Let E be a non-empty set. Let's consider the inclusion relation over P(E): (∀x, y ∈ P(E))(X ≤ Y ⇔ X ⊂ Y )

Question

Let E be a non-empty set. Let's consider the inclusion relation over P(E): (∀x, y ∈ P(E))(X ≤ Y ⇔ X ⊂ Y )

(a) Show that it is an order relation.

(b) Show that it is total if, and only if, E = {a}



I know that I have to prove that it is reflexive, antisymmetric and transitive, but how do I do that? And how do I show that it is total when E = {a}?

Answer 1

P(E) is the powerser of E.

(a) Let a,b and c be elements of E. Define A = {a}, AB = {a,b} and ABC = {a,b,c}.

Per inclusion relation R on P(E): (∀ X,Y ∈ P(E), X ≤ Y ⇔ X ⊂ Y ),

A ⊂ AB ⊂ AC => A ≤ AB ≤ABC.

Generally, let X, Y and Z be elements of P(E). R is a partial (not total) ordering of the elements of P(E). However, if X ⊂ Y, then there is an order relationship defined between X and Y, namely X ≤ Y.

It is reflexive because X ⊂ X is always true, so X ≤ X.

It is antisymmetric because if X ≠ Y and X ⊂ Y, then Y ⊄ X, so if X ≠Y and X ≤ Y, then Y is not ≤ X.

It is transitive because for X ⊂ Y and Y ⊂ Z it is true that X ⊂ Z. So, if X ≤ Y and Y ≤ Z, then X ≤ Z.

(b) --> for E is total if and only if E = {a}, i.e., E contains only 1 element, is proved by showing all other values of E, i.e., E contains more than 1 element, has subsets of E that are not subsets of each other.

<-- is proved by showing all the elements of P(E) are ordered by (∀x, y ∈ P(E))(X ≤ Y ⇔ X ⊂ Y )

for E = {a}, P(E) = { ∅,{a}} and the order relations are ∅ = ∅, ∅ < {a}, {a} = {a}, so E = {a} is total.