Well-Founded Sets and Complete Induction

A very general induction principle holds for the class of partially ordered sets having a well-founded ordering. Given a partial order≤on a setA, thestrict order <associated with≤is defined as follows:

x < yif and only ifx≤y andx)=y.

A partially ordered set < A,≤> is well-founded iff it has no infinite decreasing sequence (xi)_i∈N, that is, sequence such that xi+1 < xi for all i≥0.

The following property of well-founded sets is fundamental.

Lemma 2.1.1 Given a partially ordered set < A,≤>, < A,≤>is a well-founded set if and only if every nonempty subset ofAhas a minimal element.

Proof: First, assume that < A,≤> is well-founded. We proceed by contradiction. LetX be any nonempty subset ofA, and assume thatX does not have a minimal element. This means that for any x∈X, there is some y∈X such thaty < x, since otherwise there would be some minimal x∈X. Since X is nonempty, there is somex0 in X. By the above remark, there is somex1∈X such thatx1< x0. By repeating this argument (using induction onN), an infinite decreasing sequence (xi) can be defined inX, contradicting the fact thatAis well-founded. Hence,X must have some minimal element.

Conversely, assume that every nonempty subset has a minimal element.

If an infinite decreasing sequence (xi) exists in A, (xi) has some minimal elementxk. But this contradicts the fact thatxk+1< xk.

The principle ofcomplete induction (orstructural induction) is now de-fined. Let (A,≤) be a well-founded poset, and let P be a property of the set A, that is, a function P :A→{false,true}. We say thatP(x) holds if P(x) =true.

Principle of Complete Induction

To prove that a propertyP holds for allz∈A, it suffices to show that, for everyx∈A,

(∗) ifxis minimal, orP(y) holds for ally < x, (∗∗) thenP(x) holds.

The statement (∗) is called theinduction hypothesis, and the implication

10 2/Mathematical Preliminaries for allx, (∗) implies (∗∗)

is called the induction step. Formally, the induction principle can be stated as:

(CI) (∀x∈A)[(∀y∈A)(y < x⊃P(y))⊃P(x)]⊃(∀z∈A)P(z) Note that ifxis minimal, then there is noy ∈Asuch that y < x, and (∀y ∈ A)(y < x⊃P(y)) is true. Hence,P(x) has to be shown to betrue for every minimal elementx. These cases are called thebase cases. Complete induction is not valid for arbitrary posets (see the problems) but holds for well-founded sets as shown in the following lemma.

Lemma 2.1.2 The principle of complete induction holds for every well-founded set.

Proof: We proceed by contradiction. Assume that (CI) is false. Then, (1) (∀x∈A)[(∀y∈A)(y < x⊃P(y))⊃P(x)]

istrueand

(2) (∀z∈A)P(z)

is false, that is,

(∃z∈A)(P(z) =false) istrue.

Hence, the subsetX ofAdefined by

X ={x∈A|P(x) =false}

is nonempty. Since Ais well founded, by lemma 2.1.1, X has some minimal elementb. Since (1) istruefor allx∈A, letting x=b,

(3) [(∀y∈A)(y < b⊃P(y))⊃P(b)]

istrue. Ifbis also minimal inA, there is noy∈A such thaty < band so, (∀y∈A)(y < b⊃P(y))

holds trivially and (3) implies thatP(b) = true, which contradicts the fact that b∈X. Otherwise, for everyy ∈A such that y < b, P(y) =true, since otherwiseywould belong toX andbwould not be minimal. But then,

(∀y∈A)(y < b ⊃ P(y))

also holds and (3) implies thatP(b) =true, contradicting the fact thatb∈X. Hence, complete induction is valid for well-founded sets.

2.1 Relations, Functions, Partial Orders, Induction 11 As an illustration of well-founded sets, we define the lexicographic or-dering. Given a partially ordered set (A,≤), thelexicographic ordering <<

onA×Ainduced by ≤is defined a follows: For all x, y, x^!, y^!∈A, (x, y)<<(x^!, y^!) if and only if either

x=x^! andy=y^!, or x < x^! or x=x^! andy < y^!.

We leave as an exercise the check that<<is indeed a partial order onA×A.

The following lemma will be useful.

Lemma 2.1.3 If< A,≤>is a well-founded partially ordered set, the lexi-cographic ordering<< onA×Ais also well founded.

Proof: We proceed by contradiction. Assume that there is an infinite decreasing sequence (< xi, yi>)i∈N in A×A. Then, either,

(1) There is an infinite number of distinctxi, or (2) There is only a finite number of distinctxi.

In case (1), the subsequence consisting of these distinct elements forms a decreasing sequence inA, contradicting the fact that≤is well founded. In case (2), there is somek such that for all i≥k, xi =xi+1. By definition of

<<, the sequence (yi)i≥k is a decreasing sequence inA, contradicting the fact that≤is well founded. Hence,<<is well founded onA×A.

As an illustration of the principle of complete induction, consider the following example in which it is shown that a function defined recursively is a total function.

EXAMPLE 2.1.1

(Ackermann’s function) The following functionA:N×N→N known as Ackermann’s function is well known in recursive function theory for its extraordinary rate of growth. It is defined recursively as follows:

A(x, y) =if x= 0 then y+ 1

else if y= 0then A(x−1,1)

else A(x−1, A(x, y−1))

It is actually not obvious that such a recursive definition defines a partial function, but this can be shown. The reader is referred to Machtey and Young, 1978; or Rogers, 1967, for more details.

We wish to prove thatA is a total function. We proceed by com-plete induction over the lexicographic ordering onN×N.

12 2/Mathematical Preliminaries The base case isx= 0, y= 0. In this case, sinceA(0, y) =y+ 1, A(0,0) is defined and equal to 1.

The induction hypothesis is that for any (m, n),A(m^!, n^!) is defined for all (m^!, n^!)<<(m, n), with (m, n))= (m^!, n^!).

For the induction step, we have three cases:

(1) If m = 0, since A(0, y) = y+ 1, A(0, n) is defined and equal to n+ 1.

(2) Ifm)= 0 and n= 0, since (m−1,1) <<(m,0) and (m−1,1) )= (m,0), by the induction hypothesis,A(m−1,1) is defined, and so A(m,0) is defined since it is equal to A(m−1,1).

(3) Ifm)= 0 andn)= 0, since (m, n−1)<<(m, n) and (m, n−1))= (m, n), by the induction hypothesis, A(m, n−1) is defined. Since (m−1, y)<<(m, z) and (m−1, y))= (m, z) no matter whatyand zare, (m−1, A(m, n−1))<<(m, n) and (m−1, A(m, n−1)))= (m, n), and by the induction hypothesis,A(m−1, A(m, n−1)) is defined. But this is preciselyA(m, n), and so A(m, n) is defined.

This concludes the induction step. Hence,A(x, y) is defined for all x, y≥0.