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Prove that ℝ has no subspaces except ℝ and {0}.

  1. Mar 19, 2013 #1
    Prove that ℝ has no subspaces except ℝ and {0}.
     
    Last edited by a moderator: Mar 20, 2013
  2. jcsd
  3. Mar 19, 2013 #2

    WannabeNewton

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    What have you tried?
     
  4. Mar 19, 2013 #3
    Ive tried using a nontrivial subspace of R and showing that it equals R but I am having touhg time doing that
     
  5. Mar 19, 2013 #4

    WannabeNewton

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    Ok that's a good way to do it. Let's call this non-zero subspace [itex]V[/itex] and let [itex]v\in V[/itex] be a non-zero vector. Can you think of a property of vector spaces that we can use here to give us a critical result?
     
  6. Mar 19, 2013 #5
    Assuming we are dealing in the realm of undergraduate linear algebra...

    Continuing on what WannabeNewton said, think about why [itex]\mathbb{R}[/itex] is a subspace of itself, and then consider some nonempty set that is not [itex]\mathbb{R}[/itex] or the set consisting of just [itex]0[/itex].

    Recall that a nonempty set is a subspace of another set if and only if it is a subset of it and it is closed under both addition and "scalar" multiplication.
     
  7. Mar 20, 2013 #6
    would i use contradiction
     
  8. Mar 20, 2013 #7
    Well, if we were to look at a subspace that is not in ℝ, it would not be closed under the same addition or multiplication that is in ℝ. And additionally, from the theorem "if a subset S of a vector space V does not contain the zero vector 0 of V, then S is not a subspace of V". From this, the 0 vector of the ℝ is the set 0 itself. THerefore, the subspace must be ℝ and {0}.

    DOes this seem right?
     
  9. Mar 20, 2013 #8

    micromass

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    Yes.

    So you have a nontrivial vector space ##V\subseteq \mathbb{R}##. You have a nonzero vector ##v##.

    You wish to prove somehow that ##V=\mathbb{R}##. So given any ##w\in \mathbb{R}##, you wish to prove that ##w\in V##.
     
  10. Mar 20, 2013 #9

    micromass

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    No, this is completely incorrect. I would wish I could tell you what was wrong, but I don't understand your proof at all.
     
  11. Mar 20, 2013 #10
    So to show that w is in the vector space V, would I assume that it is not. Then, go on to show that if it is not it is not a part of the real numbers. However, under the assumption I said w is apart of the real numebrs, therefore a contradiciton.
     
  12. Mar 20, 2013 #11

    WannabeNewton

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    You don't need contradiction here mate. Did you try what I said? Let me be a bit more direct: let ##V\subseteq \mathbb{R}## be a non zero subspace of ##\mathbb{R}## over the field ##\mathbb{R}## (This is crucial! I'm assuming your assignment was to prove this assuming the field was ##\mathbb{R}## correct?). Let ##v\in V## be non zero. Remember that ##v## is still technically a non zero real number. How can you use the closure of a vector space under scalar multiplication to help you finish the proof?
     
  13. Mar 20, 2013 #12
    Well, the closure property holds for v, because since v is a part of the reals, any real number times v will still be apart of the reals. Therefore, the only subspace of the reals are the reals and {0}?
     
  14. Mar 20, 2013 #13

    micromass

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    Could you expand more? Why does the "therefore" hold??

    For example, can you tell me why ##(-2,2)## is not a subspace of ##\mathbb{R}##?
     
  15. Mar 20, 2013 #14
    I'm confused as to why (-2,2) is not a subspace. I am lost.
     
  16. Mar 20, 2013 #15
    Is (-2,2) not a sunspace because all the scalers are of the natural numbers?
     
  17. Mar 20, 2013 #16

    Dick

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    You are talking gibberish. The scalars are all real numbers. Since 1 is in your 'sunspace' the k*1 should be in it for all real k. That's the scalar product. True or false?
     
  18. Mar 20, 2013 #17
  19. Mar 20, 2013 #18

    Dick

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    Explain why you think so. If k=3 then 3*1=3. 3 is not in (-2,2), is it?
     
  20. Mar 20, 2013 #19
    No 3 is not. The sunspace (-2,2) does not contain all of the real numbers of R. For example 3.
     
  21. Mar 20, 2013 #20

    Dick

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    I hope you are saying that (-2,2) is not a subspace, and I hope you know why. Can you explain why? Don't just say "it doesn't contain all real numbers". Give a good reason why a subspace that's not {0} would have to.
     
    Last edited: Mar 20, 2013
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