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Math Help - Convergent sequence implies subspace

  1. #1
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    Convergent sequence implies subspace

    Show that the set of convergent sequences is an infinite-dimensional subspace of the vector space of all sequences of real numbers.

    Sorry, but I'm really lost on this one, any hints? Thanks.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Show that the set of convergent sequences is an infinite-dimensional subspace of the vector space of all sequences of real numbers.

    Sorry, but I'm really lost on this one, any hints? Thanks.
    Show that the set \bold{C} of convergent sequences is an infinite-dimensional subspace of the vector space \bold{S} of all sequences of real numbers.

    To show it is a subspace you need to prove three things:

    1. The zero vector of \bold{S} is in \bold{C}
    2. \bold{C} is closed under addition
    3. for any scalar a in the underlying field of \bold{S} and \bold{u} \in \bold{C} that a \bold{u} \in \bold{C}

    In addition to these you will need to show that it is not of finite dimension.

    RonL

    RonL
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  3. #3
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    proof.

    Let S be the set of convergent sequences.

    Now, the sequence {0} converges to 0, implies {0} is in S.

    Assume the sequences of {x} and {y} be in S, with {x} converges to x and {y} converges to y.

    {x} + {y} = {x+y} converges to x + y, so it is in S.

    Let r be a scalar of real number, r{x} = {rx} converges to (r)x, so it is in S.

    Thus S is a subspace.

    Is this right?
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    proof.

    Let S be the set of convergent sequences.

    Now, the sequence {0} converges to 0, implies {0} is in S.

    Assume the sequences of {x} and {y} be in S, with {x} converges to x and {y} converges to y.

    {x} + {y} = {x+y} converges to x + y, so it is in S.

    Let r be a scalar of real number, r{x} = {rx} converges to (r)x, so it is in S.

    Thus S is a subspace.

    Is this right?
    Yes, you stillneed to show that it is not finite dimensional

    RonL
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  5. #5
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    I don't quite understand how to show it. A convergent sequence can of course contain infinite coordinates as long as they all converge, right?
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  6. #6
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    Quote Originally Posted by tttcomrader View Post
    I don't quite understand how to show it. A convergent sequence can of course contain infinite coordinates as long as they all converge, right?
    Let e_i be the sequence that is zero except for the i th term which is 1. These are obviously convergent and so in C

    Suppose the subspace were finite dimensional of dimension N.

    Obviously e_1, .. e_N are linearly independent and therefore constitute a basis for the subspace. But e_{N+1} cannot be written as a linear combination of them Contradicting our assumption that the subspace was of finite dimension.

    RonL
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  7. #7
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    I understand the proof until the e_{N+1} part. For exactly which rule does it violated in order to draw a contradiction?

    I'm sorry, I know I'm pretty weak in understanding this stuff here.

    Thanks!
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  8. #8
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    Quote Originally Posted by tttcomrader View Post
    I understand the proof until the e_{N+1} part. For exactly which rule does it violated in order to draw a contradiction?

    I'm sorry, I know I'm pretty weak in understanding this stuff here.

    Thanks!
    It depends on what you know about finite dimensional vector spaces.

    A space is of dimension N then any set of N linear independent vectors span the space and no smaller set will span the space. Because e_{N+1} is not in span(e_1, .., e_N) either it is not a convergent sequence (but it is) or span(e_1, .., e_N) \ne C, which contradics our assumption.

    RonL
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