dec 5, 2007
  6. Dec 5, 2007 #5
    JasonRox's idea is good, take a basis of V and delete one element. The span of that would have to be a maximal subspace.

    But I'm assuming that mathboy wants to use Zorn's lemma. In that case choose any v in V, and let A be the collection of all subspaces not containing v. This time the upper bound of any chain will be a proper subspace. The maximal element of A would be a maximal subspace of V.
     
    Last edited: Dec 5, 2007
  7. Dec 5, 2007 #6

    morphism

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    Oops! I should learn to read! Thanks for pointing that out. :smile:
     
  8. Dec 5, 2007 #7

    Office_Shredder

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    I don't think the problem implies there IS a basis of V (unless it turns out all vector spaces have a basis, and I just don't know that yet)
     
  9. Dec 5, 2007 #8

    JasonRox

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    Is V finite-dimensional? Is the book assuming that?

    Do you know what finite-dimensional is?
     
  10. Dec 5, 2007 #9
    Every vector space V has a basis, whether it is finite-dimensional or not. In mathboy's problem V can be infinite-dimensional and the result is still true.

    If you want to prove that V has a basis if V is infinite-dimensional, you would have to use Zorn's lemma as well. Ultimately, mathboy's problem rests on Zorn's Lemma.


    My approach to mathboy's problem is: Choose any v in V, and let A be the collection of all subspaces not containing v and then use Zorn's lemma. But I'm trying to figure out if there is a better partially ordered set to use, because my A seems a little clumsy (though I believe it would still get the job done).
     
  11. Dec 5, 2007 #10

    JasonRox

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    Of course I know this!

    Ok, a vector space has a basis {v_1,...}, now delete one vector from there and span that that set. What do you get?

    Voila!
     
  12. Dec 5, 2007 #11
    Thanks guys. I forgot to say that I have to use Zorn's Lemma. But I know how to proceed now. I will use the collection of all proper subspaces that does not contain some fixed v in V.
     
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