Lattice disjoint

In mathematics, specifically in order theory and functional analysis, two elements x and y of a vector lattice X are lattice disjoint or simply disjoint if ${\displaystyle \inf \left\{|x|,|y|\right\}=0}$, in which case we write ${\displaystyle x\perp y}$, where the absolute value of x is defined to be ${\displaystyle |x|:=\sup \left\{x,-x\right\}}$.^[1] We say that two sets A and B are lattice disjoint or disjoint if a and b are disjoint for all a in A and all b in B, in which case we write ${\displaystyle A\perp B}$.^[2] If A is the singleton set ${\displaystyle \{a\}}$ then we will write ${\displaystyle a\perp B}$ in place of ${\displaystyle \{a\}\perp B}$. For any set A, we define the disjoint complement to be the set ${\displaystyle A^{\perp }:=\left\{x\in X:x\perp A\right\}}$.^[2]

Two elements x and y are disjoint if and only if ${\displaystyle \sup\{|x|,|y|\}=|x|+|y|}$. If x and y are disjoint then ${\displaystyle |x+y|=|x|+|y|}$ and ${\displaystyle \left(x+y\right)^{+}=x^{+}+y^{+}}$, where for any element z, ${\displaystyle z^{+}:=\sup \left\{z,0\right\}}$ and ${\displaystyle z^{-}:=\sup \left\{-z,0\right\}}$.

Disjoint complements are always bands, but the converse is not true in general. If A is a subset of X such that ${\displaystyle x=\sup A}$ exists, and if B is a subset lattice in X that is disjoint from A, then B is a lattice disjoint from ${\displaystyle \{x\}}$.^[2]

For any x in X, let ${\displaystyle x^{+}:=\sup \left\{x,0\right\}}$ and ${\displaystyle x^{-}:=\sup \left\{-x,0\right\}}$, where note that both of these elements are ${\displaystyle \geq 0}$ and ${\displaystyle x=x^{+}-x^{-}}$ with ${\displaystyle |x|=x^{+}+x^{-}}$. Then ${\displaystyle x^{+}}$ and ${\displaystyle x^{-}}$ are disjoint, and ${\displaystyle x=x^{+}-x^{-}}$ is the unique representation of x as the difference of disjoint elements that are ${\displaystyle \geq 0}$.^[2] For all x and y in X, ${\displaystyle \left|x^{+}-y^{+}\right|\leq |x-y|}$ and ${\displaystyle x+y=\sup\{x,y\}+\inf\{x,y\}}$.^[2] If y ≥ 0 and x ≤ y then x⁺ ≤ y. Moreover, ${\displaystyle x\leq y}$ if and only if ${\displaystyle x^{+}\leq y^{+}}$ and ${\displaystyle x^{-}\leq x^{-1}}$.^[2]

• Solid set
• Locally convex vector lattice
• Vector lattice

• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 3. New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.