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Thread: 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 $\displaystyle U$ and $\displaystyle V$ are both subspaces of the vector space $\displaystyle W$, then $\displaystyle U \cup V$ is not a vector space unless one of the subspaces $\displaystyle U$ or $\displaystyle V$ are contained in the other.

    $\displaystyle \text{\emph{Proof.}}$

    Let $\displaystyle U$ and $\displaystyle V$ be vector spaces. Clearly if either $\displaystyle U$ or $\displaystyle V$ is
    contained in the other, then $\displaystyle V \cup U$ 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 $\displaystyle U \cup V $ is a vector space and neither $\displaystyle U$ nor $\displaystyle V$ are contained
    within the other. Then obviously, since neither is a subset of the
    other, there exist elements $\displaystyle \mathbf{v} \in V$ and $\displaystyle \mathbf{u} \in U$ such that
    $\displaystyle \mathbf{v} \notin U$ and $\displaystyle \mathbf{u} \notin V$. Consider the vector $\displaystyle \mathbf{v} + \mathbf{u} $. Since we are assuming that $\displaystyle U \cup V $ is a vector space, $\displaystyle \mathbf{v}
    + \mathbf{u} \in U \cup V$ so that $\displaystyle \mathbf{v} + \mathbf{u} \in U $ or $\displaystyle \mathbf{v} + \mathbf{u} \in V $.
    If $\displaystyle \mathbf{v} + \mathbf{u} \in V $, then by closure $\displaystyle (\mathbf{v} + \mathbf{u}) + (-\mathbf{v})$ $\displaystyle = \mathbf{u}$
    $\displaystyle \in V$ since $\displaystyle (-\mathbf{v}) \in V$. This obviously contradicts that $\displaystyle \mathbf{u} $
    $\displaystyle \notin V$.A similar argument can be made if $\displaystyle \mathbf{v} + \mathbf{u} \in U$, so $\displaystyle U $
    $\displaystyle \cup V $ is not a vector space. Therefore, for $\displaystyle 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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