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 $\inf \left\ = 0$, in which case we write $x \perp y$, where the absolute value of ''x'' is defined to be $, x, := \sup \left\$.
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 $\$ then we will write $a \perp B$ in place of $\ \perp B$.
For any set ''A'', we define the disjoint complement to be the set $A^ := \left\$.

Characterizations

Two elements ''x'' and ''y'' are disjoint if and only if $\sup\ = , x , + , y ,$.
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\$ and $z^ := \sup \left\$.

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 $\$.

Representation as a disjoint sum of positive elements

For any ''x'' in ''X'', let $x^ := \sup \left\$ and $x^ := \sup \left\$, 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'', $\left, x^ - y^ \ \leq , x - y ,$ and $x + y = \sup\ + \inf\$.
If ''y ≥ 0'' and ''x'' ≤ ''y'' then ''x''⁺ ≤ ''y''.
Moreover, $x \leq y$ if and only if $x^ \leq y^$ and $x^ \leq x^$.

See also

* Solid set
* Locally convex vector lattice
* Vector lattice

References

Sources

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{{Ordered topological vector spaces

Functional analysis