2. A closer look at induction
2.6. Increasing lists of finite sets
We shall next study (again using induction) another basic feature of finite sets.
We recall that “list” is just a synonym for “tuple”; i.e., a list is ak-tuple for some k ∈N. Note that tuples and lists are always understood to be finite.
Definition 2.45. Let Sbe a set of integers. Anincreasing list ofSshall mean a list (s1,s2, . . . ,sk) of elements of S such thatS = {s1,s2, . . . ,sk} and s1 < s2 < · · · <
sk.
In other words, if S is a set of integers, then an increasing list of S means a list such that
• the setS consists of all elements of this list, and
• the elements of this list are strictly increasing.
For example,(2, 4, 6)is an increasing list of the set {2, 4, 6}, but neither(2, 6) nor (2, 4, 4, 6) nor (4, 2, 6) nor (2, 4, 5, 6) is an increasing list of this set. For another ex-ample,(1, 4, 9, 16)is an increasing list of the set
i2 | i∈ {1, 2, 3, 4} ={1, 4, 9, 16}. For yet another example, the empty list()is an increasing list of the empty set∅.
Now, it is intuitively obvious that any finite set S of integers has a unique in-creasing list – we just need to list all the elements of S in increasing order, with no repetitions. But from the viewpoint of rigorous mathematics, this needs to be proven. Let us state this as a theorem:
Theorem 2.46. LetSbe a finite set of integers. Then,Shas exactly one increasing list.
Before we prove this theorem, let us show some auxiliary facts:
Proposition 2.47. Let S be a set of integers. Let (s1,s2, . . . ,sk) be an increasing list of S. Then:
(a)The set Sis finite.
(b)We have |S|=k.
(c)The elements s1,s2, . . . ,sk are distinct.
Proof of Proposition 2.47. We know that (s1,s2, . . . ,sk) is an increasing list of S. In other words, (s1,s2, . . . ,sk) is a list of elements of S such that S = {s1,s2, . . . ,sk} and s1 <s2<· · · <sk (by the definition of an “increasing list”).
From S = {s1,s2, . . . ,sk}, we conclude that the set S has at most k elements.
Thus, the setSis finite. This proves Proposition 2.47 (a).
We have s1 < s2 < · · · < sk. Hence, if u and v are two elements of {1, 2, . . . ,k} such that u < v, then su < sv (by Corollary 2.20, applied to ai = si) and there-fore su 6= sv. In other words, the elements s1,s2, . . . ,sk are distinct. This proves Proposition 2.47(c).
Thek elementss1,s2, . . . ,sk are distinct; thus, the set {s1,s2, . . . ,sk} has sizek. In other words, the setShas size k(sinceS={s1,s2, . . . ,sk}). In other words, |S|=k.
This proves Proposition 2.47(b).
Proposition 2.48. The set ∅ has exactly one increasing list: namely, the empty list ().
Proof of Proposition 2.48. The empty list () satisfies ∅ ={}. Thus, the empty list () is a list(s1,s2, . . . ,sk) of elements of ∅such that ∅ ={s1,s2, . . . ,sk} and s1 < s2 <
· · · < sk (indeed, the chain of inequalities s1 < s2 < · · · < sk is vacuously true for the empty list (), because it contains no inequality signs). In other words, the empty list () is an increasing list of ∅ (by the definition of an increasing list). It remains to show that it is the only increasing list of∅.
Let (s1,s2, . . . ,sk) be any increasing list of ∅. Then, Proposition 2.47 (b) (ap-plied to S = ∅) yields |∅| = k. Hence, k = |∅| = 0, so that (s1,s2, . . . ,sk) = (s1,s2, . . . ,s0) = ().
Now, forget that we fixed(s1,s2, . . . ,sk). We thus have shown that if(s1,s2, . . . ,sk) is any increasing list of∅, then (s1,s2, . . . ,sk) = (). In other words, any increasing list of ∅ is (). Therefore, the set ∅ has exactly one increasing list: namely, the empty list() (since we already know that() is an increasing list of∅). This proves Proposition 2.48.
Proposition 2.49. Let Sbe a nonempty finite set of integers. Letm =maxS. Let (s1,s2, . . . ,sk)be any increasing list of S. Then:
(a)We have k≥1 andsk =m.
(b)The list (s1,s2, . . . ,sk−1) is an increasing list ofS\ {m}.
Proof of Proposition 2.49. We know that (s1,s2, . . . ,sk) is an increasing list of S. In other words, (s1,s2, . . . ,sk) is a list of elements of S such that S = {s1,s2, . . . ,sk} and s1 <s2<· · · <sk (by the definition of an “increasing list”).
Proposition 2.47 (b) yields |S| = k. Hence, k = |S| > 0 (since S is nonempty).
Thus, k ≥ 1 (since k is an integer). Therefore, sk is well-defined. Clearly, k ∈ {1, 2, . . . ,k} (sincek ≥1), so thatsk ∈ {s1,s2, . . . ,sk} =S.
We haves1< s2 <· · · <sk and thus s1 ≤s2 ≤ · · · ≤ sk. Hence, Proposition 2.18 (applied to ai =si) shows that ifuand vare two elements of {1, 2, . . . ,k} such that u≤v, then
su ≤sv. (76)
Thus, we have (sk ≥tfor each t∈ S) 57. Hence, sk is an element s ∈ S that satisfies (s≥t for eacht ∈S) (since sk ∈ S). In other words, sk is a maximum of S (by the definition of a maximum). Since we know that S has at most one maximum (by Proposition 2.31), we thus conclude that sk is themaximum ofS. In other words, sk = maxS. Hence, sk = maxS = m. This completes the proof of Proposition 2.49(a).
(b) From s1 < s2 < · · · < sk, we obtain s1 < s2 < · · · < sk−1. Furthermore, the elements s1,s2, . . . ,sk are distinct (according to Proposition 2.47 (c)). In other
57Proof. Lett∈ S. Thus,t∈S={s1,s2, . . . ,sk}. Hence,t=sufor some u∈ {1, 2, . . . ,k}. Consider thisu. Now,uandkare elements of{1, 2, . . . ,k}such thatu≤k(sinceu∈ {1, 2, . . . ,k}). Hence, (76) (applied tov=k) yieldssu ≤sk. Hence,sk≥su =t(sincet=su), qed.
words, for any two distinct elementsu andv of{1, 2, . . . ,k}, we have
su 6=sv. (77)
Hence,sk ∈ {/ s1,s2, . . . ,sk−1} 58. Now,
S
|{z}
={s1,s2,...,sk}
={s1,s2,...,sk−1}∪{sk}
\
m
|{z}=sk
= ({s1,s2, . . . ,sk−1} ∪ {sk})\ {sk}
={s1,s2, . . . ,sk−1} \ {sk} ={s1,s2, . . . ,sk−1} (since sk ∈ {/ s1,s2, . . . ,sk−1}). Hence, the elements s1,s2, . . . ,sk−1 belong to the set S\ {m} (since they clearly belong to the set {s1,s2, . . . ,sk−1} = S\ {m}). In other words,(s1,s2, . . . ,sk−1) is a list of elements ofS\ {m}.
Now, we know that(s1,s2, . . . ,sk−1) is a list of elements of S\ {m} such that S\ {m} ={s1,s2, . . . ,sk−1} and s1 <s2 <· · · <sk−1. In other words,(s1,s2, . . . ,sk−1) is an increasing list ofS\ {m}. This proves Proposition 2.49 (b).
We are now ready to prove Theorem 2.46:
Proof of Theorem 2.46. We shall prove Theorem 2.46 by induction on|S|:
Induction base: Theorem 2.46 holds under the condition that |S| = 0 59. This completes the induction base.
Induction step: Let g ∈ N. Assume that Theorem 2.46 holds under the condition that|S|= g. We shall now show that Theorem 2.46 holds under the condition that
|S| =g+1.
We have assumed that Theorem 2.46 holds under the condition that|S| = g. In other words,
ifS is a finite set of integers satisfying |S| =g, then Shas exactly one increasing list
. (78)
Now, letSbe a finite set of integers satisfying|S|= g+1. We want to prove that Shas exactly one increasing list.
The setS is nonempty (since |S| = g+1 > g ≥ 0). Thus, S has a maximum (by Theorem 2.35). Hence, maxSis well-defined. Setm =maxS. Thus,m=maxS ∈ S
58Proof. Assume the contrary. Thus, sk ∈ {s1,s2, . . . ,sk−1}. In other words, sk = su for some u∈ {1, 2, . . . ,k−1}. Consider thisu. We haveu∈ {1, 2, . . . ,k−1} ⊆ {1, 2, . . . ,k}.
Now,u ∈ {1, 2, . . . ,k−1}, so thatu ≤k−1< kand thusu6=k. Hence, the elementsu and kof {1, 2, . . . ,k}are distinct. Thus, (77) (applied to v = k) yieldssu 6=sk =su. This is absurd.
This contradiction shows that our assumption was wrong, qed.
59Proof. Let S be as in Theorem 2.46, and assume that |S| = 0. We must show that the claim of Theorem 2.46 holds.
Indeed,|S|=0, so thatSis the empty set. Thus,S=∅. But Proposition 2.48 shows that the set∅has exactly one increasing list. SinceS=∅, this rewrites as follows: The setShas exactly one increasing list. Thus, the claim of Theorem 2.46 holds. This completes our proof.
(by (67)). Therefore, |S\ {m}| = |S| −1 = g (since |S| = g+1). Hence, (78) [Proof of (79): If j+1 ≤ 1, then the chain of inequalities (79) is vacuously true (since it contains no inequality signs). Thus, for the rest of this proof of (79), we WLOG assume that we don’t havej+1≤1. Hence,j+1>1, so thatj>0 and thus j≥1 (sincejis an integer). Hence,tjis well-defined. We havej∈ {1, 2, . . . ,j}(since j ≥ 1) and thus tj ∈ t1,t2, . . . ,tj = S\ {m} ⊆ S. Hence, (68) (applied to t = tj) yields maxS≥tj. Hence,tj ≤maxS=m. Moreover,tj ∈ {/ m} (sincetj ∈S\ {m});
in other words, tj 6= m. Combining this with tj ≤ m, we obtain tj < m = tj+1. Combining the chain of inequalities t1 < t2 < · · · < tj with the single inequality tj < tj+1, we obtain the longer chain of inequalities t1 < t2 < · · · < tj < tj+1. In
is an increasing list ofS(by the definition of an “increasing list”). Hence, the set S has at least one increasing list (namely,
t1,t2, . . . ,tj+1
).
We shall next show that t1,t2, . . . ,tj+1
is the only increasing list of S. Indeed, let (s1,s2, . . . ,sk) be any increasing list of S. Then, Proposition 2.49 (a) shows that k ≥ 1 and sk = m. Also, Proposition 2.49(b) shows that the list (s1,s2, . . . ,sk−1) is an increasing list ofS\ {m}.
But recall thatS\ {m}has exactly one increasing list. Thus, in particular,S\ {m} hasat most one increasing list. In other words, any two increasing lists of S\ {m} are equal. Hence, the lists (s1,s2, . . . ,sk−1) and t1,t2, . . . ,tj
must be equal (since both of these lists are increasing lists ofS\ {m}). In other words,(s1,s2, . . . ,sk−1) =
t1,t2, . . . ,tj
. In other words,k−1= jand
(si =ti for eachi ∈ {1, 2, . . . ,k−1}). (81) Fromk−1= j, we obtaink =j+1. Hence, tk =tj+1 =m. Next, we claim that
si =ti for eachi∈ {1, 2, . . . ,k}. (82) [Proof of (82): Leti ∈ {1, 2, . . . ,k}. We must prove thatsi =ti. Ifi ∈ {1, 2, . . . ,k−1}, then this follows from (81). Hence, for the rest of this proof, we WLOG assume that we don’t have i ∈ {1, 2, . . . ,k−1}. Hence, i ∈ {/ 1, 2, . . . ,k−1}. Combining i∈ {1, 2, . . . ,k} withi ∈ {/ 1, 2, . . . ,k−1}, we obtain
i ∈ {1, 2, . . . ,k} \ {1, 2, . . . ,k−1} ={k}.
In other words,i=k. Hence,si =sk =m =tk (since tk =m). In view of k =i, this rewrites as si =ti. This proves (82).]
From (82), we obtain (s1,s2, . . . ,sk) = (t1,t2, . . . ,tk) = t1,t2, . . . ,tj+1
(since k = j+1).
Now, forget that we fixed(s1,s2, . . . ,sk). We thus have proven that if(s1,s2, . . . ,sk) is any increasing list of S, then (s1,s2, . . . ,sk) = t1,t2, . . . ,tj+1
. In other words, any increasing list of S equals t1,t2, . . . ,tj+1
. Thus, the set S has at most one increasing list. Since we also know that the setShasat leastone increasing list, we thus conclude thatS has exactly one increasing list.
Now, forget that we fixed S. We thus have shown that
ifS is a finite set of integers satisfying |S| =g+1, thenS has exactly one increasing list
.
In other words, Theorem 2.46 holds under the condition that |S| = g+1. This completes the induction step. Hence, Theorem 2.46 is proven by induction.
Definition 2.50. Let S be a finite set of integers. Theorem 2.46 shows that S has exactly one increasing list. This increasing list is called the increasing list ofS. It is also calledthe list of all elements of S in increasing order (with no repetitions). (The latter name, of course, is descriptive.)
The increasing list ofShas length |S|. (Indeed, if we denote this increasing list by (s1,s2, . . . ,sk), then its length is k = |S|, because Proposition 2.47 (b) shows that |S|=k.)
For each j∈ {1, 2, . . . ,|S|}, we define the j-th smallest element of S to be thej-th entry of the increasing list of S. In other words, if(s1,s2, . . . ,sk) is the increasing list of S, then the j-th smallest element of Sis sj. Some say “j-th lowest element ofS” instead of “j-th smallest element ofS”.
Remark 2.51. (a) Clearly, we can replace the word “integer” by “rational num-ber” or by “real numnum-ber” in Proposition 2.18, Corollary 2.19, Corollary 2.20, Definition 2.45, Theorem 2.46, Proposition 2.47, Proposition 2.48, Proposition 2.49 and Definition 2.50, because we have not used any properties specific to integers.
(b) If we replace all the “<” signs in Definition 2.45 by “>” signs, then we obtain the notion of a decreasing list of S. There are straightforward analogues of Theorem 2.46, Proposition 2.47, Proposition 2.48 and Proposition 2.49 for de-creasing lists (where, of course, the analogue of Proposition 2.49 uses minS in-stead of maxS). Thus, we can state an analogue of Definition 2.50 as well. In this analogue, the word “increasing” is replaced by “decreasing” everywhere, the word “smallest” is replaced by “largest”, and the word “lowest” is replaced by “highest”.
(c) That said, the decreasing list and the increasing list are closely related:
If S is a finite set of integers (or rational numbers, or real numbers), and if (s1,s2, . . . ,sk) is the increasing list of S, then (sk,sk−1, . . . ,s1) is the decreasing list of S. (The proof is very simple.)
(d) Let S be a nonempty finite set of integers (or rational numbers, or real numbers), and let (s1,s2, . . . ,sk) be the increasing list of S. Proposition 2.49 (a) (applied to m = maxS) shows that k ≥ 1 and sk = maxS. A similar argument can be used to show that s1 = minS. Thus, the increasing list of S begins with the smallest element of S and ends with the largest element of S (as one would expect).