Answers to Self-Study Questions

Test Yourself #1

x M y = z iff
1. (z ⊆ x) and (z ⊆ y), and
2. ∀ w in D such that (w ⊆ x) and (w ⊆ y): w = ⊆ z
Note that (1) says that z is a lower bound, and (2) says that it is the greatest lower bound.

Test Yourself #2

Reflexive: (∀ x in D, x ⊆ x)
Let x = (a,b) ∈ D1 x D2. Then (a ⊆_D1 a) and (b ⊆_D2 b) since ⊆_D1 and ⊆_D2 are reflexive. Therfore, ((a,b) ⊆_{D1 x D2} (a,b)).

Anti-symmetric: (∀ x, y in D, if ((x ⊆ y) and (y ⊆ x)) then x = y)
Let ((x₁,y₁) ∈ D1 x D2) and ((x₂, y₂) ∈ D1 x D2.)
If ((x₁,y₁) ⊆_{D1 x D2} (x₂, y₂)) and ((x₂,y₂) ⊆_{D1 x D2} (x₁, y₁)), then (x₁ ⊆_D1 x2) and (x₂ ⊆_D1 x1) . And so, (x1 = x2) since ⊆_D1 is anti-symmetric. In the same way, (y1 = y2).
Then clearly, ((x1,y1) = (x2,y2)).

Transitive: (∀ x,y,z in D, if ((x ⊆ y) and (y ⊆ z)) then x ⊆ z)
Let ((x₁,y₁) ∈ D1 x D2), ((x₂, y₂) ∈ D1 x D2.), and ((x₃, y₃) ∈ D1 x D2.)
If ((x₁,y₁) ⊆_{D1 x D2} (x₂, y₂)) and ((x₂,y₂) ⊆_{D1 x D2} (x₃, y₃)), then (x₁ ⊆_D1 x2) and (x₂ ⊆_D1 x3). And so, (x1 = ⊆_D1 x3) since ⊆_D1 is transitive.
In the same way, (y1 = ⊆_D2 y3).
And so, ((x1,y1) ⊆_{D1 x D2} (x3,y3)).

Test Yourself #3

Question 1:

We can not determine whether or not f is monotonic from that information.
Note that non-trivial chains in D are all of the form {⊥,n} where n is some integer. Then, since f(3) = ⊥, if f(⊥) ≠ ⊥ then f is not monotonic. In fact, it is clear that if f(⊥) = ⊥ then f is monotonic.

Question 2:

Note that (<2,⊥> ⊆ <2,3>) and so the set {<2,⊥>, <2,3>} is a chain in D x D.
Yet, (f(<2,⊥>) ⊈ f(<2,3>)) and so f is not monotonic.