# Is $W$ a subspace of $\mathbb{R}^3$?

I wanted to determine if $W$ is a subspace of $\mathbb{R}^3$:

$W=\{(x,y,z)\in\mathbb{R}^3:x\leq y\leq z\}$.

I believe that it is a subspace of $\mathbb{R}^3$ since I think it satisfies all the conditions (contains the zero vector, is closed under addition and scalar multiplication), is this assumption correct?

Thank You!

• It's not closed under scalar mutliplication.
– user296602
Jul 5, 2016 at 5:11
• Is it because this inequality limits you only to numbers less than or equal to zero, meaning that if you multiply it, it will surpass the given space? Jul 5, 2016 at 5:17
• You have to allow for multiplication by negative scalars as well as non-negative scalars. Jul 5, 2016 at 6:11

If $(x,y,z)\in\mathbb{R}^3$ satisfies the inequality, multiplying it by the scalar $-1\in\mathbb{R}$ will reverse the inequality.
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$ is an element of $W$ since $1\leq 2\leq 3$ (it satisfies $x\leq y\leq x$)
Is $-1\cdot \begin{bmatrix}1\\2\\3\end{bmatrix}$ an element of $W$?
Is it true that $-1\leq -2\leq -3$?