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Vector space

  1. Apr 30, 2010 #1
    If [tex]S\subseteq V[/tex] and V is a vector space, then S is a vector space.

    Assume S isn't a vector space. Since S isn't a vector space, then V isn't a vector space; however, V is a vector space. By contradiction, S is a subspace.

    Correct?
     
  2. jcsd
  3. Apr 30, 2010 #2
    Why is this true?
     
  4. Apr 30, 2010 #3

    Mark44

    Staff: Mentor

    Suppose V = R3 and S = {(x, y, z)| x + y - 2z = 3}. S is clearly a subset of V, but is S a subspace of V?
     
  5. Apr 30, 2010 #4
    I said assume S isn't a vector space.

    We have [tex]P\rightarrow Q\equiv P\and\sim Q[/tex]

    I don't know why the latex is looking messed up but that is supposed to say P and not Q is equivalent to P implies Q
     
  6. Apr 30, 2010 #5

    Mark44

    Staff: Mentor

    Which one is A? You started out with S and V.
     
  7. Apr 30, 2010 #6
    I meant S.
     
  8. Apr 30, 2010 #7

    Mark44

    Staff: Mentor

    P ==> Q <===> ~P ==> ~Q
    but you haven't shown that, given that S isn't a subspace, somehow this implies that V is not a subspace. In other words, if S is not a subspace, why does it necessarily follow that V is not a subspace? In fact, you are really concluding the opposite.

    I think you are mixing up a proof by the contrapositive with a proof by contradiction.

    See post #3.
     
  9. Apr 30, 2010 #8
    I was using de morgan's laws which says p implies q is equiv to ~p or q which is equiv to p and ~ q.

    That is why I assumed q (S isn't a vector space).
     
  10. Apr 30, 2010 #9

    Mark44

    Staff: Mentor

    Based on your original post, here are P and Q.
    P: [itex]S\subseteq V[/itex] and V is a vector space
    Q: S is a vector space

    Forget deMorgan - look at post #3.
     
  11. Apr 30, 2010 #10
    I don't know if this way change anything but it should be worded like so:

    If [tex]S\subseteq V[/tex] of a vector space V, then S is a vector space.
     
  12. Apr 30, 2010 #11
    Take Mark's counterexample (or some other counterexample you can think of), that is the best approach.
     
  13. Apr 30, 2010 #12
    The answer is supposed to be true so taking a counterexample seems counter-intuitive.
     
  14. Apr 30, 2010 #13
    S is defined as an improper subset of V; so if V is a vector space, S must be as well.
     
  15. Apr 30, 2010 #14
    For all we know, it could be a proper subset too. That was just the notation I used.
     
  16. Apr 30, 2010 #15
    Then it is false. There is one vector that S can exclude yet still retain it's subset status. Think of the definition of a vector space.
     
  17. Apr 30, 2010 #16
    Look at post #3, which has been said a few times now.
     
  18. Apr 30, 2010 #17
    The answer was true so I am not to sure about finding a counterexample as being the correct method.
     
  19. Apr 30, 2010 #18
    But a counterexample *does* exist. Therefore, the statement is false (assuming it was meant to be universally true). The statement is true if and only if S also satisfies the closure axioms of a vector space under the operations of addition and multiplication as defined by V (assuming S is a nonempty subset of V).
     
    Last edited: Apr 30, 2010
  20. Apr 30, 2010 #19

    Mark44

    Staff: Mentor

    Not true. Not all subsets of vector spaces are themselves subspaces. Here's another example, with V = R2. Let S = {(x, y} | y = 1}.
    [itex]S~\subseteq V[/itex], and V is a vector space, but S is not a subspace.
     
  21. May 1, 2010 #20
    I said S was an improper subset.
     
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