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Math Help - union of vector space

  1. #1
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    union of vector space

    I could use some help with this problem,
    Suppose that V is a vector space over any field F and that U and W are subspaces of V.
    Show that the only time the union of U and W is a subspace of V is when either U is a subset of W or W is a subset of U.
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  2. #2
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    Suppose that neither is a subset of the other.
    This means a \in U\backslash V\quad \& \quad b \in V\backslash U.
    If U \cup V is a subspace then  a + b = c \in \left( {U \cup V} \right).
    But if c \in U then c - a \in U or b \in U.
    That is a contradiction. Similarly if c \in V we get a contradiction.
    Thus one of U or V is a subset of the other.

    Clearly, if one of U or V is a subset of the other then their union is a subspace.
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  3. #3
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    Hello math helpers...
    I cant understand exactly why that is a contradiction...maybe if you explained me with less formal language I could get it.
    Thank you, and sorry for annoying you all..
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  4. #4
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    Suppose that neither is a subset of the other.
    This means

    What is meat here is that there is an element 'a' which is in U but not in V
    Similarly there is an element 'b' which is in V but not in U
    This statement follows directly from the fact that U is not a subset of V and V is not a subset of U

    If is a subspace then

    both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.

    But if then or
    That is a contradiction. Similarly if we get a contradiction.
    Thus one of U or V is a subset of the other.

    As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!

    Same can be argued for V.

    Hence the result. Thanks
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  5. #5
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    Suppose that neither is a subset of the other.
    This means

    What is meant here is that there is an element 'a' which is in U but not in V
    Similarly there is an element 'b' which is in V but not in U
    This statement follows directly from the fact that U is not a subset of V and V is not a subset of U

    If is a subspace then

    both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.

    But if then or
    That is a contradiction. Similarly if we get a contradiction.
    Thus one of U or V is a subset of the other.

    As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!

    Same can be argued for V.

    Hence the result. Thanks
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  6. #6
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    Suppose that neither is a subset of the other.
    This means

    What is meant here is that there is an element 'a' which is in U but not in V
    Similarly there is an element 'b' which is in V but not in U
    This statement follows directly from the fact that U is not a subset of V and V is not a subset of U

    If is a subspace then

    both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.

    But if then or
    That is a contradiction. Similarly if we get a contradiction.
    Thus one of U or V is a subset of the other.

    As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!

    Same can be argued for V.

    Hence the result. Thanks
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  7. #7
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    sorry about the multiple posts above. browser got stuck and made multiple submits, i guess.
    @admin: if there is a way i can delete them plz do let me know. i cldn't find an option under edit. thanks

    Quote Originally Posted by aman_cc View Post
    Suppose that neither is a subset of the other.
    This means

    What is meant here is that there is an element 'a' which is in U but not in V
    Similarly there is an element 'b' which is in V but not in U
    This statement follows directly from the fact that U is not a subset of V and V is not a subset of U

    If is a subspace then

    both 'a' and 'b' (defined above) will belong to which we assume to be a sub-space. As subspace is closed under addition so a+b =c will also belong to the union of U and V.

    But if then or
    That is a contradiction. Similarly if we get a contradiction.
    Thus one of U or V is a subset of the other.

    As c belong to the union it has to belong to (at least) one of U,V. Let the chosen one be U. Now c and a both belong to U. Thus c-a=b will again below to U (under the closure and inverse axiom). But we always maintained b doesn't belong to U. Hence contradiction !!

    Same can be argued for V.

    Hence the result. Thanks
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  8. #8
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    I realize it has been a while, but I was thinking about the same problem trying to prove that it either has to be U is a subset of V or V is a subset of U. However, I am confused. Are a,b,c, vectors?
    Also, could you explain a little more about how you got to a+b=c and how that actually gives us the desired result.

    Thanks!
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  9. #9
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    a, b and c are vectors. If you assume that neither is a subspace of the other one, then there exist a and b as described above.

    Since all live in a vector space, you can add a and b, call the result c, so define c=a+b.
    Now we assume that the union of U and V is a subspace. Then since a,b \in (U \cup V), their sum (that is c) is also inside U \cup V.

    Clear?
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