Subspace of $R^2$, $S = \{\begin{bmatrix}x\\y\end{bmatrix} : x \le y \}$

Question

I'm a bit confused at why the set $S$ in the header is not considered a subspace. It seems to fulfill all the requirements.

It's closed under scalar multiplication, since any scalar multiple will be applied to both $x$ and $y,$ and $x$ will remain $\le y.$

It's closed under addition for the same reason.

It contains the zero element, with $x = y = 0$ being a part of the set. Why is this not a subspace of $R^2$?

Answer 1

$$1 \le 2$$

$$-1 \ge -2$$

Do you see why it is not a subspace?