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]

Characterizations

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 ${\displaystyle z^{-}:=\sup \left\{-z,0\right\}}$.

Properties

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]

See also

• Solid set
• Locally convex vector lattice
• Vector lattice

References

Sources

• 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.