Proof check: prove that any nonzero subspace of $\mathbb{R}^1$ is $\mathbb{R}^1$ itself. Let $U, V$ be nonzero subspaces of $\mathbb{R}^1$ s.t. $U \cup V = \mathbb{R}^1$ , and let $k\vec u$ be some vector in $U$ with $k \in \mathbb{R}$. But, $k$ is an arbitrary scalar, and $\vec u$ is an arbitrary nonzero one-dimensional vector, so by the definition of a vector space, $k\vec u \in V$. Then, $U$ and $V$ are the same, so $U \cup U = \mathbb{R}^1$, and therefore $U = \mathbb{R}^1$.
Would this work, or are there elements of rigour lacking? Thank you!
 A: *

*"Let $U, V$ be nonzero subspaces of $\mathbb{R}^1$ s.t. $U \cup V = \mathbb{R}^1$"

The title only speaks of a nonzero subspace of $\mathbb{R}^1$. It's a priori possible that $\mathbb{R}^1$ is not a union of two subspaces, but nevertheless is has a nonzero subspace. (For example the plane $\mathbb{R}^2$ is not the union of two subspaces, but still has nonzero subspaces.)
To start the proof correctly you have to translate the question more directly.


*"let $k\vec u$ be some vector in $U$ with $k \in \mathbb{R}$."

This is fine, but you should mention that $\vec{u}$ is some vector in $U$. The way you write this, makes it seem as though $k\vec{u}$ is some vector in $U$, which is not the intended definition, I think.


*"But, $k$ is an arbitrary scalar, and $\vec u$ is an arbitrary nonzero one-dimensional vector, so by the definition of a vector space, $k\vec u \in V$."

This doesn't logically follow. There's nothing in the definition of a vector space that says the scalar product of an arbitrary vector in a subspace needs to belong to another subspace.


*"Then, $U$ and $V$ are the same, so $U \cup U = \mathbb{R}^1$, and therefore $U = \mathbb{R}^1$."

This also doesn't logically follow. Noting that one vector has a scalar product that belongs to another vector space, is insufficient to conclude they are the same. You have to prove a stronger claim than that.
Hint for the correct proof:
Start by assuming there exists a subspace $0 \subsetneq U \subsetneq \mathbb{R}^1$. Erase $V$ entirely from the proof and from your mind.
