Results 1 to 5 of 5

Thread: Subspace of a Vector space: Why not the other way around?

  1. #1
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    308
    Thanks
    16

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

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


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

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

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

    2nd one:
    I've the same question regarding $\displaystyle C^1(-\infty, \infty)$ and $\displaystyle C^m(-\infty, \infty)$. Can we say $\displaystyle C^1(-\infty, \infty)$ is a subspace of $\displaystyle C^m(-\infty, \infty)$? If no then why not?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1
    Quote Originally Posted by x3bnm View Post
    My book says:
    The continuous functions $\displaystyle C(-\infty, \infty)$ on the interval $\displaystyle (-\infty, \infty)$ form a subspace of $\displaystyle F(-\infty, \infty)$
    since they are closed under addition and scalar multiplication. To take this step further, for each positive
    integer $\displaystyle m$, the functions with continuous $\displaystyle m$th derivatives on $\displaystyle (-\infty, \infty)$ form a subspace of $\displaystyle C^1(-\infty, \infty)$. So $\displaystyle C^m(-\infty, \infty)$ is a subspace of $\displaystyle C^1(-\infty, \infty)$


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

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

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

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

    Maybe you should define the symbols.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    308
    Thanks
    16
    Quote Originally Posted by Plato View Post
    Is the space $\displaystyle F(-\infty, \infty)$ the set of real function?
    If so isn't it true that $\displaystyle C(-\infty, \infty)\subset F(-\infty, \infty)$ and not the other way?

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

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

    But can we say(described in above post) $\displaystyle C^1(-\infty, \infty)$ be a subspace of $\displaystyle C^m(-\infty, \infty)$?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2010
    Posts
    193
    I think you're missing a major part of the definition of a subspace (namely, the SUB part).

    The definition is: $\displaystyle W$ is a subspace of $\displaystyle V$ means that both $\displaystyle V,W$ are vector spaces, AND THAT $\displaystyle W\subseteq V$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    308
    Thanks
    16
    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: $\displaystyle W$ is a subspace of $\displaystyle V$ means that both $\displaystyle V,W$ are vector spaces, AND THAT $\displaystyle W\subseteq V$.
    Thanks for explaining.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Subspace of a vector space
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 15th 2011, 09:57 AM
  2. Subspace of a vector space
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Dec 1st 2011, 01:16 PM
  3. Vector space and its subspace
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Nov 8th 2011, 04:01 AM
  4. vector space and subspace
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: Oct 11th 2011, 02:05 PM
  5. [SOLVED] Subspace of a vector space
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: May 2nd 2010, 08:07 AM

/mathhelpforum @mathhelpforum