16.1 Ordered Vector Spaces
Definition 16.1.1 (Ordered Vector Space).label Let $E$ be a vector space over $K \in \RC$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a ordered vector space if
- (LO1)
For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
- (LO2)
For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
The set $C = \bracs{x \in E|x \ge 0}$ is the positive cone of $E$.
Definition 16.1.2 (Ordered Topological Vector Space).label Let $(E, \le)$ be an ordered vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then the triple $(E, \topo, \le)$ is an ordered topological vector space if the positive cone $C = \bracs{x \in E|x \ge 0}$ is closed.
Proposition 16.1.3.label Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then
- (1)
$\sup(A + B) = \sup(A) + \sup(B)$.
- (2)
$\sup(A) = -\inf (-A)$
Proof. (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
(2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$.$\square$
Definition 16.1.4 (Interval).label Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
is the order interval with endpoints $x$ and $y$.
Definition 16.1.5 (Order Bounded).label Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is order bounded if there exists $x, y \in E$ such that $A \subset [x, y]$.
Definition 16.1.6 (Order Complete).label Let $(E, \le)$ be an ordered vector space, then $E$ is order complete if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
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