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More Examples 275

Figure 9

of which was easy to solve. If we had started with a larger grid, we might have
had to repeat the splitting many times before reaching easy 2 × 2 problems.
Recursion and its relationship to mathematical induction is the subject of our
next section.

Exercises

1. Complete the proof in Example 6.2.2 by doing the following proofs. (We
∗
use the same notation here as in the example.)
(a) Prove that R is a partial order on A .

(b) Prove that T is a total order on A and R ⊆ T .
2. Suppose R is a partial order on a set A, B ⊆ A, and B is finite. Prove
that there is a partial order T on A such that R ⊆ T and ∀x ∈ B∀y ∈
A(xT y ∨ yT x). Note that, in particular, if A is finite we can let B = A,
and the conclusion then means that T is a total order on A. Thus, this
gives an alternative approach to the proof of the theorem in Example
6.2.2. (Hint: Use induction on the number of elements in B. For the
induction step, assume the conclusion holds for any set B ⊆ A with n
elements, and suppose B is a subset of A with n + 1 elements. Let b be
any element of B and let B = B \{b}, a subset of A with n elements. By

inductive hypothesis, let T be a partial order on A such that R ⊆ T

and ∀x ∈ B ∀y ∈ A(xT y ∨ yT x). Now let A 1 ={x ∈ A | xT b} and

A 2 = A \ A 1 , and let T = T ∪ (A 1 × A 2 ). Prove that T has all the

required properties.)
3. Suppose R is a total order on a set A. Prove that every finite, nonempty
set B ⊆ A has an R-smallest element.
∗
4. (a) Suppose R is a relation on A, and ∀x ∈ A∀y ∈ A(xRy ∨ yRx). (Note
that this implies that R is reflexive.) Prove that for every finite,
nonempty set B ⊆ A there is some x ∈ B such that ∀y ∈ B((x, y) ∈
R ◦ R). (Hint: Imitate Example 6.2.1.)