Given with
there must exist disjoint with
with and not both empty.
If , then and can be chosen such that also:
Proof
For each let be its indicator.
As , there are some not-all-zero such that
Thus define and .
Now the sets and must have the same support.
Consequently
Now suppose and let
(where is the indicator of the complement of )
Note that
so lie in some plane of dimension in
This plane cannot contain the origin,
so lie in a dimensional subspace of .
Thus find not-all-zero such that:
Pick and and conclude