2. A closer look at induction
2.5. Induction on a derived quantity: maxima of sets
+ (m+1) = m(m+1)
2 + (m+1) = m(m+1) +2(m+1) 2
= (m+2) (m+1)
2 = (m+1) (m+2)
2 = (m+1) ((m+1) +1) 2
(since m+2 = (m+1) +1). In other words, (64) holds for n = m+1. This completes the induction step. Hence, (64) is proven by induction. This proves Proposition 2.28.
2.5. Induction on a derived quantity: maxima of sets
2.5.1. Defining maxima
We have so far been applying the Induction Principle in fairly obvious ways: With the exception of our proof of Proposition 2.16, we have mostly been doing induc-tion on a variable (n or k or i) that already appeared in the claim that we were proving. But sometimes, it is worth doing induction on a variable that does not explicitly appear in this claim (which, formally speaking, means that we introduce a new variable to do induction on). For example, the claim might be saying “Each nonempty finite set S of integers has a largest element”, and we prove it by in-duction on |S| −1. This means that instead of directly proving the claim itself, we rather prove the equivalent claim “For each n ∈ N, each nonempty finite set S of integers satisfying |S| −1 = n has a largest element” by induction on n. We shall show this proof in more detail below (see Theorem 2.35). First, we prepare by discussing largest elements of sets in general.
Definition 2.29. Let Sbe a set of integers (or rational numbers, or real numbers).
Amaximumof Sis defined to be an elements ∈ Sthat satisfies (s ≥t for eacht ∈ S).
In other words, a maximum ofSis defined to be an element ofSwhich is greater or equal to each element of S.
(The plural of the word “maximum” is “maxima”.)
Example 2.30. The set{2, 4, 5} has exactly one maximum: namely, 5.
The set N = {0, 1, 2, . . .} has no maximum: If k was a maximum of N, then we would have k≥k+1, which is absurd.
The set {0,−1,−2, . . .} has a maximum: namely, 0.
The set ∅ has no maximum, since a maximum would have to be an element of∅.
In Theorem 2.35, we shall soon show that every nonempty finite set of integers has a maximum. First, we prove that a maximum is unique if it exists:
Proposition 2.31. Let S be a set of integers (or rational numbers, or real num-bers). Then, Shasat most one maximum.
Proof of Proposition 2.31. Let s1 and s2 be two maxima of S. We shall show that s1 =s2.
Indeed,s1is a maximum ofS. In other words,s1is an elements∈ Sthat satisfies (s≥t for eacht∈ S) (by the definition of a maximum). In other words, s1 is an element ofS and satisfies
(s1 ≥tfor each t∈ S). (65)
The same argument (applied to s2 instead of s1) shows that s2 is an element of S and satisfies
(s2 ≥tfor each t∈ S). (66)
Now, s1 is an element of S. Hence, (66) (applied to t = s1) yields s2 ≥ s1. But the same argument (with the roles of s1 and s2 interchanged) shows that s1 ≥ s2. Combining this withs2 ≥s1, we obtain s1 =s2.
Now, forget that we fixed s1 and s2. We thus have shown that if s1 and s2 are two maxima of S, then s1 =s2. In other words, any two maxima ofS are equal. In other words,S hasat most onemaximum. This proves Proposition 2.31.
Definition 2.32. Let Sbe a set of integers (or rational numbers, or real numbers).
Proposition 2.31 shows that S has at most one maximum. Thus, if S has a maximum, then this maximum is the unique maximum of S; we shall thus call it the maximumofS orthe largest element ofS. We shall denote this maximum by maxS.
Thus, if S is a set of integers (or rational numbers, or real numbers) that has a maximum, then this maximum maxS satisfies
maxS∈ S (67)
and
(maxS≥tfor each t∈ S) (68)
(because of the definition of a maximum).
Let us next show two simple facts:
Lemma 2.33. Let x be an integer (or rational number, or real number). Then, the set {x} has a maximum, namely x.
Proof of Lemma 2.33. Clearly, x ≥ x. Thus, x ≥ tfor each t ∈ {x} (because the only t∈ {x} isx). In other words, x is an elements∈ {x} that satisfies
(s≥t for eacht∈ {x})(since x ∈ {x}).
But recall that a maximum of {x} means an element s ∈ {x} that satisfies (s≥t for eacht∈ {x})(by the definition of a maximum). Hence, x is a maximum of {x} (since x is such an element). Thus, the set {x} has a maximum, namely x.
This proves Lemma 2.33.
Proposition 2.34. LetPandQbe two sets of integers (or rational numbers, or real numbers). Assume that Phas a maximum, and assume that Q has a maximum.
Then, the set P∪Qhas a maximum.
Proof of Proposition 2.34. We know that P has a maximum; it is denoted by maxP.
We also know that Q has a maximum; it is denoted by maxQ. The sets P and Q play symmetric roles in Proposition 2.34 (since P∪Q = Q∪P). Thus, we can WLOG assume that maxP ≥ maxQ (since otherwise, we can simply swap Pwith Q, without altering the meaning of Proposition 2.34). Assume this.
Now, (67) (applied to S = P) shows that maxP ∈ P ⊆ P∪Q. Furthermore, we claim that
(maxP≥tfor each t∈ P∪Q). (69) [Proof of (69): Let t∈ P∪Q. We must show that maxP≥t.
We have t ∈ P∪Q. In other words, t ∈ Por t ∈ Q. Hence, we are in one of the following two cases:
Case 1: We havet∈ P.
Case 2: We havet∈ Q.
(These two cases might have overlap, but there is nothing wrong about this.) Let us first consider Case 1. In this case, we have t ∈ P. Hence, (68) (applied to S= P) yields maxP≥t. Hence, maxP≥t is proven in Case 1.
Let us next consider Case 2. In this case, we havet ∈ Q. Hence, (68) (applied to S = Q) yields maxQ ≥ t. Hence, maxP ≥ maxQ ≥t. Thus, maxP ≥ t is proven in Case 2.
We have now proven maxP≥tin each of the two Cases 1 and 2. Since these two Cases cover all possibilities, we thus conclude that maxP ≥ t always holds. This proves (69).]
Now, maxPis an elements∈ P∪Qthat satisfies(s≥t for eacht∈ P∪Q)(since maxP ∈ P∪Qand (maxP≥t for eacht∈ P∪Q)).
But recall that a maximum of P∪Q means an element s ∈ P∪Q that satisfies (s≥t for eacht∈ P∪Q) (by the definition of a maximum). Hence, maxP is a maximum of P∪Q (since maxP is such an element). Thus, the set P∪Q has a maximum. This proves Proposition 2.34.
2.5.2. Nonempty finite sets of integers have maxima
Theorem 2.35. Let S be a nonempty finite set of integers. Then, S has a maxi-mum.
First proof of Theorem 2.35. First of all, let us forget that we fixed S. So we want to prove that ifSis a nonempty finite set of integers, then Shas a maximum.
For eachn ∈N, we let A(n) be the statement
ifS is a nonempty finite set of integers satisfying |S| −1=n, then S has a maximum
. We claim thatA(n) holds for alln ∈N.
Indeed, let us prove this by induction onn:
Induction base: IfSis a nonempty finite set of integers satisfying|S| −1=0, then S has a maximum50. But this is exactly the statement A(0). Hence, A(0) holds.
This completes the induction base.
Induction step: Let m ∈ N. Assume that A(m) holds. We shall now show that A(m+1) holds.
We have assumed thatA(m) holds. In other words,
if Sis a nonempty finite set of integers satisfying |S| −1=m, then Shas a maximum
(70) (because this is what the statementA(m) says).
Now, let Sbe a nonempty finite set of integers satisfying |S| −1=m+1. There exists somet ∈ S(sinceSis nonempty). Consider this t. We have(S\ {t})∪ {t} = S∪ {t}=S (sincet∈ S).
Fromt∈ S, we obtain|S\ {t}| =|S| −1=m+1>m≥0 (sincem∈ N). Hence, the setS\ {t} is nonempty. Furthermore, this set S\ {t} is finite (since S is finite) and satisfies |S\ {t}| −1 = m (since |S\ {t}| = m+1). Hence, (70) (applied to S\ {t}instead ofS) shows thatS\ {t}has a maximum. Also, Lemma 2.33 (applied to x = t) shows that the set {t} has a maximum, namely t. Hence, Proposition 2.34 (applied to P = S\ {t} and Q = {t}) shows that the set (S\ {t})∪ {t} has a maximum. Since (S\ {t})∪ {t} = S, this rewrites as follows: The set S has a maximum.
Now, forget that we fixed S. We thus have shown that if S is a nonempty finite set of integers satisfying |S| −1 = m+1, then S has a maximum. But this is precisely the statement A(m+1). Hence, we have shown that A(m+1) holds.
This completes the induction step.
Thus, we have proven (by induction) that A(n) holds for all n ∈ N. In other words, for alln∈ N, the following holds:
if Sis a nonempty finite set of integers satisfying |S| −1=n, thenS has a maximum
(71)
50Proof. LetSbe a nonempty finite set of integers satisfying|S| −1=0. We must show thatShas a maximum.
Indeed, |S| = 1 (since |S| −1 = 0). In other words, S is a 1-element set. In other words, S={x}for some integerx. Consider thisx. Lemma 2.33 shows that the set{x}has a maximum.
In other words, the setShas a maximum (sinceS={x}). This completes our proof.
(because this is whatA(n) says).
Now, let S be a nonempty finite set of integers. We shall prove that S has a maximum.
Indeed, |S| ∈ N (since S is finite) and |S| > 0 (since S is nonempty); hence,
|S| ≥ 1. Thus, |S| −1 ≥ 0, so that |S| −1 ∈ N. Hence, we can define n ∈ N by n = |S| −1. Consider this n. Thus, |S| −1 = n. Hence, (71) shows that S has a maximum. This proves Theorem 2.35.
2.5.3. Conventions for writing induction proofs on derived quantities
Let us take a closer look at the proof we just gave. The definition of the statement A(n) was not exactly unmotivated: This statement simply says that Theorem 2.35 holds under the condition that |S| −1 = n. Thus, by introducing A(n), we have
“sliced” Theorem 2.35 into a sequence of statementsA(0),A(1),A(2), . . ., which then allowed us to prove these statements by induction on n even though no “n”
appeared in Theorem 2.35 itself. This kind of strategy applies to various other problems. Again, we don’t need to explicitly define the statement A(n) if it is simply saying that the claim we are trying to prove (in our case, Theorem 2.35) holds under the condition that |S| −1 = n; we can just say that we are doing
“induction on|S| −1”. More generally:
Convention 2.36. Let B be a logical statement that involves some variables v1,v2,v3, . . .. (For example, B can be the statement of Theorem 2.35; then, there is only one variable, namely S.)
Letqbe some expression (involving the variablesv1,v2,v3, . . . or some of them) that has the property that whenever the variablesv1,v2,v3, . . . satisfy the assump-tions ofB, the expressionq evaluates to some nonnegative integer. (For example, if B is the statement of Theorem 2.35, then q can be the expression |S| −1, be-cause it is easily seen that ifS is a nonempty finite set of integers, then |S| −1 is a nonnegative integer.)
Assume that you want to prove the statement B. Then, you can proceed as follows: For each n∈ N, define A(n) to be the statement saying that51
(the statementB holds under the condition that q =n). Then, prove A(n) by induction onn. Thus:
• Theinduction base consists in proving that the statementB holds under the condition that q=0.
• Theinduction step consists in fixing m ∈ N, and showing that if the state-ment B holds under the condition that q =m, then the statementB holds under the condition that q =m+1.
Once this induction proof is finished, it immediately follows that the statement B always holds.
This strategy of proof is called “induction on q” (or “induction over q”). Once you have specified what q is, you don’t need to explicitly define A(n), nor do you ever need to mentionn.
Using this convention, we can rewrite our above proof of Theorem 2.35 as fol-lows:
First proof of Theorem 2.35 (second version). It is easy to see that |S| −1 ∈ N 52. Hence, we can apply induction on|S| −1 to prove Theorem 2.35:
Induction base: Theorem 2.35 holds under the condition that |S| −1 =0 53. This completes the induction base.
Induction step: Let m∈ N. Assume that Theorem 2.35 holds under the condition that |S| −1= m. We shall now show that Theorem 2.35 holds under the condition that|S| −1=m+1.
We have assumed that Theorem 2.35 holds under the condition that|S| −1=m.
In other words,
if Sis a nonempty finite set of integers satisfying |S| −1=m, then Shas a maximum
. (72) Now, let Sbe a nonempty finite set of integers satisfying |S| −1=m+1. There exists somet ∈ S(sinceSis nonempty). Consider this t. We have(S\ {t})∪ {t} = S∪ {t}=S (sincet∈ S).
Fromt∈ S, we obtain|S\ {t}| =|S| −1=m+1>m≥0 (sincem∈ N). Hence, the setS\ {t} is nonempty. Furthermore, this set S\ {t} is finite (since S is finite) and satisfies |S\ {t}| −1 = m (since |S\ {t}| = m+1). Hence, (72) (applied to S\ {t}instead ofS) shows thatS\ {t}has a maximum. Also, Lemma 2.33 (applied to x = t) shows that the set {t} has a maximum, namely t. Hence, Proposition 2.34 (applied to P = S\ {t} and Q = {t}) shows that the set (S\ {t})∪ {t} has a maximum. Since (S\ {t})∪ {t} = S, this rewrites as follows: The set S has a maximum.
Now, forget that we fixed S. We thus have shown that if S is a nonempty finite set of integers satisfying |S| −1 = m+1, then S has a maximum. In other words, Theorem 2.35 holds under the condition that |S| −1 = m+1. This completes the induction step. Thus, the induction proof of Theorem 2.35 is complete.
51We assume that no variable named “n” appears in the statementB; otherwise, we need a different letter for our new variable in order to avoid confusion.
52Proof.We have|S| ∈N(sinceSis finite) and|S|>0 (sinceSis nonempty); hence,|S| ≥1. Thus,
|S| −1∈N, qed.
53Proof. LetSbe as in Theorem 2.35, and assume that|S| −1=0. We must show that the claim of Theorem 2.35 holds.
Indeed, |S| = 1 (since |S| −1 = 0). In other words, S is a 1-element set. In other words, S={x}for some integerx. Consider thisx. Lemma 2.33 shows that the set{x}has a maximum.
In other words, the setShas a maximum (sinceS={x}). In other words, the claim of Theorem 2.35 holds. This completes our proof.
We could have shortened this proof even further if we didn’t explicitly state (72), but rather (instead of applying (72)) said that “we can apply Theorem 2.35 toS\ {t} instead ofS”.
Let us stress again that, in order to prove Theorem 2.35 by induction on |S| −1, we had to check that|S| −1∈ NwheneverSsatisfies the assumptions of Theorem 2.35.54 This check was necessary. For example, if we had instead tried to proceed by induction on |S| −2, then we would only have proven Theorem 2.35 under the condition that |S| −2 ∈ N; but this condition isn’t always satisfied (indeed, it misses the case whenSis a 1-element set).
2.5.4. Vacuous truth and induction bases
Can we also prove Theorem 2.35 by induction on |S| (instead of |S| −1)? This seems a bit strange, since|S|can never be 0 in Theorem 2.35 (because Sis required to be nonempty), so that the induction base would be talking about a situation that never occurs. However, there is nothing wrong about it, and we already do talk about such situations oftentimes (for example, every time we make a proof by contradiction). The following concept from basic logic explains this:
Convention 2.37. (a) A logical statement of the form “if A, then B” (where A and B are two statements) is said to be vacuously true if A does not hold. For example, the statement “if 0 = 1, then every set is empty” is vacuously true, because 0 = 1 is false. The statement “if 0 = 1, then 1 = 1” is also vacuously true, although its truth can also be seen as a consequence of the fact that 1 = 1 is true.
By the laws of logic, a vacuously true statement is always true! This may sound counterintuitive, but actually makes perfect sense: A statement “if A, then B” only says anything about situations where A holds. If A never holds, then it therefore says nothing. And when you are saying nothing, you are certainly not lying.
The principle that a vacuously true statement always holds is known as “ex falso quodlibet” (literal translation: “from the false, anything”) or “principle of explosion”. It can be restated as follows: From a false statement, any statement follows.
(b) Now, let X be a set, and let A(x) and B(x) be two statements de-fined for each x ∈ X. A statement of the form “for each x ∈ X satisfying A(x), we have B(x)” will automatically hold if there exists no x ∈ X satis-fying A(x). (Indeed, this statement can be rewritten as “for each x ∈ X, we have(ifA(x), thenB(x))”; but this holds because the statement “if A(x), then B(x)” is vacuously true for each x ∈ X.) Such a statement will also be called vacuously true.
54In our first version of the above proof, we checked this at the end; in the second version, we checked it at the beginning of the proof.
For example, the statement “ifn ∈Nis both odd and even, thenn =n+1” is vacuously true, since no n∈ Ncan be both odd and even at the same time.
(c) Now, let X be the empty set (that is, X = ∅), and let B(x) be a statement defined for each x ∈ X. Then, a statement of the form “for each x ∈ X, we have B(x)” will automatically hold. (Indeed, this statement can be rewritten as “for eachx ∈ X, we have(ifx ∈ X, thenB(x))”; but this holds because the statement
“if x ∈ X, thenB(x)” is vacuously true for eachx ∈ X, since its premise (x∈ X) is false.) Again, such a statement is said to bevacuously true.
For example, the statement “for each x∈ ∅, we have x6=x” is vacuously true (because there exists no x∈ ∅).
Thus, if we try to prove Theorem 2.35 by induction on|S|, then the induction base becomes vacuously true. However, the induction step becomes more complicated, since we can no longer argue thatS\ {t}is nonempty, but instead have to account for the case when S\ {t} is empty as well. So we gain and we lose at the same time. Here is how this proof looks like:
Second proof of Theorem 2.35. Clearly, |S| ∈N(sinceSis a finite set). Hence, we can apply induction on|S| to prove Theorem 2.35:
Induction base: Theorem 2.35 holds under the condition that |S| = 0 55. This completes the induction base.
Induction step: Let m∈ N. Assume that Theorem 2.35 holds under the condition that|S| =m. We shall now show that Theorem 2.35 holds under the condition that
|S| =m+1.
We have assumed that Theorem 2.35 holds under the condition that|S| =m. In other words,
ifS is a nonempty finite set of integers satisfying |S| =m, then Shas a maximum
. (73) Now, letS be a nonempty finite set of integers satisfying |S| =m+1. We want to prove thatS has a maximum.
There exists some t ∈ S (since S is nonempty). Consider this t. We have (S\ {t})∪ {t} = S∪ {t} = S (since t ∈ S). Lemma 2.33 (applied to x = t) shows that the set {t} has a maximum, namelyt.
We are in one of the following two cases:
Case 1: We haveS\ {t} =∅.
Case 2: We haveS\ {t} 6=∅.
Let us first consider Case 1. In this case, we haveS\ {t} = ∅. Hence, S ⊆ {t}. Thus, eitherS = ∅or S = {t} (since the only subsets of {t} are∅and {t}). Since
55Proof. Let S be as in Theorem 2.35, and assume that |S| = 0. We must show that the claim of Theorem 2.35 holds.
Indeed,|S|=0, so thatSis the empty set. This contradicts the assumption thatSbe nonempty.
From this contradiction, we conclude that everything holds (by the “ex falso quodlibet” princi-ple). Thus, in particular, the claim of Theorem 2.35 holds. This completes our proof.
S = ∅ is impossible (because S is nonempty), we thus have S = {t}. But the set {t} has a maximum. In view of S = {t}, this rewrites as follows: The set S has a maximum. Thus, our goal (to prove that Shas a maximum) is achieved in Case 1.
Let us now consider Case 2. In this case, we have S\ {t} 6= ∅. Hence, the set S\ {t} is nonempty. From t ∈ S, we obtain |S\ {t}| = |S| −1 = m (since |S| = m+1). Furthermore, the set S\ {t} is finite (sinceS is finite). Hence, (73) (applied toS\ {t}instead of S) shows thatS\ {t} has a maximum. Also, recall that the set {t} has a maximum. Hence, Proposition 2.34 (applied toP =S\ {t} and Q={t}) shows that the set (S\ {t})∪ {t} has a maximum. Since (S\ {t})∪ {t} = S, this rewrites as follows: The setShas a maximum. Hence, our goal (to prove thatShas a maximum) is achieved in Case 2.
We have now proven that S has a maximum in each of the two Cases 1 and 2.
Therefore, Salways has a maximum (since Cases 1 and 2 cover all possibilities).
Now, forget that we fixedS. We thus have shown that ifSis a nonempty finite set of integers satisfying|S|=m+1, thenShas a maximum. In other words, Theorem 2.35 holds under the condition that|S| =m+1. This completes the induction step.
Thus, the induction proof of Theorem 2.35 is complete.
2.5.5. Further results on maxima and minima
We can replace “integers” by “rational numbers” or “real numbers” in Theorem 2.35; all the proofs given above still apply then. Thus, we obtain the following:
Theorem 2.38. Let S be a nonempty finite set of integers (or rational numbers, or real numbers). Then, Shas a maximum.
Hence, if S is a nonempty finite set of integers (or rational numbers, or real numbers), then maxS is well-defined (because Theorem 2.38 shows that S has a maximum, and Proposition 2.31 shows that this maximum is unique).
Moreover, just as we have defined maxima (i.e., largest elements) of sets, we can define minima (i.e., smallest elements) of sets, and prove similar results about them:
Definition 2.39. Let Sbe a set of integers (or rational numbers, or real numbers).
Aminimumof Sis defined to be an elements ∈ Sthat satisfies (s ≤t for eacht ∈ S).
In other words, a minimum ofS is defined to be an element ofSwhich is less or equal to each element of S.
(The plural of the word “minimum” is “minima”.)
Example 2.40. The set{2, 4, 5} has exactly one minimum: namely, 2.
The set N={0, 1, 2, . . .} has exactly one minimum: namely, 0.
The set {0,−1,−2, . . .}has no minimum: Ifkwas a minimum of this set, then we would have k≤k−1, which is absurd.
The set∅ has no minimum, since a minimum would have to be an element of
∅.
The analogue of Proposition 2.31 for minima instead of maxima looks exactly as one would expect it:
Proposition 2.41. Let S be a set of integers (or rational numbers, or real num-bers). Then, Shasat most one minimum.
Proof of Proposition 2.41. To obtain a proof of Proposition 2.41, it suffices to replace every “≥” sign by a “≤” sign (and every word “maximum” by “minimum”) in the proof of Proposition 2.31 given above.
Definition 2.42. Let S be a set of integers (or rational numbers, or real num-bers). Proposition 2.41 shows that S hasat most one minimum. Thus, ifS has a minimum, then this minimum is the unique minimum of S; we shall thus call it the minimum of S or the smallest element ofS. We shall denote this minimum by minS.
The analogue of Theorem 2.38 is the following:
Theorem 2.43. Let S be a nonempty finite set of integers (or rational numbers, or real numbers). Then, Shas a minimum.
Proof of Theorem 2.43. To obtain a proof of Theorem 2.43, it suffices to replace every
“≥” sign by a “≤” sign (and every word “maximum” by “minimum”) in the proof of Theorem 2.38 given above (and also in the proofs of all the auxiliary results that were used in said proof).56
Alternatively, Theorem 2.43 can be obtained from Theorem 2.38 by applying the latter theorem to the set {−s | s ∈ S}. In fact, it is easy to see that a number x is the minimum of S if and only if −x is the maximum of the set {−s | s∈ S}. We leave the details of this simple argument to the reader.
We also should mention that Theorem 2.43 holds without requiring that S be finite, if we instead require that Sconsist of nonnegative integers:
Theorem 2.44. Let S be a nonempty set of nonnegative integers. Then, S has a minimum.
ButS does not necessarily have a maximum in this situation; the nonnegativity requirement has “broken the symmetry” between maxima and minima.
56To be technically precise: not every “≥” sign, of course. The “≥” sign in “m≥ 0” should stay
56To be technically precise: not every “≥” sign, of course. The “≥” sign in “m≥ 0” should stay