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

1. ## 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 $m$th 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?

2. Originally Posted by x3bnm
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 $m$th 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.

3. Originally Posted by Plato
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 $C^m(-\infty, \infty)$?

4. 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 $W\subseteq V$.

5. Originally Posted by topspin1617
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.