If $V$ is a subspace of $\mathbb{R}^1$ and $\vec v \in V$ and is nonzero, how do I prove $\mathbb{R}^1$ is a subspace of $\text{span}({\vec v})$? Now, I know that I'm supposed to somehow show how it is that $\vec v$ spans $\mathbb{R}^1$, but this is precisely what I don't know how to do.
I know that if $\mathbb{R}^1$ is a subspace of $\text{span}({\vec v})$, then any $k\vec v$ will be able to attain any $\vec r$ in $\mathbb{R}^1$. But how exactly do I write this down?
The point of this proof is to eventually prove that any nonzero subspace of $\mathbb{R}^1$ has to be $\mathbb{R}^1$ itself. 
 A: If $v \neq 0$ then any $x \in \mathbb R$ can be written as $cv$ where $c=\frac x v$ so $x$ belongs to $span (v)$. 
A: I see that you understand the bigger picture, namely "any nonzero subspace of $\mathbb{R}^1$ has to be $\mathbb{R}^1$ itself". This means, of course, that $\vec{v}$ cannot be $0$.
So how should we express this? In short, we rely on the more general result that: 


*

*A subspace $U$ of a vector space $V$ (where $\dim V = n$) has dimension $d \in \{0,1,\dots,n\}$ .

*If $U$ is a subspace of a vector space $V$ and $\dim U = \dim V$, then $U = V$.


Understanding the first is as simple as seeing that any subspace $U$ is spanned by the basis of $V$, so there exists a way to reduce it down to a basis of $U$. This implies that any subspace of $\mathbb{R}^1$ either has dimension $0$ (the trivial subspace) or dimension $1$.
Now, we look at the second result. One can easily show that any linearly independent set of vectors of size $n$ in a space of dimension $n$ is spanning. Thus, consider the basis of the vector space $U$. It has length $n$ and is linearly independent, so it spans $V$ (and thus is a basis for $V$). Since the same list of vectors is a basis for $U$ and $V$, $U = V$. 
