Hoffman and Kunze Subspace question So I'm trying to do this exercise that is asking to prove that $\Bbb R^1$ only has the subspaces 0 and itself. I understand how both of these are subspace and visually I can see how they are the only subspaces but I can't come with an idea on how to show that these are the ONLY subspaces. Any hints or ideas? Also the only theorems provided on this section is that the intersection of two subspaces is a subspace, a subspace is closed is under multiplication and addition, and the span is the set of all linear combination. So please no suggestions using dimension theorems and stuff like that. 
 A: Since this exercise appears before the definition of basis in Hoffman & Kunze, one is expected to use only the definition of a subspace and the related theorems to complete the exercise. In particular, Theorem 3 (on page 37) is relevant:

Theorem 3. The subspace spanned by a non-empty subset $\mathrm{S}$ of a vector space $\mathrm{V}$ is the set of all linear combinations of vectors in $\mathrm{S}$.

So, let $V$ be a subspace of $R^1$. If $V$ is the zero subspace, then there is nothing to show. So, assume that $V$ is not the zero subspace. Then, there is a nonzero vector $v \in V$. Consider the subspace $W$ spanned by the singleton set $\{ v \}$. By Theorem 3 above, $W = \{ c v : c \in R^1 \}$. But, every $r \in R^1$ is an element of $W$ because $r = (r/v) \cdot v$, which makes sense since $v \neq 0$. Thus, $R^1 \subseteq W \subseteq V \subseteq R^1$. Hence, $V = R^1$.
Thus, every subspace of $R^1$ is either the zero subspace or $R^1$.
A: See $\Bbb R$ is $1$ dimensional. Now we know any subspace of a vector space has a basis, so does $\Bbb R$. Now being $1$ dimensional any subspace will have dimension $0$ or $1$. Having 
dimension $0$ the subspace is $0$ & having dimension $1$ the subspace is $Sp \{c\}=\{cx| x\in \Bbb R\}=\Bbb R$. 
