Page 282 - HOW TO PROVE IT: A Structured Approach, Second Edition

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268 Mathematical Induction
How can we use the inductive hypothesis to reach our goal? The inductive
hypothesis tells us that if we had a subset of A with n elements, then it would
have a minimal element. To apply it, we need to find a subset of A with n
elements. Our arbitrary set B is a subset of A, and we have assumed that it has
n + 1 elements. Thus, a simple way to produce a subset of A with n elements
would be to remove one element from B. It is not clear where this reasoning
will lead, but it seems to be the simplest way to make use of the inductive
hypothesis. Let’s give it a try.
Let b be any element of B, and let B = B \{b}. Then B is a subset of A

with n elements, so by the inductive hypothesis, B has a minimal element.
This is an existential statement, so we immediately introduce a new variable,

say c, to stand for a minimal element of B .
Our goal is to prove that B has a minimal element, which is also an existential
statement, so we should try to come up with a minimal element of B. We only
know about two elements of B at this point, b and c, so we should probably
try to prove that one of these is a minimal element of B. Which one? Well,
it may depend on whether one of them is smaller than the other according to
the partial order R. This suggests that we may need to use proof by cases. In
our proof we use the cases bRc and ¬bRc. In the first case we prove that b is
a minimal element of B, and in the second case we prove that c is a minimal
element of B. Note that to say that something is a minimal element of B is a
negative statement, so in both cases we use proof by contradiction.
Solution

Theorem. Suppose R is a partial order on a set A. Then every finite, nonempty
set B ⊆ A has an R-minimal element.
Proof. We will show by induction that for every natural number n ≥ 1, every
subset of A with n elements has a minimal element.
Base case: n = 1. Suppose B ⊆ A and B has one element. Then B ={b} for
some b ∈ A. Clearly ¬∃x ∈ B(x = b), so certainly ¬∃x ∈ B(xRb ∧ x = b).
Thus, b is minimal.
Induction step: Suppose n ≥ 1, and suppose that every subset of A with n
elements has a minimal element. Now let B be an arbitrary 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, we can choose a minimal element

c ∈ B .
Case 1. bRc. We claim that b is a minimal element of B. To see why,
suppose it isn’t. Then we can choose some x ∈ B such that xRb and x = b.
Since x = b, x ∈ B . Also, since xRb and bRc, by transitivity of R it follows

that xRc. Thus, since c is a minimal element of B , we must have x = c. But

then since xRb we have cRb, and we also know bRc, so by antisymmetry of R