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Math Help - Subspace of a Vector space: Why not the other way around?

  1. #1
    Senior Member x3bnm's Avatar
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    Subspace of a Vector space: Why not the other way around?

    My book says:
    The continuous functions C(-\infty, \infty) on the interval (-\infty, \infty) form a subspace of F(-\infty, \infty)
    since they are closed under addition and scalar multiplication. To take this step further, for each positive
    integer m, the functions with continuous mth derivatives on (-\infty, \infty) form a subspace of C^1(-\infty, \infty). So C^m(-\infty, \infty) is a subspace of C^1(-\infty, \infty)


    By definition:
    W is a subspace of V if and only if W is closed under addition and closed under scalar multiplication.
    Now I've two questions.

    1st one:
    Why C(-\infty, \infty) is a subspace of F(-\infty, \infty) and not the other way around?

    I mean I understand because they are closed under addition and scalar multiplication. But so is F(-\infty, \infty). So can we say F(-\infty, \infty) is a subspace of C(-\infty, \infty)?

    2nd one:
    I've the same question regarding C^1(-\infty, \infty) and C^m(-\infty, \infty). Can we say C^1(-\infty, \infty) is a subspace of C^m(-\infty, \infty)? If no then why not?
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  2. #2
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    Quote Originally Posted by x3bnm View Post
    My book says:
    The continuous functions C(-\infty, \infty) on the interval (-\infty, \infty) form a subspace of F(-\infty, \infty)
    since they are closed under addition and scalar multiplication. To take this step further, for each positive
    integer m, the functions with continuous mth derivatives on (-\infty, \infty) form a subspace of C^1(-\infty, \infty). So C^m(-\infty, \infty) is a subspace of C^1(-\infty, \infty)


    By definition:
    W is a subspace of V if and only if W is closed under addition and closed under scalar multiplication.
    Now I've two questions.

    1st one:
    Why C(-\infty, \infty) is a subspace of F(-\infty, \infty) and not the other way around?

    I mean I understand because they are closed under addition and scalar multiplication. But so is F(-\infty, \infty). So can we say F(-\infty, \infty) is a subspace of C(-\infty, \infty)?

    2nd one:
    I've the same question regarding C^1(-\infty, \infty) and C^m(-\infty, \infty). Can we say C^1(-\infty, \infty) is a subspace of C^m(-\infty, \infty)? If no then why not?
    Is the space F(-\infty, \infty) the set of real function?
    If so isn't it true that C(-\infty, \infty)\subset F(-\infty, \infty) and not the other way?

    Maybe you should define the symbols.
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  3. #3
    Senior Member x3bnm's Avatar
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    Quote Originally Posted by Plato View Post
    Is the space F(-\infty, \infty) the set of real function?
    If so isn't it true that C(-\infty, \infty)\subset F(-\infty, \infty) and not the other way?

    Maybe you should define the symbols.
    Yes F(-\infty, \infty) is a set of real function. I understand what you're saying.

    C^1(-\infty, \infty) = set of continuous first derivative
    C^m(-\infty, \infty) = set of continuous m derivatives

    But can we say(described in above post) C^1(-\infty, \infty) be a subspace of C^m(-\infty, \infty)?
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    I think you're missing a major part of the definition of a subspace (namely, the SUB part).

    The definition is: W is a subspace of V means that both V,W are vector spaces, AND THAT W\subseteq V.
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  5. #5
    Senior Member x3bnm's Avatar
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    Quote Originally Posted by topspin1617 View Post
    I think you're missing a major part of the definition of a subspace (namely, the SUB part).

    The definition is: W is a subspace of V means that both V,W are vector spaces, AND THAT W\subseteq V.
    Thanks for explaining.
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