I'm reading a paper on a Banach space with an order induced by a cone.
Let $(\mathbb{X}, || \cdot||)$ be a real Banach space.
We define a subset $P$ of $\mathbb{X}$ by $P := \{ x \in \mathbb{X} : x\geq 0\}$.
My question is what does $ x\geq 0$ mean, knowing that $\mathbb{X}$ is quite abstract and could be any real Banach space.
Given an open cone $K$ in a Banach space (i.e. a subset closed with respect to summation and multiplication by non negative factors $\lambda\geq 0$, and with zero intersection with it's opposite $-K\cap K = \{0\}$) you can define the order relation by $a>b$ iff $a-b\in K$.
In an analogous way, given a closed cone $C$ you can define the order relation by $a\geq b$ iff $a-b\in C$.
So if you assume the cone to be closed then you'll simply have $P=C$, otherwhise $P$ will be the closure of $K$.
To rephrase it to answer your question: $x\geq 0$ means for $x$ to be in the closure of the cone you use to define your partial order on $\mathbb{X}$.
Edit:
Note that we'd like to define a continuous partal order (i.e. such that if $a\geq b_\lambda$ for each element of a net $\{b_\lambda\}_{\lambda\in\Lambda}$ converging to $b$, then $a\geq b$), and in order to have this property, you need your defining cone to be closed.