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 $\inf \{|x|,|y|\}=0$, in which case we write $x\perp y$, where the absolute value of x is defined to be $|x|:=\sup \{x,-x\}$.Template:Sfn 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 $A\perp B$.Template:Sfn If A is the singleton set $\{a\}$ then we will write $a\perp B$ in place of $\{a\}\perp B$. For any set A, we define the disjoint complement to be the set $A^{\perp }:=\{x\in X:x\perp A\}$.Template:Sfn

Characterizations

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

Properties

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

Representation as a disjoint sum of positive elements

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

See also

• Solid set
• Locally convex vector lattice
• Vector lattice

Sources

• Template:Cite book