As we already mentioned, Lemma 2.4 and its analogue for vertex-infection provide two examples of inclusion-preserving shade maps Shade. An example of an inclusion-reversing shade map comes from the theory of posets:
Example 4.9. Let E be a poset. For any F⊆E, we define F↓ ={e∈ E | there exists an f ∈ Fwith e < f} and
ShadeF =E\F↓.
Then, this map Shade : P(E) → P(E) is an inclusion-reversing shade map.
Proof of Example 4.9 (sketched). We shall prove the following three statements:
Statement 0: If A and B are two subsets of E such that A ⊆ B, then ShadeB ⊆ShadeA.
Statement 1:IfF ∈ P(E)andu∈ E\ShadeF, then Shade(F∪ {u}) = ShadeF.
Statement 2:IfF ∈ P(E)andu∈ E\ShadeF, then Shade(F\ {u}) = ShadeF.
[Proof of Statement 0: Let A and B be two subsets of E such that A ⊆ B. We must show that ShadeB⊆ShadeA.
Indeed, A ⊆ B entails A↓ ⊆ B↓ (by the definitions of A↓ and B↓). Hence, E\A↓ ⊇ E\B↓, so that E\B↓ ⊆ E\ A↓. But the definition of Shade yields ShadeA = E\A↓ and ShadeB = E\B↓. Hence, ShadeB = E\B↓ ⊆ E\A↓ = ShadeA. This proves Statement 0.]
[Proof of Statement 1: Let F ∈ P(E) and u∈ E\ShadeF. We must prove that Shade(F∪ {u}) =ShadeF.
We haveu∈ E\ShadeF= F↓ (since ShadeF =E\F↓). In other words, there exists a g∈ F with u< g(by the definition of F↓). Consider this g.
We must prove that Shade(F∪ {u}) = ShadeF. In other words, we must prove that(F∪ {u})↓ = F↓(since Shade(F∪ {u}) = E\(F∪ {u})↓and ShadeF= E\ F↓). Since F↓ ⊆ (F∪ {u})↓ is rather obvious (by the definitions of F↓ and (F∪ {u})↓), we only need to prove that (F∪ {u})↓ ⊆F↓.
So letv ∈ (F∪ {u})↓. We must show thatv ∈ F↓.
We have v ∈ (F∪ {u})↓. In other words, there exists an f ∈ F∪ {u} with v < f (by the definition of (F∪ {u})↓). Consider this f. If f ∈ F, then we immediately obtain v ∈ F↓ (since v < f and f ∈ F), and thus our proof is complete. Hence, we WLOG assume that f ∈/ F. Hence, f = u (since f ∈ F∪ {u} but f ∈/ F) and thus v < f = u < g. This entails v ∈ F↓ (by the definition of F↓, since g∈ F).
Forget that we fixed v. We thus have shown that v ∈ F↓ for each v ∈ (F∪ {u})↓. In other words, (F∪ {u})↓ ⊆ F↓. As we explained above, this completes the proof of Statement 1.]
[Proof of Statement 2: Let F ∈ P(E) and u∈ E\ShadeF. We must prove that Shade(F\ {u}) =ShadeF.
We haveu∈ E\ShadeF= F↓ (since ShadeF =E\F↓). In other words, there exists a g ∈ F with u < g (by the definition of F↓). Consider this g. Note that u< gentails g6=u, so that g∈ F\ {u} (since g∈ F).
We must prove that Shade(F\ {u}) = ShadeF. In other words, we must prove that(F\ {u})↓ = F↓(since Shade(F\ {u}) = E\(F\ {u})↓and ShadeF = E\F↓). Since (F\ {u})↓ ⊆ F↓ is rather obvious (by the definitions of F↓ and (F\ {u})↓), we only need to prove that F↓ ⊆(F\ {u})↓.
So letv ∈ F↓. We must show thatv∈ (F\ {u})↓.
We have v ∈ F↓. In other words, there exists an f ∈ F with v < f (by the definition of F↓). Consider this f. If f ∈ F\ {u}, then we immediately obtain v ∈ (F\ {u})↓ (since v < f and f ∈ F\ {u}), and thus our proof is complete.
Hence, we WLOG assume that f ∈/ F\ {u}. Hence, f = u (since f ∈ F but f ∈/ F\ {u}) and thus v < f = u < g. This entails v ∈ (F\ {u})↓ (by the definition of(F\ {u})↓, since g∈ F\ {u}).
Forget that we fixed v. We thus have shown that v ∈ (F\ {u})↓ for each v∈ F↓. In other words, F↓ ⊆(F\ {u})↓. As we explained above, this completes the proof of Statement 2.]
Now, we have proved Statements 1 and 2. Thus, the map Shade : P(E) → P(E) satisfies the two axioms in Definition 4.2. In other words, this map is a shade map. Moreover, this map is inclusion-reversing (by Statement 0). This proves Example 4.9.
Another example of a shade map comes from discrete geometry:
Example 4.10. Let Abe an affine space overR. IfSis a finite subset of A, then anontrivial convex combinationofSwill mean a point of the form ∑
s∈S
λss∈ A, where the coefficientsλs are nonnegative reals smaller than 1 and satisfying
s∑∈S
λs =1.
Fix a finite subset Eof A. For any F ⊆E, we define
ShadeF={e ∈ E | e isnota nontrivial convex combination ofF}. Then, this map Shade : P(E) → P(E) is an inclusion-reversing shade map.
Proof of Example 4.10 (sketched). We shall prove the following three statements:
Statement 0: If X and Y are two subsets of E such that X ⊆ Y, then ShadeY⊆ShadeX.
Statement 1:IfF ∈ P(E)andu∈ E\ShadeF, then Shade(F∪ {u}) = ShadeF.
Statement 2:IfF ∈ P(E)andu∈ E\ShadeF, then Shade(F\ {u}) = ShadeF.
[Proof of Statement 0: Let X and Y be two subsets of E such that X ⊆ Y. We must prove that ShadeY⊆ShadeX.
Indeed, we have X ⊆ Y. Thus, any nontrivial convex combination of X is a nontrivial convex combination ofY. Therefore, anye∈ Ethat isnota nontrivial convex combination ofY also cannot be a nontrivial convex combination of X.
In other words, ShadeY ⊆ ShadeX (by the definition of Shade). This proves Statement 0.]
Assume the contrary. Thus, pis a nontrivial convex combination of F∪ {u}. In other words, p = ∑
s∈F∪{u}
µss for some nonnegative reals µs smaller than 1 and satisfying ∑
s∈F∪{u}
µs =1. Consider these reals µs.
We haveu ∈ E\ShadeF. Hence, u∈ Eand u ∈/ ShadeF. From u∈/ ShadeF, we see that u is a nontrivial convex combination of F. In other words, u =
f∑∈F
λff for some nonnegative realsλf smaller than 1 and satisfying ∑
f∈F Clearly, the coefficients µf +µuλf on the right hand side of this equality are nonnegative reals. Moreover, they are easily seen to be smaller than 1 13.
13Proof. Assume the contrary. Thus, µf +µuλf > 1 for some f ∈ F. Consider this f. Note that µf < 1 (since all coefficients µs are smaller than 1) and λf < 1 (for similar reasons).
Now,µf +µuλf >1 >µf (sinceµf <1), so thatµuλf >0. This entailsµuλf 6=0 and thus
Finally, they satisfy
Hence, (10) shows that p is a nontrivial convex combination of F. In other words, p ∈/ ShadeF. This contradicts p ∈ ShadeF. This contradiction shows that our assumption was false. This completes the proof of Statement 1.]
[Proof of Statement 2: Let F ∈ P(E) and u∈ E\ShadeF. We must prove that Shade(F\ {u}) =ShadeF. If u ∈/ F, then this is obvious (since F\ {u} = F in this case); thus, we WLOG assume thatu ∈ F.
We haveu ∈ E\ShadeF. Hence, u∈ Eand u ∈/ ShadeF. From u∈/ ShadeF, we see that u is a nontrivial convex combination of F. In other words, u =
f∑∈F
λff for some nonnegative realsλf smaller than 1 and satisfying ∑
f∈F
1−λu is a well-defined positive real.
We have
on the right hand side of this equality are non-negative reals (since 1−λu > 0). Moreover, they are easily seen to be smaller
Since all theµs are nonnegative, this subsum thus must be 6 ∑
s∈F∪{u} µf+µuλf >1. This contradiction shows that our assumption was wrong, qed.
than 1 14. Finally, they satisfy
14Proof. Assume the contrary. Thus, λf
1−λu >1 for some f ∈ F\ {u}. Consider this f. From λf
1−λu >1, we obtainλf >1−λu (since 1−λu >0), thusλf +λu>1.
However, f 6= u (since f ∈ F\ {u}), so that λf +λu is a subsum of the sum ∑
g∈Fλg. Since all the λg are nonnegative, this subsum thus must be 6 ∑
g∈F
g∈Fλg (since all theλg are nonnegative) and therefore λf +λu+λh 6 ∑
g∈F
λg = 1 = λf +λu, so thatλh 6 0 and thereforeλh = 0 (sinceλhis a nonnegative real). Thus, we have shown that all elementsh∈ Fexcept for (possibly) f and usatisfyλh=0. Thus, we had to rename the summation index f ashsince f means something else now.) Thus,
u=
∑
h∈F
λhh=λff+λuu,
so that u−λuu =λff. In view ofu−λuu= (1−λu)uand λf =1−λu, this rewrites as (1−λu)u = (1−λu)f. We can cancel 1−λu from this equality (since 1−λu > 0), and thus obtainu = f. This contradicts f 6= u. This contradiction shows that our assumption was wrong, qed.
Shade(F\ {u}). Therefore,
Shade(F\ {u}) = Shade
(F\ {u})∪ {u}
| {z }
=F (sinceu∈F)
=ShadeF.
This proves Statement 2.]
Now, we have proved Statements 1 and 2. Thus, the map Shade : P(E) → P(E) satisfies the two axioms in Definition 4.2. In other words, this map is a shade map. Moreover, this map is inclusion-reversing (by Statement 0). This proves Example 4.10.
As a contrast to Example 4.10, let us mention a not-quite-example (satisfying only one of the two axioms in Theorem 4.3):
Example 4.11. LetVbe a vector space overR. IfSis a finite subset ofV, then anontrivial conic combination ofS will mean a vector of the form ∑
s∈S
λss∈ V, where the coefficientsλsare nonnegative reals with the property that at least two elementss ∈ Ssatisfy λs >0.
Fix a finite subset EofV. For any F⊆ E, we define
ShadeF={e ∈ E | e isnota nontrivial conic combination ofF}. It can be shown that this map Shade : P(E) → P(E) satisfies Axiom 1 in Definition 4.2. In general, it does not satisfy Axiom 2. Thus, it is not a shade map in general.