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 $$\inf \left\{ |x|, |y| \right\} = 0$$, in which case we write $$x \perp y$$, where the absolute value of x is defined to be $$|x| := \sup \left\{ x, - x \right\}$$. 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 $$A \perp B$$. 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 $$A^{\perp} := \left\{ x \in X : x \perp A \right\}$$.

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 $$\left(x + y \right)^{+} = x^{+} + y^{+}$$, where for any element z, $$z^{+} := \sup \left\{ z, 0 \right\}$$ and $$z^{-} := \sup \left\{ -z, 0 \right\}$$.

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

Representation as a disjoint sum of positive elements
For any x in X, let $$x^{+} := \sup \left\{ x, 0 \right\}$$ and $$x^{-} := \sup \left\{ -x, 0 \right\}$$, where note that both of these elements are $$\geq 0$$ 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 $$\geq 0$$. For all x and y in X, $$\left| x^{+} - y^{+} \right| \leq | x - y |$$ and $$x + y = \sup\{ x, y \} + \inf\{ x, y \}$$. If y ≥ 0 and x ≤ y then x+ ≤ y. Moreover, $$x \leq y$$ if and only if $$x^{+} \leq y^{+}$$ and $$x^{-} \leq x^{-1}$$.