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278 Mathematical Induction

Figure 11

is, regions that share an edge) are different colors. (Figure 12 shows an
example in the case n = 4.)

Figure 12

15. What’s wrong with the following proof that if A ⊆ N and 0 ∈ A then
A = N?

Proof. We will prove by induction that ∀n ∈ N(n ∈ A).
Base case: If n = 0, then n ∈ A by assumption.
Induction step: Let n ∈ N be arbitrary, and suppose that n ∈ A. Since
n was arbitrary, it follows that every natural number is an element of A,
and therefore in particular n + 1 ∈ A.

16. Suppose f : R → R. What’s wrong with the following proof that for
every finite, nonempty set A ⊆ R there is a real number c such that
∀x ∈ A( f (x) = c)?

Proof. We will prove by induction that for every n ≥ 1, if A is any subset
of R with n elements then ∃c ∈ R∀x ∈ A( f (x) = c).
Base case: n = 1. Suppose A ⊆ R and A has one element. Then A =
{a}, for some a ∈ R. Let c = f (a). Then clearly ∀x ∈ A( f (x) = c).
Induction step: Suppose n ≥ 1, and for all A ⊆ R,if A has n ele-
ments then ∃c ∈ R∀x ∈ A( f (x) = c). Now suppose A ⊆ R and A has
n + 1 elements. Let a 1 be any element of A, and let A 1 = A \{a 1 }.
Then A 1 has n elements, so by inductive hypothesis there is some c 1 ∈ R