# Which of the following subsets of $\mathbb{R}^2$ exist?

(a) A line that is not a subspace.

(b) A subspace that is not a line.

(c) A subset that is not a subspace but is the union of two subspaces.

(d) A subset that is not a subspace but is the intersection of two subspaces.

Actually, I'm asked to graph these or prove they don't exist, but I'm simply looking for the intuition on how to come up with a solution. I know (d) doesn't exist, because every intersection of two subspaces is itself a subspace. And for (b), we can just take all of $\mathbb{R}^2$, as every subspace is a subset of itself. But I don't know what to do with (a) and (c). It seems an axiomatic approach would be best. For example, with (a), if we consider the line $L:y=1$, then we would need to show that $L$ does not meet at least one requirement for being a subspace (nonempty, closed under vector addition/scalar multiplication) in order for this to be an example. However I'm not sure how to even go about checking closure under vector addition for the equation of a line. Also, intuitively, a line in $\mathbb{R}^2$ should be a subspace.

• I think if you realize that (0,0) is an element of every subspace, it should help your intuition – Mark Jan 7 '18 at 0:13
• Oh, that explains that (a) doesn't exist. I used that fact in my proof of (d), but didn't realize it helped with (a) as well. Now for (c)... – Atsina Jan 7 '18 at 0:14
• For (a), consider any line not passing through the origin. For (c), consider the union of any two distinct lines through the origin. – Xander Henderson Jan 7 '18 at 0:15
• For c), a subspace is necessarily convex. Meaning if two points are in the subspace, the line segment between them is also there. This is because a line segment is expressible as a linear combination of your two endpoints. – Mark Jan 7 '18 at 0:15

a. An example for this is any line that does not pass through the origin $(0,0)$.

b. For this notice that $v\in A$, where $A$ is a subspace implies $cv\in A$ for all $c\in\Bbb{R}$.

c. Take any two distinct straight lines passing through the origin, say, $y=kx$ and $y=mx$ with $k\neq m$, then $t(x,kx)+(1-t)(x,mx)$ does not lie on any of these straight lines for $t\neq0,1$ and $x\neq 0$. So their union is not a subspace.

d. As you have already said, the intersection of two subspaces has to be a subspace.

(a) A line that is not a subspace: exists, a line not through the origin.

(b) A subspace that is not a line: exists, a plane.

(c) A subset that is not a subspace but is the union of two subspaces: exists, two (distinct) lines (through the origin).

(d) A subset that is not a subspace but is the intersection of two subspaces: does not exists.

• Certainly for (c) it needs to be two lines that pass through the origin. Otherwise they are not subspaces themselves. But why is the union of two lines not a subspace? – Atsina Jan 7 '18 at 0:22
• the union of two (distinct) lines through the origin do not satisfy the sub set properties, eg (1,0)+(0,1)=(1,1) – gimusi Jan 7 '18 at 0:24
• If we assume subspace must be a proper subset the R^2 is a plane and has no subset that is a plane. – fleablood Jan 7 '18 at 0:29
• @fleablood sorry I don't get your point – gimusi Jan 7 '18 at 0:44
• A plane can not be a proper subset of $\mathbb R^2$. (But the trivial space of just the origin point can be). – fleablood Jan 7 '18 at 0:48

A subspace of $\mathbb{R}^2$ is a vector space, so it must have a basis. A maximal basis of $\mathbb{R}^2$ has 2 elements. So all subspaces of $\mathbb{R}^2$ are either 0, 1, or 2 dimensional.

There is one 0 dimensional subspace: The origin.

Each 1 dimensional subspace is spanned by a particular unit vector: and any vector spans a line through the origin.

A 2 dimensional subspace of $\mathbb{R}^2$ is just $\mathbb{R}^2$ itself.

For (a), the axiomatic approach implies that, if A is a (nonempty) subspace, then $(0,0)$ must be in $A$ (since, if $x\in A$, than $\lambda x \in A$, and with $\lambda=0$ and the definitions of multiplication by a scalar, it comes $0.x=(0,0)\in A$). So, each line $L$ such that $(0,0)\notin L$, will not be a subset.

For (c), take two independent vectors $x$ and $y$, then $C=\mathrm{vect}(x)\cup \mathrm{vect}(y)$ is not a subset and yet, union of two subsets. (if $C$ were a subset, since $x,y\in C$, we would have $\mathrm{vect}(x,y)\in C$, but since $x$ and $y$ are independent, $\mathrm{vect}(x,y)=\mathbb{R}^2$, which is contradictory).