Proving the Subspace Condition for R1: Is S Either {0} or R1?

[Mdhiggenz]

Homework Statement

Prove that if S is a subspace of R¹, then either S={0} or S=R¹.

Trying to come up with a proof I dissected each statement, I know that in order for S to be

a subspace the zero vector must lie within the subset. So I know S={0} is true. I then

checked an arbitary vector x₁ which lies on R¹ to make sure it

was closed under scalar multiplication, and addition, and that checked out as well.

Not sure if I am on the right track.

Thanks

Homework Equations

The Attempt at a Solution

 

[jbunniii]

You need to specify what field of scalars you are using. The statement is true if the field is $\mathbb{R}$, false if it is, for example, $\mathbb{Q}$.

Assuming you are using $\mathbb{R}$ for the field of scalars, use the fact that a subspace must be closed under scalar multiplication.

 

[Karnage1993]

Your ideas are right, but it's not really a proof unless you do everything step-by-step.
For example, for the first possible set: $S = \{0\}$, show that all three properties of a subspace are satisfied.

 

[jbunniii]

Your ideas are right, but it's not really a proof unless you do everything step-by-step.
For example, for the first possible set: $S = \{0\}$, show that all three properties of a subspace are satisfied.

Why is that necessary? The problem statement tells you that $S$ is a subspace. All you need to do is show that if it's not $\{0\}$ then it must be all of $\mathbb{R}^1$.

So: if $S$ is not $\{0\}$, then $S$ contains a nonzero element, say $s \neq 0$. Since $S$ is a subspace, it is closed under scalar multiplication. Thus it must contain $\alpha s$ for every $\alpha \in F$ where $F$ is the scalar field (presumably $\mathbb{R}$). Therefore...?

 

[Mdhiggenz]

why is that necessary? The problem statement tells you that $s$ is a subspace. All you need to do is show that if it's not $\{0\}$ then it must be all of $\mathbb{r}^1$.

So: If $s$ is not $\{0\}$, then $s$ contains a nonzero element, say $s \neq 0$. Since $s$ is a subspace, it is closed under scalar multiplication. Thus it must contain $\alpha s$ for every $\alpha \in f$ where $f$ is the scalar field (presumably $\mathbb{r}$). Therefore...?

s=r1?