Axler Exercise 2.C.2: Subspaces of $\mathbb{R}^2$ This is a question of how to order the necessary elements of the proof and perhaps exactly what would suffice. This proof sounds like it requires several "biconditional" statements (e.g., $U$ is a one-dimensional subspace of $\mathbb{R}^2$ if and only if it is a line through the origin), but some of them are tougher to state.

Show that the subspaces of $\mathbb{R}^2$ are precisely $\{0\}$, $\mathbb{R}^2$, and all lines in $\mathbb{R}^2$ through the origin.

I'll show the sketch of the proof I have in mind.
If $U \subset \mathbb{R}^2$, then $\dim U \leq \mathbb{R}^2$, so $\dim U \leq 2$ since $(1,0)$, $(0,1)$ is a basis for $\mathbb{R}^2$.
If $U$ is zero-dimensional, then $U = \{0\}$. This is because the unique basis for a zero-dimensional subspace of $\mathbb{R}^2$ is the empty list, which is the unique list with no elements. Further, $\{0\}$ is vacuously a subspace: it contains the additive identity and is vacuously closed under addition and scalar multiplication.
(This sounds like a jump because I'm treating bases, in the style of Axler, as ordered lists instead of sets. It's an axiom of set theory that the empty set is the unique set with no elements, but I don't have any such axioms for lists, but it is surely the case that there is exactly one "empty list." It's so obvious I can't imagine how else I could prove it. I don't know if this can be reworded in terms of the language of set theory.)
If $U$ is a two-dimensional subspace of $\mathbb{R}^2$, then $U = \mathbb{R}^2$ by exercise 2.C.1. Indeed, let $u_1, u_2$ be a basis for $U$. Then $u_1, u_2$ is linearly independent in $U$ and hence in $\mathbb{R}^2$, so it can be extended to a basis of $\mathbb{R}^2$, but as every basis of $\mathbb{R}^2$ contains the same number of elements, this extension is trivial, adjoining no elements, so $u_1, u_2$ is also a basis for $\mathbb{R}^2$, so any $v \in \mathbb{R}^2$ can be written as $a_1 u_1 + a_2 u_2$ for $a_1, a_2 \in U$, and hence $v \in U$, so $U = \mathbb{R}^2$. So the only two-dimensional subspace of $\mathbb{R}^2$ is $\mathbb{R}^2$ itself. Of course, $\mathbb{R}^2$ admits the structure of a vector space and is hence a subspace of itself.
The one-dimensional subspace case is the hardest case. So let $U$ be a one-dimensional subspace of $\mathbb{R}^2$. Then there exists $v \neq 0$ in $U$. If there is no such $v$, then $U = \{0\}$, which is zero-dimensional. Then for any $a \in \mathbb{R}$, we have $av \in U$, so $U$ contains a line, $L$, through the origin containing $v$. So $U \supset L$. Furthermore, if $w \in L$, then $w = cv$ for some $s$, so $w \in U$ by closure of $U$ under scalar multiplication, so $L \subset U$. Therefore, $L = U$. Furthermore, any line $L$ is a subspace of $\mathbb{R}^2$. Any such line has the form $L := \{av \mid a \in \mathbb{R}\}$. Taking $a = 0$, $L$ contains the zero vector. If $av, a'v \in L$, then $av + a'v = (a + a')v \in L$. Finally, if $c \in \mathbb{R}$ and $av \in L$, then $c(av) = (ca)v \in L$. So $L$ is a subspace.

This proof is much more verbose than it should be, but it raises a few questions.

*

*Is there any further explanation needed for there be a single "unique" empty list? There's such an axiom for set theory, but Axler uses lists instead of sets to describe bases, spanning sets, and linearly independent sets.


*Is it correct that I not only need to show that every subspace of $\mathbb{R}^2$ has such a form (singleton set containing $0$, line through the origin, or all of $\mathbb{R}^2$) but also that those three things are actually subspaces? If I start with a subspace and show that it's equal to some set, does that already require that the set is a subspace itself? I'm a bit confused on the "biconditional" component of this proof.


*I feel like I repeated myself quite a lot when showing that $U = L$. I used in some sense the fact from geometry that "any two points determine a unique line," so I'm taking $L$ to be the unique line joining $0$ and $v$. I could also argue by contradiction and suppose that $U \supsetneqq L$; then there exists $u \in U \setminus L$, which $u, v$ are linearly independent, and any linearly independent list of two elements $\mathbb{R}^2$ must be a basis for $\mathbb{R}^2$, so $U$ has to be two-dimensional in that case, so if $U$ is one-dimensional, there is no such $u$. It doesn't seem necessary to go down this route, though.
 A: When you say that $U$ is $0$-dimensional, it means that every basis of $U$ has $0$ elements. This means that $\emptyset$ is a basis of $U$. So by definition of basis, $U = \text{span}(\emptyset) = \{0\}$. For defining bases, sets are adequate. You need a "list" when the order is important. Formally, a list of vectors in $V$ is an element of $V^N$, where $N$ is the length of the list. When $N = 0$, we have $V^0 = \{()\}$. So the only list of length $0$ is $()$.
Yes you need to show that every subspace of $\mathbb{R}^2$ is one of the mentioned forms and that every mentioned form is a subspace of $\mathbb{R}^2$.
You should not and do not need to use geometry here. Your proof only proved that $L \subset U$, so it is incomplete.
Your proof for $\dim(U) = 0$ and $\dim(U) = 2$ isn't too much more verbose than it needs to be.
Your proof that $\dim(U) = 1 \implies U = \mathbb{R}^2$ is too verbose. All you need to note is that if $\dim(U) = 1$, then $U = \text{span}(v)$, where $\{v\}$ is a basis for $U$. Since $\text{span}(v) = \{cv : c \in \mathbb{R}\}$ is, by definition of line, the line passing through the origin containing $v$, you are done.
