In mathematics, specifically in functional analysis and order theory, an ordered topological vector space, also called an ordered TVS, is a topological vector space (TVS) X that has a partial order ≤ making it into an ordered vector space whose positive cone ${\displaystyle C:=\left\{x\in X:x\geq 0\right\))$ is a closed subset of X.[1] Ordered TVSes have important applications in spectral theory.

## Normal cone

 Main article: Normal cone (functional analysis)

If C is a cone in a TVS X then C is normal if ${\displaystyle {\mathcal {U))=\left[{\mathcal {U))\right]_{C))$, where ${\displaystyle {\mathcal {U))}$ is the neighborhood filter at the origin, ${\displaystyle \left[{\mathcal {U))\right]_{C}=\left\{\left[U\right]:U\in {\mathcal {U))\right\))$, and ${\displaystyle [U]_{C}:=\left(U+C\right)\cap \left(U-C\right)}$ is the C-saturated hull of a subset U of X.[2]

If C is a cone in a TVS X (over the real or complex numbers), then the following are equivalent:[2]

1. C is a normal cone.
2. For every filter ${\displaystyle {\mathcal {F))}$ in X, if ${\displaystyle \lim {\mathcal {F))=0}$ then ${\displaystyle \lim \left[{\mathcal {F))\right]_{C}=0}$.
3. There exists a neighborhood base ${\displaystyle {\mathcal {B))}$ in X such that ${\displaystyle B\in {\mathcal {B))}$ implies ${\displaystyle \left[B\cap C\right]_{C}\subseteq B}$.

and if X is a vector space over the reals then also:[2]

1. There exists a neighborhood base at the origin consisting of convex, balanced, C-saturated sets.
2. There exists a generating family ${\displaystyle {\mathcal {P))}$ of semi-norms on X such that ${\displaystyle p(x)\leq p(x+y)}$ for all ${\displaystyle x,y\in C}$ and ${\displaystyle p\in {\mathcal {P))}$.

If the topology on X is locally convex then the closure of a normal cone is a normal cone.[2]

### Properties

If C is a normal cone in X and B is a bounded subset of X then ${\displaystyle \left[B\right]_{C))$ is bounded; in particular, every interval ${\displaystyle [a,b]}$ is bounded.[2] If X is Hausdorff then every normal cone in X is a proper cone.[2]

## Properties

• Let X be an ordered vector space over the reals that is finite-dimensional. Then the order of X is Archimedean if and only if the positive cone of X is closed for the unique topology under which X is a Hausdorff TVS.[1]
• Let X be an ordered vector space over the reals with positive cone C. Then the following are equivalent:[1]
1. the order of X is regular.
2. C is sequentially closed for some Hausdorff locally convex TVS topology on X and ${\displaystyle X^{+))$ distinguishes points in X
3. the order of X is Archimedean and C is normal for some Hausdorff locally convex TVS topology on X.