Characterization of Preordered Vector Spaces

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a vector space over $\GF$.

Let $\succeq$ be a preordering on $X$.


Then $\struct {X, \succeq}$ is a preordered vector space if and only if there exists a convex cone $P \subseteq X$ such that $\succeq$ is the preordering on $X$ induced by $P$.


Proof

Necessary Condition

Suppose that $\struct {X, \succeq}$ is a preordered vector space.

Let:

$P = \set {x \in X : x \succeq 0}$

Let $\succeq^P$ be the preordering on $X$ induced by $P$.

We want to show that, for $x, y \in X$, we have $x \succeq y$ if and only if $x \succeq^P y$.

Let $x, y \in X$.

By the definition of the preordering on $X$ induced by $P$, we have:

$x \succeq^P y$ if and only if $x - y \in P$

By the definition of $P$:

$x - y \in P$ if and only if $x - y \succeq 0$

Since $\struct {X, \succeq}$ is a preordered vector space, we obtain:

$x - y \succeq 0$ if and only if $x \succeq y$

Hence $\succeq$ is the preordering on $X$ induced by $P$.

$\Box$

Sufficient Condition

Let $P \subseteq X$ be a convex cone.

Let $\succeq^P$ be the preordering on $X$ induced by $P$.

We show that $\struct {X, \succeq^P}$ is a preordered vector space.

We need to establish:

$(1) \quad$ for all $x, x', y, y' \in X$ such that $x \succeq x'$ and $y \succeq y'$, we have $x + y \succeq^P x' + y'$
$(2) \quad$ for all $\alpha \in \R_{\ge 0}$ and $x, x' \in X$ with $x \succeq^P x'$, we have $\alpha x \succeq^P \alpha x'$.

Let $x, y, x', y' \in X$ be such that $x \succeq^P x'$ and $y \succeq^P y'$.

Then $x - x' \in P$ and $y - y' \in P$.

From the definition of a convex cone, we have $\paren {x - x'} + \paren {y - y'} \in P$.

That is, $\paren {x + y} - \paren {x' + y'} \in P$.

So we have $x + y \succeq^P x' + y'$, establishing $(1)$.

Now let $\alpha \in \R_{\ge 0}$ and $x, x' \in X$ be such that $x \succeq^P x'$.

Then we have $x - x' \in P$.

Since $P$ is a cone and $\alpha \in \R_{\ge 0}$, we have $\alpha \paren {x - x'} \in P$.

That is, $\alpha x - \alpha x' \in P$.

So we have $\alpha x \succeq^P \alpha x'$.

This establishes $(2)$ and hence shows that $\struct {X, \succeq^P}$ is a preordered vector space.

$\blacksquare$


Sources

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