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Math Help - Vector Space proof

  1. #1
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    Vector Space proof

    Hey,

    Can you check to see if this problem is correct? Thank you!

    Prove that if, If U and V are both subspaces of the vector space W, then U \cup V is not a vector space unless one of the subspaces U or V are contained in the other.

    \text{\emph{Proof.}}

    Let U and V be vector spaces. Clearly if either U or V is
    contained in the other, then V \cup U is a vector space because
    their union would simply be the space that the other one in
    contained in. Now, suppose for the sake of contradiction that  U \cup V is a vector space and neither U nor V are contained
    within the other. Then obviously, since neither is a subset of the
    other, there exist elements \mathbf{v} \in V and \mathbf{u} \in U such that
    \mathbf{v} \notin U and \mathbf{u} \notin V. Consider the vector  \mathbf{v} + \mathbf{u} . Since we are assuming that  U \cup V is a vector space,  \mathbf{v}<br />
+ \mathbf{u} \in U \cup V so that  \mathbf{v} + \mathbf{u} \in U or  \mathbf{v} + \mathbf{u} \in V .
    If  \mathbf{v} + \mathbf{u} \in V , then by closure (\mathbf{v} + \mathbf{u}) + (-\mathbf{v})  = \mathbf{u}
    \in V since (-\mathbf{v}) \in V. This obviously contradicts that \mathbf{u}
    \notin V.A similar argument can be made if  \mathbf{v} + \mathbf{u} \in U, so  U
     \cup V is not a vector space. Therefore, for  U \cup V to be a
    vector space, one of the spaces must be contained within the other.
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  2. #2
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    looks good to me.
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