Page 283 - HOW TO PROVE IT: A Structured Approach, Second Edition

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More Examples 269

it follows that b = c. This is clearly impossible, since c ∈ B = B \{b}. Thus,
b must be a minimal element of B.
Case 2. ¬bRc. We claim in this case that c is a minimal element of B.To
see why, suppose it isn’t. Then we can choose some x ∈ B such that xRc

and x = c. Since c is a minimal element of B , we can’t have x ∈ B ,sothe

only other possibility is x = b. But then since xRc we must have bRc, which
contradicts our assumption that ¬bRc. Thus, c is a minimal element of B.
Note that an infinite subset of a partially ordered set need not have a minimal
element, as we saw in part 1 of Example 4.4.5. Thus, the assumption that B is
finite was needed in our last theorem. This theorem can be used to prove another
interesting fact about partial orders, again using mathematical induction:

Example 6.2.2. Suppose A is a finite set and R is a partial order on A. Prove
that R can be extended to a total order on A. In other words, prove that there
is a total order T on A such that R ⊆ T .
Scratch work

We’ll only outline the proof, leaving many details as exercises. The idea is
to prove by induction that ∀n ∈ N∀A∀R[(A has n elements and R is a partial
order on A) →∃T (T is a total order on A and R ⊆ T )]. The induction step is
similar to the induction step of the last example. If R is a partial order on a set A
with n + 1 elements, then we remove one element, call it a, from A, and apply
the inductive hypothesis to the remaining set A = A \{a}. This will give us

a total order T on A , and to complete the proof we must somehow turn this
into a total order T on A such that R ⊆ T . The relation T already tells us how

to compare any two elements of A , but it doesn’t tell us how to compare a

to the elements of A . This is what we must decide in order to define T, and

the main difficulty in this step of the proof is that we must make this decision
in such a way that we end up with R ⊆ T . Our resolution of this difficulty in
the following proof involves choosing a carefully in the first place. We choose
a to be an R-minimal element of A, and then when we define T, we make a
smaller in the T ordering than every element of A . We use the theorem in the

last example, with B = A, to guarantee that A has an R-minimal element.
Solution
Theorem. Suppose A is a finite set and R is a partial order on A. Then there
is a total order T on A such that R ⊆ T.
Proof. We will show by induction on n that every partial order on a set with
n elements can be extended to a total order. Clearly this suffices to prove the
theorem.