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Help with subspace

  1. Sep 5, 2004 #1
    I have been trying this problem for hours. I can't believe I can't get it. The question is "Find a subset U of R^2 such that U is closed under scalar multiplication but is not a subspace of R^2". I know that for U to be a subspace 0 must be an element of U and U has to be closed under scalar multiplication and vector addition. If U is closed under scalar multiplication then it must contain O vector right? So I have to think of a subset that is not closed under addition, but is closed under multiplication right? I can not think of any. Does anyone have an idea of one that satifies these properties?
     
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  3. Sep 5, 2004 #2

    Hurkyl

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    Can you think of any subsets of R^2 that aren't closed under addition?

    Study your examlpes of sets closed under multiplication. Can you identify any properties they have in common? Can you then try to come up with a new example without that property?
     
  4. Sep 5, 2004 #3
    I can think of subsets of R^2 that aren't closed under addition, but the fact that the set is closed under scalar multiplicaton always messes things up for me. For example {(x1,x2): x1,x2 don't=0}. I know that this isn't closed under addition because (x1,x2)+(-x1,-x2)=0 vector. But this also isn't closed under scalar multiplicaiton because 0 is an element of the field R, and 0(x1,x2)=(0,0). I have thought of examples that aren't closed under addition, but they always violate the fact that they must be closed under multiplication.
     
  5. Sep 5, 2004 #4

    Hurkyl

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    Hrm. Well, is there a way to take an arbitrary set and turn it into a set that is closed under multiplication?
     
  6. Sep 5, 2004 #5

    Fredrik

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    The set {(x,y)|xy=0}, satisfies the required properties. This is just the set you suggested plus the element (0,0).

    In order to be closed under scalar multiplication, the set has to be a union of straight lines passing through (0,0). The only proper subsets of that kind that are closed under addition are the ones that consist of only one such line.
     
    Last edited: Sep 5, 2004
  7. Sep 6, 2004 #6
    Thanks Fred. I knew it had to be something real simple that I was completely over looking.
     
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