Let $ V = \mathbb{R}^3 $, considering $ W = \{ (x,y,z) \in V: x \leq y \leq z \} $. Is W a subspace of V?
To proof that I have to demonstrate:
1) W is nonempty, that means $W \ne \emptyset$
2) if $ x \in W$ and $ y \in W$, then $x+y \in W$
3) if $\alpha \in \mathbb{C}$ and $x \in W$, then $\alpha x \in W$
I don't know how to proof that using the inequalities above.
How can I apply the three conditions?
If you are asking how you would show each of these, typically the way one shows a purported subspace is not empty is the show that (0, 0, 0) is in the sunset. Certainly it is true that $0\le 0\le 0$. To show that "if (a, b, c) and (x, y, z) are in the set then (a+ x, b+ y, c+ z) is in the set" argue that "if (a, b, c) is in the set then $a\le b\le c$ and if (x, y, z) is in the set then $x\le y\le z$. So $a+ x\le b+ x$ and $b+ x\le b+ y$. Therefore $a+ x\le b+ y$. Also $b+ y\le c+ y$ and $c+ y\le c+ z$ so $b+ y\le c+ z$. Together they say that $a+ x\le b+ y\le c+ z$ which shows that (a+ x, b+ y, c+ z) is in the set.
As you seem to recognize, it is the third one "if $\alpha\in C$ and $x\in W$ then $\alpha X\in W$" is NOT true. You also realize that is is not true for $\alpha$ negative.
To prove that general statement is NOT true, it is sufficient to give a "counter example"- a single example where the statement is not true. In this case, as "mathcounterexamples.net" says, taking $\alpha= -1$ is sufficient. With $\alpha= -1$ and $X= (1, 2, 3)$ so that $x\le y\le z$ then $\alpha X= -1(1, 2, 3)= (-1, -2, -3)$ but $-1\le -2\le -3$ is NOT true.