A question about ordered vector spaces

Question

Relevant Definitions:

Definition:
Let $X$ be a set. A partial order $\leq$ on $X$ is a relation that is reflexive, anti-symmetric, and transitive.

Definition:
Let $\leq$ be partial order on a real vector space $E.$ The pair $(E, \leq)$ is said to be an ordered vector space (OVS) if:\

1. For all $f,g,h \in E$ one has, $f \leq g \implies f+h \leq g+h$
2. For all $f,g \in E$ with $f \leq g$ and for all $\alpha \in \mathbb{R}, \alpha \geq 0$ one has $f \leq g \implies \alpha f \leq \alpha g.$

Definition
Let $E$ be a real vector space. $C \subset E, C \neq \emptyset$ is said to be a convex cone with vertex at the origin if

1. For all $x,y \in C$ we have $x+y \in C$
2. For all $x \in C, \lambda \geq 0$ we have $\lambda x \in C.$

Definition: Let $E$ be a real vector space. Let $C \subset E$ be a convex cone with its vertex at the origin. $C$ is said to be a strict cone if:

$$\forall \;\; x \in C \setminus \{0\} \;\;\text{one has} -x \notin C.$$

Definition: Let $(E,\leq)$ be a real ordered vector space. Define

$$E_+=\{x \in E: 0 \leq x\}$$

The set $E_+$ is called the cone of positive elements.

Proposition: $E_+$ is a strict cone.

Proof: First let $x,y \in E_+$. Then $0 \leq x$ and $0 \leq y$ so that

$$0\leq y = 0+y \leq x+y$$

by the first property of an OVS applied to the inequality $0 \leq x$ (adding $y$ to both sides). Now let $x$ be in $C$ and let $\lambda$ be a non-negative real number. Then by the second property of an OVS we have

$$0 \leq x \implies 0=\lambda 0 \leq \lambda x,$$

which proves that $E_+$ is a cone with its vertex at the origin. To see that $E_+$ is strict, let $x \in C \setminus \{0\}.$ Then $x \geq 0 \implies -x \leq 0$ by property 1) of OVS.

My Question: What I want to argue here is that since $-x \neq 0$ we must have that $-x <0$ which completes the proof. However, I do not see why this is evident from the definition of a partial order. My understanding is that there are relationships between what I think are termed strict vs non-strict partial orders, and it may be that one inducing the other is required here. Is there a way to establish the desired result using only the definitions above or am I perhaps misunderstanding what the text I am working with means by partial order. For reference, the text I am working with calls $\leq$ an order and then makes statements of the form, if $\leq$ is total then... so I assumed they meant partial order. The text I am working makes no distinction between $\leq$ and $<$ or $=$. Thank you in advanced.

Answer 1

You must use antisymmetry of the order $\leq$. Namely, you're assuming that $x\in C\smallsetminus \{0\}$, and you want to show that $-x\in C$. Assume by contradiction that $-x\in C$, so that $-x \geq 0$. By adding $x$ to both sides, it follows that $x \leq 0$. But $x\in C$ means that $x\geq 0$. Both conditions $x\geq 0$ and $x\leq 0$ together imply that $x=0$ (this is antisymmetry of $\leq$), which is a contradiction.

Answer 2

Your answer lies in the definition of $<$, which is not a new relation really, but a conjunction of the $\neq$ and $\leq$ relations.

So by definition, if $v\leq w$ and $v\neq w$ then $v<w$.

You might think that its not sufficient to define it this way, but for partial orders this definition satisfies an important property:

$$\forall_x,y[x<y \land x\geq y \implies \perp]$$

The proof for this is simple:

Assume $x<y \land x\geq y$ then

$$\implies x\neq y \land x\leq y \land x\geq y$$

$$\implies x\neq y \land x=y $$

$$\implies \perp $$

Where the second to last step was thanks to the anti-symmetry of a partial order.