Download Induction and the Well-Ordering Principle Capturing All The Whole

Induction and the Well-Ordering Principle
Capturing All The Whole Numbers: The Principle of Induction
“We now have a fairly clear picture of how whole numbers are defined in terms of sets. One
technical issue that I need to address is a point I raised at the beginning of this discussion: How do
we know we have really ‘captured’ the whole numbers with our set-based definition? As I mentioned
before, we were able to provide criteria for deciding whether our set-based definition of ‘ordered
pair’ was sufficient; and we followed a similar path in an earlier lesson when we went through the
construction of the reals from the rationals.
“I mentioned earlier that the whole numbers have two characterizing features. The first, as we
discussed, is the existence of a successor function. The analog for ω to the successor function s on
W, defined by s(n) = n + 1, is ss, defined by ss(an ) = an ∪ {an } = an+1 .
“The second characterizing feature is more subtle. To see what is needed, let’s step back for a
moment and consider what we need to do to ‘comprehend’ the set W. The fact is, being an infinite
set, W is difficult to conceive of all at once. The infinitude of W becomes more manageable if we
organize it into a sequence, starting at 0 and repeatedly obtaining the next term by adding 1. That
is, we think of the sequence
0, s(0), s(s(0)), s(s(s(0))), . . .
as being a complete enumeration of the elements of W. But how can we be sure that this sequence
is going to capture all the elements of W? Is it possible that as the sequence unfolds, some elements
of W are unexpectedly left out? For example, suppose we define a ‘successor function’ s0 on the
positive rational numbers, defined in the same way as s — namely, for each rational q, s0 (q) = q + 1.
Now, is it true that the set of positive rationals is enumerated by the sequence 0, s(0), s(s(0)), . . .?
“Of course not,” I replied. “Many of the positive rationals will be left out of this sequence.”
“Right,” he said. “So it is quite reasonable to ask how we can establish with certainty that
the sequence 0, s(0), s(s(0)), . . . really is a complete list of the elements of W.
“In a similar way, we would like to have a criterion for deciding whether any given set of whole
numbers actually contains all the whole numbers. For instance, let’s revisit the Whole Numbers As
Sets Theorem for a moment. In part (4) of that theorem, we asserted that rank(an ) = n for every
whole number n. Our explanation for this was informal. To give a more formal proof of this fact
requires that we have a criterion to determine whether a given subset of W is in fact equal to W.
In this example, suppose we let A denote the set of all whole numbers n for which rank(an ) = n. In
order for part (4) of the Theorem to be true, it must be the case that A = W. But by what logic
are we permitted to conclude that this is true? Informally, we may argue in a manner like this:
‘You can see that rank(ai ) = i for i = 0, 1, 2 since in those cases Vi = ai , and
then, by writing down the elements, you can see that a3 ⊆ V3 and a4 ⊆ V4 . In
the same way, an ⊆ Vn for all n.’
1
“In reality, though, this kind of reasoning is making an assumption that the infinite sequence
of statements
a0 ⊆ V0
a1 ⊆ V 1
a2 ⊆ V 2
•
•
•
can all be proven to hold true just by proving that the first few of them are true.
“The principle that we need in order to prove such things is known as the Principle of Induction.
The Principle of Induction provides a means of ‘certifying’ a sequence of whole numbers as a ‘full
enumeration’ of W. It says that you may conclude that a subset A of the whole numbers in fact
equals W if the following two conditions are met:
(a) 0 belongs to A
(b) for any n ∈ A, the successor s(n) is also in A
Using the Principle of Induction, we can verify the intuition that the set consisting of 0, s(0),
s(s(0)), . . ., is indeed all of W. We can also verify that rank(an ) = n for every n ∈ W. Let’s record
our Induction Principle.”
He wrote:
Principle of Induction Suppose A ⊆ W, and suppose A has the following two properties:
(1) (Basis Step) 0 ∈ A.
(2) (Induction Step) If n ∈ A, then s(n) ∈ A.
Then, A = W.
“In my experience teaching mathematics, I have found that students are sometimes bewildered
by the Principle of Induction. The abstract statement of the principle may seem reasonable enough,
but applying the principle can seem mystifying at first.
“So, I would like to show you a simple application of the Principle of Induction and let you
practice with some additional exercises. Then I will give you some hints to help you apply induction
to establish the truth of the two propositions we were just discussing.
“To start out, let’s prove the following.”
Proposition. For all whole numbers n, n < 2n .
“If you try a few values for n, you will very quickly become convinced of the truth of the
2
Proposition. For instance,
1 < 21 = 2
3 < 23 = 8
10 < 210 = 1024.
So, it may seem obvious that the Proposition is true. But, how can we be sure that the expected
result holds true for all whole numbers, even for numbers that are too big to write down? This is
the value of the Principle of Induction.
“Let’s begin our proof using this principle by deciding which set will play the role of A; this is
the set we will want to prove to be equal to W. Let’s let A denote the set of all n for which n < 2n .
A = {n | n < 2n }.
“Clearly, then, once we prove A = W, we will have established the truth of the Proposition.
To prove A = W, we begin with the Basis Step. To show 0 ∈ A, we must show that 0 < 20 . But
this is obvious, since 20 = 1.
“Next, we perform the Induction Step. For this, we must show that for any n, if n belongs to
A, then n + 1 also belongs to A. So, let n ∈ W and assume n ∈ A. This means that n < 2n . We
must show that n + 1 ∈ A; that is, we must show that
n + 1 < 2n+1
(∗)
“If n happens to be 0, notice that (∗) does hold true, since 0 + 1 < 20+1 = 2. So for the rest
of the Induction Step, let’s assume n > 0. We make two observations:
(A) Since 1 ≤ n, we may add n to both sides of the inequality to obtain n + 1 ≤ n + n.
(B) By the induction hypothesis, n < 2n . If we multiply both sides of the inequality by 2, we
obtain 2n < 2 · 2n = 2n+1 .
“Putting these two observations together, we get:
n + 1 ≤ n + n = 2n < 2 · 2n = 2n+1 ,
as required.”
“At first, this style of proof may seem like magic (or worse) since, at the Induction Step, we
seem to be assuming the very thing we are trying to prove. We are trying to prove that n < 2n for
each n, and then at the Induction Step, we assume that n < 2n is true. This certainly seems fishy.
“To clarify what is going on, let’s change the problem slightly. I am going to invent a miniature
Induction Principle—a principle that is valid but not usually stated in this way —which will allow
me to prove a statement holds true for the first four whole numbers, 0, 1, 2, 3. Here it is.”
3
He wrote:
Miniature Induction Principle. Suppose A ⊆ {0, 1, 2, 3}, and suppose A has the following two
properties:
(1) (Basis Step) 0 ∈ A
(2) (Induction Step) for any n = 0, 1, 2, if n ∈ A then s(n) ∈ A.
Then A = {0, 1, 2, 3}.
“Let’s see what this mini-principle is saying when we let
A = {n | 0 ≤ n ≤ 3 and n < 2n }.
“Showing A = {0, 1, 2, 3} would establish that n < 2n for all n = 0, 1, 2, 3. The mini-principle
declares that, in order to prove A = {0, 1, 2, 3}, we need to prove the following four statements:
(a) 0 ∈ A
(b) if 0 ∈ A then 1 ∈ A
(c) if 1 ∈ A then 2 ∈ A
(d) if 2 ∈ A then 3 ∈ A
“Statements (b), (c), (d) are an elaboration of the Induction Step: Showing that for any
n = 0, 1, 2, if n ∈ A then s(n) ∈ A is accomplished by showing (b), (c), and (d) all at once. The
claim of the mini-principle is that, to show A = {0, 1, 2, 3}, all you have to do is prove (a)-(d). Does
this make sense? Is this really enough?
“Let’s check that the mini-principle’s claim that this is enough really does make sense. Suppose
we can indeed prove (a)-(d). Since we can prove (a), we know 0 ∈ A. Since we can prove (b), we
know that if 0 ∈ A, then 1 ∈ A. Putting (a) and (b) together, we may now conclude that1 1 ∈ A.
0∈A
if 0 ∈ A then 1 ∈ A
1∈A
“Next, we have also proved (c). But since we now know that 1 ∈ A, we can put this fact
together with (c) to conclude that 2 ∈ A.
1
The three lines in the display are intended to be read in the following way: “FROM 0 ∈ A
AND if 0 ∈ A then 1 ∈ A CONCLUDE THAT 1 ∈ A .”
4
1∈A
if 1 ∈ A then 2 ∈ A
2∈A
“And finally, since we now know 2 ∈ A, we can put this together with (d) to conclude that
3 ∈ A.
2∈A
if 2 ∈ A then 3 ∈ A
3∈A
“This shows why the mini-principle is reasonable. The Induction Step is stating that three
different if..then statements hold true. If you can prove the Basis Step, and can prove that all these
if..then statements hold, then it follows that A = {0, 1, 2, 3}.
“Now, if we return to the actual Induction Principle, we will see the same logic at work. To
prove that n < 2n holds for all n, it suffices to prove that
0 < 20 and that
if 0 < 20 then 1 < 21 (from these two statements we may prove 1 < 21 ) and that
if 1 < 21 then 2 < 22 (from these three statements we may prove 2 < 22 ) and that
if 2 < 22 then 3 < 23 (from these four statements we may prove 3 < 23 ) and that
•
•
•
if n < 2n then n + 1 < 2n+1 (from these statements we may prove n + 1 < 2n+1 ) and that
•
•
•
“We can’t formally string together this infinite list of if..then statements into a single assertion
in the way we have suggested here — sentences always have to be of finite length. But the idea
is the same — if you can establish the Induction Step, you have in effect established all of these
if..then sentences. The Principle of Induction says that if you can do this and also establish the
Base Case, then the conclusion must be that A = W.
5
“As I mentioned before, Induction is used to verify that the sequence 0, s(0), s(s(0)), . . . includes
all whole numbers, and that rank(an ) = n for all n ∈ W. I would like you to try your hand at
proving these on your own. I have given you some easier warm-up exercises on induction — once
you are comfortable with the warm-ups, try these two exercises, and use the hints I’ve provided.
“A related point I want to mention here has to do with the way the Principle of Induction is
stated. What I have given you is the ‘strongest’ form of this principle, but there is a slightly weaker
form that can — and in certain contexts, must — be used in place of the version we’re using. In
this weaker version, one uses the concept of a ’formula’ in place of a subset of W. A formula P (n)
about whole numbers is simply a statement in the language of ordinary arithmetic, where n is used
as a placeholder for an arbitrary whole number. Examples of such formulas are:”
• n < 2n
• n is an even number
• 1 + 2 + 3 + . . . + n = n(n + 1)/2.
“Here is the alternative prinicple of induction:”
He wrote:
Elementary Principle of Induction. Suppose P (n) is a formula about whole numbers. Then
P (n) holds true for all whole numbers if the following two clauses can be established:
(1) (Basis Step) P (0) holds.
(2) (Induction Step) Assuming P (n) holds, P (n + 1) also holds.
“As an example, let’s prove the Proposition we have been looking at using this alternative
form of induction; that is, let’s prove that, for all n ∈ W, n < 2n once again. Here, we are not
trying to show that the set consisting of all n for which n < 2n is all of W. Instead, we are trying
to show that some formula P (n) always holds true. What should be our formula P (n)?”
I saw that one of the examples he’d given us would work — I mentioned the formula “n < 2n ”.
“Right,” Dr. Wu said. Now, for the Basis Step, we need to establish that P (0) holds. What
does P (0) say?”
“It says that 0 < 20 ,” I responded.
“And, clearly, that is a true statement, since 0 < 1. We have established the Basis Step. For
the Induction Step, let’s assume P (n) holds true. What exactly does P (n) assert?”
“It says that n < 2n ,” I replied.
“And we must prove on that basis that P (n + 1) holds; what exactly does P (n + 1) assert?”
I could see that we needed to replace n by n + 1, so I said that it asserts n + 1 < 2n+1 .
“Good. Now I hope you can see that the reasoning we used before, to go from the assertion
n < 2n to the assertion n + 1 < 2n+1 can be used equally well here.” He paused to see if this point
was clear.
6
“Therefore,” he continued, “we have established the Induction Step, and the result is proved.
I’ll let you try some examples with this formulation in the exercises. There are two main reasons this
version is used in mathematical foundations. First, when the concepts of arithmetic are formalized
into an axiomatic system (a first-order system, as we’ll discuss later), this version must be used
because there is no way to state the stronger version in the language of the formal system. Secondly,
as we learn about ‘counting past the whole numbers’ so that we can enumerate big sets — using
infinite ordinal numbers — we will find yet another induction principle arise, and that principle
will of necessity make use of the ‘formula’ approach to induction. In any case, we will elaborate on
these points later in the course.”
The Well-Ordering Principle
“Next I would like to take a look at an equivalent form of the Principle of Induction — a form
that is often easier to work with, especially as you first become acquainted with the ideas.”
He wrote:
Well-Ordering Property Every non-empty subset of N has a least element.
“This property may seem too obvious to be important. It certainly seems clear that if you are
given a nonempty set of whole numbers, one of the numbers in the set is the smallest. For instance,
consider the following sets; in each case I have listed the elements of the set in increasing order.
What is the least element in each case?”
E = {0, 2, 4, 6, 8, . . .}
X = {1, 17, 33, 49, . . .}
Y = {100, 200, 300, 400, . . .}
It was obvious that the smallest element of E is 0, of X is 1, and of Y is 100.
“Good,” he said. “As it happens, the Well-Ordering Principle is particularly important because
it is equivalent to the Principle of Induction. The main thing that will concern us is that it is indeed
a fact about W, so let’s prove it using Induction; I’ll let you take a look at the other direction of the
equivalence — showing that the Principle of Induction can be proven from it — in the exercises.
A Technical Moment: A Proof of the Well-Ordering Principle.
“Before defining our set A for the induction, I want to define an auxiliary notion. Suppose
B is any set of whole numbers. Let’s call B n-good if B contains a whole number in the set
{0, 1, 2, . . ., n}. For instance, which of the following sets are 4-good?”
X1 = {1, 3, 8, 12, 17}
X2 = {100, 200, 300, . . ., 100n, . . .}
X3 = {4, 8, 12, 16, 20, . . .}
7
This seemed easy enough. X1 actually has two numbers in {0, 1, 2, 3, 4} — namely 1 and 3 —
so it is 4-good. X3 has the number 4 in this set, so it is also 4-good. But X2 ’s smallest element is
100, which lies outside of {0, 1, 2, 3, 4}, so X2 is not 4-good.
“Right,” Dr. Wu said after hearing my response. Now, what if we could prove that, for every
n, every n-good set has a least element? Would this establish the Well-Ordering Property?”
This was a twist. I tried to think how the logic should go: Let’s assume Dr. Wu’s little
theorem is right, that every n-good set has a least element. I’ll try to prove the Well-Ordering
Property. Suppose B is a non-empty set. Why must B have a least element? Hmmm. I have to
use Dr. Wu’s little theorem somehow. If I could show that B is n-good for some n...Oh, but wait.
If B is non-empty, it must contain some number—call it k. So B has to be k-good since k lies in
{0, 1, 2, . . ., k}. By Dr. Wu’s little theorem, B must have a least element. Cool. I explained to Dr.
Wu.
“Exactly right,” he said. “So, I just need to establish my little theorem about n-good sets —
that every n-good set has a least element. We are ready to define our set A to this end:
A = {n | n ∈ W and every n-good set has a least element}.
“We prove that A = W by Induction. First, let’s see why 0 ∈ A. Suppose B is a 0-good set.
We must show B has a least element. Being 0-good means that B has an element in {0} — but this
means 0 ∈ B. Since no number in W is smaller than 0 and 0 is already in B, we have discovered
that B does indeed have a smallest element; namely, 0.
“For the Induction Step, we will assume that n ∈ A. This means that all n-good sets have a
least element. We must prove that n + 1 ∈ A; that is, given an n + 1-good set B, we must show that
B has a least element. Since B is n + 1-good, it has an element x in the set {0, 1, 2, . . ., n, n + 1}.
There are two possibilities. The first possibility is that x actually happens to lie in {0, 1, 2, . . ., n}.
If this happens to be true, what can we say about B? What kind of set must B actually be?”
By the definition, if B has an element x with x ∈ {0, 1, . . ., n}, B must be n-good.
“Good,” Dr. Wu acknowledged. “But, in our induction argument, we are allowed to assume
that every n-good set has a smallest element—this is the induction hypothesis. Therefore, in this
case, we may conclude that B does have a least element.
“The second possibility is that, although B is n + 1-good, B does not contain an element x in
the set {0, 1, 2, . . ., n}. But now, the only way B could contain an element in {0, 1, 2, . . ., n, n + 1}
but not in {0, 1, 2, . . ., n} is if this element is n + 1. What does this tell us? Here, we know that
n + 1 ∈ B, but no number smaller than n + 1 belongs to B. What kind of number must n + 1 be?”
“It must be the smallest number in B. n + 1 is the least element of B,” I said.
“Right,” he said. “Therefore, in this second case we have also established that B has a least
element. Thus, in both cases — when the n + 1-good set B has an element in {0, 1, 2, . . . , n}, and
when it does not have such an element— we have established that B has a least element. Therefore,
8
n+1 ∈ A. And so, by the Principle of Induction, A = W, showing that every n-good set has a least
element. And, as we observed at the beginning, this allows us to conclude that the Well-Ordering
Property holds true.
Using the Well-Ordering Property In Place of Induction.
“I must tell you that, though this proof might seem a bit tricky, the Well Ordering Principle
itself is very easy to use. Usually, it is intuitively clearer for students at first than the Induction
Principle. Let me show you what I mean by proving that for all n, n < 2n once again, this time
using the Well-Ordering Principle.
“This time, let’s prove the result by arguing indirectly. That is, let’s assume that for some
n, n 6< 2n , which means that for some n, n ≥ 2n . Let’s try to arrive at a contradiction from this
hypothesis. To do this, let’s define a set B to be {n | n ∈ W and n ≥ 2n }. Notice that B 6= ∅ since
we are assuming that, for at least some n, n ≥ 2n . Therefore, we can apply the Well-Ordering
Principle. Let’s let m denote the least element of B.
“First of all, notice that m 6= 0 because it is obviously true that 0 < 20 . So, we know for sure
that m > 0. Let’s consider k = m − 1. Since k < m, what must be true about k?”
Since m was the least element of B, it must be that k 6∈ B. So, it must be true that k < 2k . I
said this to Dr. Wu.
“That’s correct. So we know k < 2k , and, multiplying both sides by 2, 2k < 2(2k ). Recall that
if we add 1 to k, we get m. So we can write down:”
m = k + 1 ≤ k + k = 2k < 2(2k ) = 2k+1 = 2m .
“So what happened?”
Dr. Wu was asking for a high-level view of the proof. Here, we used the fact that k was small
enough so that it still satisfied the property k < 2k . But the number just 1 larger than k is m, a
number that supposedly does not have this property. Dr. Wu used the fact that k < 2k to argue
that k + 1 < 2k+1 , which shows that m < 2m . But this contradicts the assumption that m ∈ B.
Therefore, the assumption that for some n, n 6< 2n is not correct. This proves the theorem!
“Excellent, Paul,” Dr. Wu exclaimed. “So you see? It is essentially the same as the induction
argument, but we perform the induction step as a step that illustrates a contradiction in an indirect
argument instead.”
9