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Theoretical groups proof 1,1

  1. Oct 26, 2009 #1

    lom

    User Avatar

    V is a vectoric space.

    [tex]W_1,W_2\subseteq V\\[/tex]
    [tex]W_1\nsubseteq W_2\\[/tex]
    [tex]W_2\nsubseteq W_1\\[/tex]
    prove that [tex]W_1 \cup W_2[/tex] is not a vectoric subspace of V.



    i dont ave the shread of idea on how to tackle it

    i only know to prove that some stuff is subspace

    but constant mutiplication

    and by sum of two coppies



    this question here differs alot
     
    Last edited: Oct 26, 2009
  2. jcsd
  3. Oct 26, 2009 #2

    Mark44

    Staff: Mentor

    Differs a lot from what? Is there any other information given in your problem? For example, you have that
    [tex]W_1\nsubseteq W_2[/tex]
    and
    [tex]W_2\nsubseteq W_1[/tex]

    but are you given anything about
    [tex]W_1 \bigcap W_2[/tex]
    ?

    BTW, the term is "vector space" not vectoric space. No such word as vectoric.
     
  4. Oct 26, 2009 #3

    lom

    User Avatar

    it differs by its pure theoretic way

    i am used to prove that f(x)=0 is a subspace
    by the constant multiplication and sum of two copies law
    i only know that
    [tex]
    W_1,W_2\subseteq V\\
    [/tex]

    i have written all the given stuff
     
  5. Oct 26, 2009 #4
    Since you are only given conditions on two subspaces, the only thing you can do next is look at the individual elements. The two conditions [itex]W_1 \not\subseteq W_2[/itex] and [itex]W_2 \not\subseteq W_1[/itex] imply the existence of what elements in these sets?
     
  6. Oct 26, 2009 #5

    lom

    User Avatar

    W1 and W2 are foreign to each other
    there is no intersection between them
     
  7. Oct 26, 2009 #6
    Are you sure that there is no intersection between them? Let's take a step back. What is the actual condition that [itex]W_1 \not\subseteq W_2[/itex]?
     
  8. Oct 26, 2009 #7

    lom

    User Avatar

    its not only
    [itex]
    W_1 \not\subseteq W_2
    [/itex]
    its both
    [itex]
    W_1 \not\subseteq W_2
    [/itex]
    and [itex]
    W_2 \not\subseteq W_1
    [/itex]


    as for what you say:
    [itex]
    W_1 \not\subseteq W_2
    [/itex]
    means that all the members of W1 are not a part W2 group
     
  9. Oct 26, 2009 #8
    No, it does not mean that all the members of W1 are not in W2. It means that there exists a member of W1 that is not in W2. Does that make sense? If so, where can you go from there?
     
  10. Oct 26, 2009 #9

    Mark44

    Staff: Mentor

    You can't conclude that from these two statements:
    [tex]W_1\nsubseteq W_2\\[/tex]
    [tex]W_2\nsubseteq W_1\\[/tex]

    It's very possible that W1 contains some elements that are in W2, but other elements that aren't in W2. Same thing for the other statement. That's why I asked if you were given that these two sets are disjoint. You said you weren't given that information, and now here you're saying that they are.
     
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