Equivalent Sets
Definition Let Eß F be sets. We say that E is equivalent to F iff there exists a bijection 0ÀEÄF . If E is equivalent to F , we write E¸FÐ or E¶F or EµF or something similar: the notation varies from book to bookÑ .
It is intuitively clear that for finite sets E¸F iff E and F have the same number of elementsÞ Therefore Ö+ß ,× ¸ Ö"ß #× (assuming that +Á, ) but Ö"ß #ß $× ¸Î Ö"ß #×Þ
Theorem The relation ¸ is an equivalence relation among sets.
Proof Let EFßG , be sets.
a) The identity mapping 0ÐBÑœB is a bijection 0ÀEÄE . Therefore E¸E , so the relation is reflexive.
b) If E¸F , then there must exist a bijection 0ÀEÄFÞ Then the function 1œ0" ÀFÄE is also a bijection, so F¸EÞ Therefore the relation is symmetric . (Therefore, to show two specific sets E+8. B are equivalent, it doesn't matter whether you show that there is a bijection from EF to or that there is a bijection from FE to .)
cÑ Suppose E¸F and F¸G . Then there are bijections 0ÀEÄF and 1ÀFÄG . Then 1‰0 ÀEÄG is a bijection (check! ) so E¸GÞ Therefore the relation is transitive.
Examples
1) Suppose +ß , −‘ and that + ,Þ Then the intervals Ð!ß "Ñ and Ð+ß ,Ñ are equivalent. We can see this using a “straight line” bijection 0 À Ð!ß "Ñ Ä Ð+ß ,Ñ À
0ÐBÑœ+Ð,+ÑB !B"
2) Suppose -ß . −‘ and that - .Þ By Example 1), Ð!ß "Ñ ¸ Ð-ß .Ñ . Since it's also true that Ð!ß "Ñ ¸ Ð+ß ,Ñ , it follows by symmetry and transitivity that Ð-ß .Ñ ¸ Ð+ß ,Ñ À any two “open intervals” in ‘ are equivalent.
3) Suppose, in Example 1), we change the domain of 0 to !ŸB"Þ Then the graph of 0Ð!ß+Ñ0 will also include the point , and the modified function is a bijection between Ò!ß "Ñ and Ò+ß ,ÑÞ Therefore Ò!ß "Ñ ¸ Ò+ß ,Ñ , and it follows that Ò-ß .Ñ ¸ Ò+ß ,Ñ , just as in Example 2).
Similarly, we can “tweak” the domain of 0 in Example 1) to show that
Ð-ß.Ó¸Ò+ß,Ó and that Ò-ß.Ó¸Ò+ß,Ó
4) Ð!ß "Ó ¸ Ò!ß "Ñ because the function 0 À Ð!ß "Ó Ä Ò!ß "Ñ given by 0ÐBÑ œ " B is a bijection. Using “domain modifications” like to those in Example 3), show that Ð-ß.Ó¸Ò+ß,Ñ.
11 11 5) (ßѸ##‘‘ because tan ÀßÑÄ ( ## is a bijection. By Example 2), this means that every interval Ð+ß ,Ñ ¸‘ Þ
6) ‘‘¸ Ð!ß ∞Ñ because the function 0 À Ä Ð!ß ∞Ñ where 0ÐBÑ œ /B ß is a bijection.
Examples
1) =¸ because the function 0 À Ö!ß "ß #ß ÞÞÞß × Ä Ö"ß #ß $ß ÞÞÞß × given by 0ÐBÑ œ B" is a bijection (check !)
2) Let 5 − . Then the function 0 À Ö"ß #ß $ß ÞÞÞ× Ä Ö5ß #5ß $5ß ÞÞÞ× given by 0Ð8Ñ œ 58 is a bijection. So, for example, ¸ Ö#ß %ß 'ß )ß ÞÞÞ× and ¸ Ö"&ß $!ß %&ß '!ß ÞÞÞ×Þ By symmetry and transitivity, Ö#ß %ß 'ß )ß ÞÞÞ× ¸ Ö"&ß $!ß %&ß '!ß ÞÞÞ× .
3) 0 À ÖÞÞÞß $ß #ß "× Ä Ö"ß #ß $ß ÞÞÞ× given by 0ÐBÑ œ B is a bijection, so the set of negative integers is equivalent to the set of positive integers.
4) As a slightly more complicated example, we can also show that ™¸ . We can see this using the bijection 0À™ Ä given by
8 # if 8 is even 0Ð8Ñ œ "8 # if 8 is odd.
(For example, 0Ð#Ñ œ "ß 0Ð%Ñ œ #ß 0Ð'Ñ œ $ß ÞÞÞß 0Ð"Ñ œ !ß 0Ð$Ñ œ "ß 0Ð&Ñ œ #ß ÞÞÞ Ñ
0 is one-to-one : Suppose 0Ð7Ñ œ 0Ð8ÑÞ
Notice that if 7 is even and 8 is odd , then 0Ð7Ñ ! and 0Ð8Ñ Ÿ ! , so 0Ð7Ñ Á 0Ð8Ñ . Similarly, it cannot be that 7 is odd and 8 even. So 7ß 8 are either both even or both odd.
78 If 7ß 8 are both even, then 0Ð7Ñ œ 0Ð8Ñ Ê##œ Ê 7 œ 8 . "7 "8 If 7ß 8 are both odd, then 0Ð7Ñ œ 0Ð8Ñ Ê ##œ Ê7œ8 . 0 is onto ™ : Let D−™ À
#D if D!, let 8œ#D− À then 0Ð8Ñœ# œDÞ "Ð"#DÑ if D Ÿ !ßlet 8 œ " #D − Ðwhy? ) À then 0Ð8Ñ œ# œ DÞ
5) We saw earlier (near the end of the notes on functions) that there is a bijection 0À Ä œ the set of all positive integers, so ¸ Þ
Using the following theorem, we can combine certain sets that we know are equivalent to conclude that other sets are also equivalent (without needing to explicitly produce a bijection between them).
E¸G and Theorem Suppose that F¸H
Then i) E‚F¸G‚H ii) If E∩Fœg and G∩Hœgß then E∪F¸G∪HÞ
Proof Since E¸G and F¸H , we know that there are bijections 0ÀEÄG and 1ÀFÄHÞ
i) Define 2ÀE‚FÄG‚H by the formula À
For Ð+ß ,Ñ − E ‚ F , let 2Ð+ß ,Ñ œ Ð0Ð-Ñß 1Ð.ÑÑ − G ‚ H
2 is onto G‚HÀ Suppose Ð-ß.Ñ−G‚H . Then -−G so since 0 is onto, there is an + − E for which 0Ð+Ñ œ -Þ Similarly, there is a , − F for which 1Ð,Ñ œ .Þ Then Ð+ß ,Ñ − E ‚ F and 2Ð+ß ,Ñ œ Ð0Ð+Ñß 1Ð,ÑÑ œ Ð-ß .ÑÞ
2 is one-to-one ÀÐ+ß,ÑÐ+ß,Ñ−E‚F Suppose "" and ## and that 2Ð+"" ß , Ñ œ 2Ð+ ## ß , Ñ. By definition of 2 , this means that Ð0Ð+"" Ñß 1Ð, ÑÑ œ Ð0Ð+ ## Ñß 1Ð, ÑÑ , and therefore 0Ð+"# Ñ œ 0Ð+ Ñ and 1Ð, "# Ñ œ 1Ð, ÑÞ Since 0 and 1 are both one-to-one, this gives us that +" œ+ # and , " œ,Þ # Therefore Ð+ß,ÑœÐ+ß,ÑÞ "" ##
ii) domainÐ0Ñ ∩ domain Ð1Ñ œ E ∩ F œ g , so 0 ∪ 1 œ 2 is a function from E ∪ F to G∪H . (There are no points where 01 and “disagree” so there is no problem with 0ÐBÑif B − E 0∪1 being a function--see Homework 9). We can write 2ÐBÑ œ 1ÐBÑif B − FÞ
2 is onto G∪HÀ if C−G∪H , then C−G or C−H . IfC−G , then Ð since 0 is onto) there is an +−E for which Cœ0Ð+Ñœ2Ð+Ñ and if C − H , then there is a , − F for which C œ 1Ð,Ñ œ 2Ð,ÑÞ
In both cases, C− range Ð2Ñ . Therefore 2 is onto.
2ÀBßC−E∪F2ÐBÑœ2ÐCÑÞ is one-to-one Suppose and
IfB−E and C−F (or vice versa ) ß then 2ÐBÑœ0ÐBÑ−G and 2ÐBÑœ2ÐCÑœ1ÐCÑ−HÞ This would mean G∩HÁg . So either B and C are in EB , or and C are in F .
If Bß C − E , then 2ÐBÑ œ 0ÐBÑ œ 0ÐCÑ œ 2ÐCÑ so, since 0 is one-to-one, B œ CÞ If Bß C − F , then 2ÐBÑ œ 1ÐBÑ œ 1ÐCÑ œ 2ÐCÑ , so B œ C since 1 is one-to-one. ñ
Examples
Ð!ß "Ñ ¸ ‘ and 1) Ð!ß "Ñ ¸ ‘
Part i)( of the Theorem gives us that Ð!ß "Ñ ‚ Ð!ß "Ñ ¸‘‘ ‚ that is, the “open box” Ð!ß "Ñ## in the plane is equivalent to the whole plane ‘ .
Using part ii) of the theorem again, we get that the “open box” Ð!ß "Ñ$$ in three-space ‘ is equivalent to all of ‘$Þ
2) Since Ð!ß "Ó ¸ Ð+ß ,Ó , we conclude that Ð!ß "Ó## ¸ Ð+ß ,Ó
Ð Ð!ß "Ó# is a “box” in the plane that contains its top and right edges, but not its left or bottom edges: draw it.)
With all these examples, we might be tempted to think that all infinite sets are equivalent but this is not true. The next theorem gives us a way to create (infinitely many) nonequivalent infinite sets. Theorem For any set EE¸ÐEÑÞ , Î c
Proof For a finite set EE8ÐEÑœ this is clear because if contains elements, then c the set of all subsets of E# contains 8 elements.
Suppose, in general, that 00ÀEÄÐEÑÞ0 is any function c We claim that cannot be onto and therefore 0E¸ÐEÑ cannot be a bijection. This will show that Î c .
If + − E , then 0Ð+Ñ −c ÐEÑ , so 0Ð+Ñ is a subset of E. Therefore it makes sense to ask whether or not +−0Ð+ÑÞ
It will help in understanding the proof to keep referring to the following illustration of what's going on:
If E œ Ö"ß #ß $× , then c ÐEÑ œ Ö g ß Ö"×ß Ö#×ß Ö$×ß Ö"ß #×ß Ö"ß $×ß Ö#ß $×ß Ö"ß #ß $× ×Þ
Suppose (say) 0ÀEÄc ÐEÑ is the function given by
0Ð"Ñ œ Ö"ß #×ß 0Ð#Ñ œ Ö"ß $× , 0Ð$Ñ œ Ö"×
Then "−0Ð"Ñß #Â0Ð#Ñ and $Â0Ð$Ñ
Let FœÖ+−EÀ+Â0Ð+Ñ× . Then F©E , so F−c ÐEÑÞ We claim that F is not in the range of 0.
In the illustration, F œ Ö#ß $× , and, as promised, F is not in the range of 0 .
Suppose there were an B−E for which 0ÐBÑœFÞ Then either B−F or BÂFÞ
If B − F , then B must satisfy the membership requirement for F : that B Â 0ÐBÑÞ But then BÂ0ÐBÑœF , so this is impossible.
If B Â F , then B fails to meet the membership requirement for F : so B − 0ÐBÑÞ But then B−0ÐBÑœF which is impossible.
Either way, the assumption that there is an B0ÐBÑœF such that leads to a contradiction , so no such B0ñ can exist. Therefore is not onto.
Example By the theorem, c¸Ð/ ).
Notice that there is a one-to-one (not onto ) map 0Àc Ä Ð ÑÐ for example, let 0Ð8Ñ œ Ö8× −c Ð Ñ Ñ0ÀEÄÐEÑ. If c is such a one-to-one function, then E ¸ 0ÒEÓ œrange Ð0Ñ ©cc ÐEÑ . So E is always equivalent to a subset of ÐEÑ but never equivalent to the whole set ccÐEÑ . Intuitively, this means that ÐEÑ is a “bigger” infinite set than E.
We can continue to apply the power set operation over and over to get ¸Ð// c ), c ÐѸÐÐÑ cc ) , cc ÐÐѸ ) / ccc ( ÐÐÑ ) ), ... and thus create a sequence of “larger” and “larger” nonequivalent infinite sets. We proved that for every set EE¸ÐEÑ , Î cc . In particular, ¸ÐÑÞ/ However, we can show that cÐ Ñis equivalent to another familiar set: Ö!ß "× . Recall that Ö!ß "× denotes the set of all functions 0 À Ä Ö!ß "×Þ Such an 0 is a sequence for which every term has value ! , or " :
if 0 − Ö!ß "× , we have values 0Ð"Ñß 0Ð#Ñß 0Ð$Ñß ÞÞÞß 0Ð8Ñß ÞÞÞ where each 0Ð8Ñ œ ! or "
In other words, Ö!ß "× is the set of all “binary sequences.”
Theorem cÐ Ñ ¸ Ö!ß "×
Proof Define a function FcÀ Ö!ß "× Ä Ð Ñ as follows:
if 0 − Ö!ß "× , then F Ð0Ñ œ Ö8 − À 0Ð8Ñ œ "× − c Ð Ñ
If 0Ð0Ñ8 is a binary sequence, then F is the subset of consisting of all those for which the 8"0>2 term of the sequence is . For example suppose has values
0Ð"Ñß 0Ð#Ñß 0Ð$Ñß 0Ð%Ñß 0Ð&Ñß 0Ð'Ñß ÞÞÞ ll ll ll ll ll ll !ß "ß "ß !ß !ß " ß ÞÞÞ
then FÐ0Ñ œ Ö#ß $ß 'ß ÞÞÞ ×Þ
F is one-to-one : if 0ß 1 − Ö!ß "× and 0 Á 1ß then there is an 8 − for which 0Ð8Ñ Á 1Ð8ÑÞ Suppose 0Ð8Ñ œ " and 1Ð8Ñ œ !Þ Then 8 −FFFF Ð0Ñ but 8  Ð1Ñ , so Ð0ÑÁ Ð1Ñ . Similarly if 0Ð8Ñ œ ! and 1Ð8Ñ œ " , then 8 ÂFF Ð0Ñ but 8 − Ð1Ñ , so FFÐ0Ñ Á Ð1Ñ.
F is onto: Suppose E−c Ð Ñ , that is, suppose E© Þ Define a binary sequence 0 as follows:
"8−Eif 0Ð8Ñ œ !8ÂEif
Then 0 − Ö!ß "× and, by definition, F Ð0Ñ œ E . ñ
Finite Sets
We have informally used the terms “finite set” and “infinite set.” We now make a precise definition.
Definition For each 5 − ß let5 œ Ö"ß #ß ÞÞÞß 5× . A set E is called finite if Eœg or if Ðb5 − Ñ E ¸5 Þ E is called infinite if E is not finite (that is, if EÁg and Ða5− ÑE¸/ 5 Þ
Definition For a finite set EÀ i) if Eœgß we say that ! is the cardinal number of E ÐEor that has cardinality 0ÑlElœ!Þ , and write
ii) if E¸5 ß we say that 5 is the cardinal number of E ÐE or that has cardinality 5Ñß and write lEl œ 5Þ
We already have a lot of intuitive ideas about how finite sets behave. But now that we have an “official” definition of “finite set,” each of these intuitive statements should be proved: this will show that the formal definition of finite set correctly “captures” the intuitive idea of finite set. It turns out that we can prove that our informal ideas about finite sets are actually theorems using the official formal definition for finite sets. (As we shall see later: we also have some intuitive ideas about infinite sets. But for infinite sets our intuition is sometimes wrong: for infinite sets, careful proofs become very important.)
Here is a list of familiar properties of finite sets. They are all proved in Section 5.1 of the textbook; many of the proofs use induction. We are not going to go over those proofs doing them all is a tedious, it takes a lot of time, and the results, in the end, are just what we thought intuitively would be true. But we will prove one important fact about finite sets, below.
Theorems About Finite Sets
1. If EE¶FF is finite and , then is finite; a finite set is never equivalent to a proper subset of itself.
2. If E is finite and B  E , then E ∪ ÖB× is finite, and lE ∪ ÖB×l œ lEl "Þ
3. If E is finite and B − E , then E ÖB× is finite and lE ÖB×l œ lEl "Þ
4. Every subset of a finite set is finite.
5. If EF and are finite:
a) lE ∪ Fl œ lEl lFl lE ∩ Fl b) If E ∩ F œ gßthen lE ∪ Fl œ lEl lFlÞ
6. If E is finite, then cc ÐEÑ is finite, and | ÐEÑl œ #lEl Þ
7. If 8 − and E"8 ß ÞÞÞß E are finite sets, then E "# ∪ E ∪ ÞÞÞ ∪ E 8 is finite. (“A finite union of finite sets is finite. ”)
There is one additional theorem about finite sets that is quite intuitive and which turns out to be unexpectedly useful.
Theorem (The Pigeonhole Principle) Suppose 8ß < − and that 0 À8< Ä Þ If 8 < , then 0 is not one-to-one. The idea behind the name is that if there are 8< “pigeons” to be put into “pigeonholes” and there are more pigeons Ð8Ñ than there are pigeonholes Ð<Ñ , then at least two pigeons must go into the same pigeonhole. The Pigeonhole Principle is also called the Dirichlet Box Principle .
Proof The proof is done by induction on . Suppose < represents a natural number and that 0À8< Ä Þ We want to prove Ða8− ) TÐ8Ñß where
TÐ8Ñ is the statement: if 8 <ß then 0 cannot be one-to-one.
Since << "8 #Þ represents a natural number so and so Therefore our induction starts with 8œ#Þ
Base case If 8 œ # , then < œ " and 0 À#" œ Ö"ß #× Ä Ö"× œ Þ Therefore 0 must be constant: 0Ð"Ñœ"œ0Ð#ÑÞSo 0 is not one-to-one.
Assume that TÐ8Ñ is true for some particular value 8 œ 5Þ That is, our induction hypothesis is
(*) If 0À5< Ä (where < is any natural number 5Ñ , then 0 is not one-to-one.
We want to prove:
If 0ÀR5" Ä < (where < is any natural number 5"Ñß then 0 is not one-to-one
We prove this by contradiction.
Suppose there is a one-to-one function 0ÀR5" Ä < , where 5"< . (**)
0 À Ö"ß #ß ÞÞÞß 5ß 5 "× Ä Ö"ß #ß ÞÞÞß <×
0Ð5 "Ñ is one of the numbers "ß #ß ÞÞÞß <ÞDelete 5 " from the first set Ð leaving 5 Ñ and 0Ð5"Ñ from the second set (leaving < Ö0Ð5"Ñ× ). Then restricting the function 0 to the smaller domain we have a one-to-one function
1 œ 0l55 À Ä < Ö0Ð5"Ñ×
The set on the right Ðœ the codomain of 1Ñ has one less element than < , so there is a bijection 2À<<" Ö0Ð5"Ñ×Ä Þ
Composing functions, we get a one -to -one function 2‰1À5<" Ä ß
Since 5"<ß we have that 5<" , and the existence of such a one-to-one function 2‰1 contradicts the induction hypothesis (*).
So no such function as (**) can exist TÐ5"Ñ which says that is true.
By PMI, TÐ8Ñ is true for all 8 − . ñ
As mentioned earlier, this “obvious” Pigeonhole Principle has lots of applications. Before we look at a few examples, we note one simple corollary. Corollary If :ß ; − and : Á ; , then :;Þ ¸Î
Proof One of :; or is larger than the other say :;Þ0ÀÄ If :; is any function, the, by the Pigeonhole Principle, 00ñ cannot be one-to-one, so cannot be a bijection.
The corollary tells us that a set E:Á;À cannot be equivalent to both :; and if otherwise, by transitivity, we would have :;¸Þ This means that the cardinal number of E is well- defined: if lEl œ :ß then lEl œ ; is impossible if ; Á :Þ
Example There are two people in Missouri who have the same number of hairs on their heads.
In 2007, the census gives the estimated Missouri population as 5,878,415. Also, various studies indicate that the average number of hairs on a human head is around 150,000, with, of course, lots of variation. (Incidentally, the average number varies with the color of the hair!) But it seems safe to say that every human head has less than 1,000,000 hairs.
Let the population of Missouri be 8 (where 8 is some number &ß )()ß %"& ). Let < œ "ß !!!ß !!!Þ Number the people as "ß #ß $ß ÞÞÞß 8Þ
Define a function 0 À Ö"ß #ß $ß ÞÞÞß 5ß ÞÞÞ8× Ä Ö"ß #ß ÞÞÞß <× where 0Ð5Ñ œ the number of hairs on 5 's head. Since 8< , the Pigeonhole Principle says that 0 cannot be one-to-one, that is, there are at least two Missourians with the same number of hairs on their heads.
In this example, the “pigeons” are the people in Missouri; you can think of the “pigeonholes” as boxes numbered "ß ÞÞÞß < . Each person is placed in the box corresponding to the number of hairs on the head.
In using the Pigeonhole Principle, sometimes some cleverness is required: the key thing (after realizing that the Pigeonhole Principle might solve the problem) is to clearly identify the “pigeons” and the “pigeonholes.” Example Suppose WR is a set of natural numbers. Show that there is a subset of W whose sum is divisible by RÞ
For instance, if W œ Ö&ß "#ß "*ß $"× , then R œ % and the subset Ö&ß "#ß $"× works, and if W œ Ö#ß "*ß $$×ß then R œ $ and the subset Ö$$× works; so does the subset Ö#ß "*ß $$×Þ
Proof Let W œ Ö+"# ß + ß ÞÞÞß +R × where each + 3 − Þ Consider the R subsets Ö+""#"#3 ×ß Ö+ ß + ×ß ÞÞÞß Ö+ ß + ß ÞÞÞß + ×ß ÞÞÞß Ö+ "#R ß + ß ÞÞÞß + ×Þ
If any one of the sums ++ÞÞÞ+"# 3 is divisible by Rß then the subset Ö+"# ß + ß ÞÞÞß + 3 × works and we are done. So assume that none of these sums is divisible by RÞ Therefore each of these sums has one of the remainders "ß #ß ÞÞÞß R " when divided by RÞ
Since there are R sums +"" ß + + # ß ÞÞÞß + " ÞÞÞ + R but only R " possible remainders, two of these sums must have the same remainder when divided by R (the Pigeonhole Principle). In other words, there are two sums for which
+"3R"35 ÞÞÞ + ´ + ÞÞÞ + ÞÞÞ +
Subtracting +"3 ÞÞÞ + from both sides, we get that
! ´R3" + ÞÞÞ + 5 so RlÐ+3" ÞÞÞ + 5 ÑÞ Therefore the subset Ö+ 3" ß ÞÞÞß + 5 × works.
Here, in Case ii), the “pigeons” are the R sets Ö+""#R ×ß ÞÞÞ ß Ö+ ß + ß ÞÞÞß + × and the “pigeonholes” are the R" possible remainders: each set is put into a “pigeonhole” determined by the remainder when its sum is divided by RÞ Example Prove that there is a natural number whose digits are all " 's and which is divisible by 7777.
Consider the 7778 numbers "ß ""ß """ß ÞÞÞß ""ÞÞÞ"""" Å this last number contains ((() digits, all œ "Þ
Assign to each number (“pigeon” ) in the list, its remainder (“ the pigeonhole” ) when divided by ((((. There are only (((( possible remainders Ð< œ !ß "ß ÞÞÞß or ((('Ñß so two of the ((() listed numbers have the same remainder (the Pigeonhole Principle).
Suppose these two are + and , , where + ,Þ Then , ´(((( + , so , + is divisible by (((( .
However, not all the digits of , + are " 's: if , œ """ ÞÞÞ """"" and + œ """"" then , + œ """ ÞÞÞ !!!!! ends in a string of ! ' =
In fact, if there are 6" 's in ,ß and there are 5" 's in + , then ,+ consists of 65 1's followed 5 ,+ by 5! 's. We can “cancel out” all the ! 's in ,+ by dividing by "!À so "!5 is a natural number all of whose digits are " . Is it still divisible by (((( ?
Notice that (((( œ ( † "" † "!" , so that , + œ (((( † 7 œ ( † "" † "!" † 7 for some 7 − Þ
, + (†""†"!"†7 ,+ Therefore "!555œ # & œ ( † "" † "!" † 4 œ (((( † 4 for some 4 − Ðwe know "!5 is a natural number; the #& 's and 's in the denominator must cancel out with #& 's and 's in the prime 7 factorization of 7À#&55 œ4−ÑÞ
,+ So "!5 is a natural number with only " 's for digits and divisible by (((( . Example ()taken from the Spring 2008 Missouri MAA Mathematics Competition Consider the set of points ™™‚œÖÐDßAÑ−ÀDA ‘## and are both integers × in ‘ . Imagine having a palette containing 4 colors. You choose what color to paint the points in ™™‚ , in any way you like.
Prove that, when you're done, there must be four points in ™ ‚ ™ which all have the same color and which form the vertices of a rectangle.
Proof (We will use the pigeonhole principle, twice.)
1) For any set of 5 points in ™™‚ , at least two must have the same color (pigeonhole principle ). For example, in any “column of 5”œ ÖÐBß !Ñß ÐBß "Ñß ÐBß #Ñß ÐBß $Ñß ÐBß %Ñ× two points must have the same color.
2) For any “column of 5”, there are 4 ways to color the first point ÐBß "Ñ ; then 4 (independent) ways to color the second point ÐBß #Ñ ; ...; and finally 4 (independent) ways to color the 5th point ÐBß &Ñ. Altogether, there are % † % † % † % † % œ %& œ "!#% possible color patterns for a “column of 5.”
3) Look at the first "!#& “columns of & ” À
Column 1 œ ÖÐ"ß !Ñß Ð"ß "Ñß Ð"ß #Ñß Ð"ß $Ñß Ð"ß %Ñ× Column# œ ÖÐ#ß !Ñß Ð#ß "Ñß Ð#ß #Ñß Ð#ß $Ñß Ð#ß %Ñ× ã Column"!#& œ ÖÐ"!#&ß !Ñß Ð"!#&ß "Ñß Ð"!#&ß #Ñß Ð"!#&ß $Ñß Ð"!#&ß %Ñ×
Two of these “columns of 5” must have exactly the same color pattern (by the Pigeonhole principle, used again). Suppose two such columns are
Column4 œ Ö Ð4ß !Ñß Ð4ß "Ñß Ð4ß #Ñß Ð4ß $Ñß Ð4ß %Ñ× Column5 œ ÖÐ5ß!ÑßÐ5ß"Ñß Ð5ß#ÑßÐ5ß$Ñß Ð5ß%Ñ×
Our first observation 1) tells us that two points in Column4 have the same color: suppose they are Ð4ß 7Ñ and Ð4ß 8Ñ Ðwhere !Ÿ78Ÿ%ÑÞ
Since Column5Ð5ß7ÑÐ5ß8Ñ has the same color pattern, and are the same color as Ð4ß 7Ñ and Ð4ß 8Ñ .
Then Ð4ß 7Ñß Ð4ß 8Ñß Ð5ß 7Ñß and Ð5ß 8Ñ are the same color and are the vertices of a rectangle. ñ
Notice that the number of colors , 4, is not important for the argument . If there had been, sayß 2008 colors available in the palette, then simply consider “columns of 2009” rather than “columns of 5”. For each column, there are #!!)#!!* possible coloring patterns. Look at the first #!!)#!!* " “columns of 9.” All the logic in the argument works in exactly the same way (but the 4 points that you find might turn out to be the vertices of a very large rectangle!) Infinite Sets
We have defined finite and infinite sets and stated some properties of finite sets. One of these properties of finite sets, the Pigeonhole Principle, turned out to be particularly useful so we digressed a little to look at how the Pigeonhole Principle could be used. We now turn our attention to infinite sets, where our intuitive ideas can lead us astray: for example, a “whole” set (like „ ) can be equivalent to one of its proper subsets Ð like œ Ö#ß %ß 'ß ÞÞÞ×ÑÞ So we have to be careful not to jump to conclusions when making statements about infinite sets.
Theorem The set E0ÀÄEÞ is infinite if and only if there exists a one-to-one function
Proof (Ê ) Suppose E is infinite . Then
EÁg so we can pick an element +" −EÞ EÁÖ+×""" since Ö+׸ and E is not finite. So EÖ+×Ág"#" so we can pick an + −EÖ+× (and +Á+Ñ#" EÁÖ+ß+×"# since Ö+ß+׸ "# # and E is not finite. So EÖ+ß+×Ág"# so we can pick an + $ −EÖ+ß+× "# (and +Á+ß+Á+$"$# )
Continue to define the ++88 's by induction. Once is defined. then
E Á Ö+"# ß + ß ÞÞÞß + 8 ×since Ö+ "# ß + ß ÞÞÞß + 8 × ¸ 8 and E is not finite. So E Ö+"# ß + ß ÞÞÞß + 8 × Á g so we can pick an + 8" − E Ö+ "# ß + ß ÞÞÞß + 8 × (and +8" Á + " ß + # ß ÞÞÞß + 8 )
In this way we have inductively defined +8−8 for all and the terms in the sequence +"# ß + ß ÞÞÞß + 8 ß ÞÞÞ are all different (each was chosen in a way that makes it different from all its predecessors). Define 0À ÄE by 0Ð8Ñœ+Þ88 Since the + 's are all different, 0 is one-to- one.
For (ÉE ), we use the contrapositive: suppose is finite and show that no such one-to-one function 0 can exist.
If Eœg , then there are no functions 0À ÄE (see earlier discussion of ]Ñ\
If E ¸55 œ Ö"ß #ß ÞÞÞß 5 } ß then there is a bijection 2 À E Ä
If there were a one-to-one function 0À ÄEß then we could restrict the domain of 01œ0lÀÄEÞ to get a “new” one-to-one function 5" 5" Then 2‰1À5" Ä 5 is one-to-one which contradicts the Pigeonhole PrincipleÞ
So, if E0ÀÄEÞñ is finite, there is no one-to-one function Note: A function 0À ÄE is a sequence in EÞ The terms of the sequence are +"# œ 0Ð"Ñß + œ 0Ð#Ñß ÞÞÞß + 8 œ 0Ð8Ñß ÞÞÞ, and if 0 is one-to-one, then all these terms are different. So the preceding theorem could be paraphrased as: E is infinite iff there exists a sequence of distinct terms+"# ß + ß ÞÞÞß + 8 ß ÞÞÞ in E .
Example The point here is just to illustrate that our official definition of “infinite” gives results that coincide with some of our intuitive ideas about what sets are infinite.
1) The function 0À‘ Ä given by 0Ð8Ñœ8 is one-to-one. By the preceding theorem, ‘ is infinite.
" 2) The function 0 À Ä Ð!ß "Ñ given by 0Ð8Ñ œ8 is one-to-one, so the interval Ð!ß "Ñ is infinite.
3) Suppose E0ÀÄEE©F is infinite, so that there is a one-to-one function . If , then we can view 00ÀÄFÞF also as being a one-to-one function So is also infinite.
The following theorem tells us that if a subset E© is “too big to be finite,” then E is equivalent to the whole set „„ . For example, consider the set œ Ö#ß %ß 'ß ÞÞÞ× © . is infinite so (according to the following theorem) „„ is equivalent to all of . In the case of , this statement is rather obvious: we can easily see that 0À„ Ä given by 0Ð8Ñœ#8 is a bijection. But we want to prove it for an arbitrary infinite subset of .
Theorem If EE©E¸Þ is infinite and , then
Proof EÁgÞ By the Well-Ordering Principle (WOP), there is a smallest element in EÀ call it +Þ" Since E is infinite, there is a smallest element, call it +ß#"#"Þ in EÖ+×ß and + Á+ If we have chosen distinct points +"8 ß ÞÞÞß + − E , then (since E is infinite) E Ö+"8 ß ÞÞÞß + × Á gßso we let + 8" be the smallest element in E Ö+ "8 ß ÞÞÞß + × Ð and + 8" will be different from all the preceding+Ñ38 's . In this way, we have defined (inductively) + for all 8−, and all the +8 's are different. Then define a function 0À ÄE by 0Ð8Ñœ+à8 this 0 is a bijection. ñ
Definition The set E is called countable if
i) E is finite, or ii) EE¸Þ is infinite and
If E¸ , we use the symbol i!! to denote the cardinal number of EÀ l lœiÞ ( “ii ” is “aleph”, the first letter of the Hebrew alphabet. “! ” is read as “aleph-zero” or “aleph null” or “aleph naught” (more British). This notation was first used by George Cantor in his groundbreaking work of the theory of infinite sets (in the later part of the 19th century). We think of i! as an infinite number: it represents the “ number of elements” in the set (and in any set equivalent to ).
Examples Countable Sets
Finite
glglœ!
Ö+ß,ß-× lÖ+ß,ß-×lœ$
Infinite
„ œ Ö#ß %ß 'ß )ß ÞÞÞ×
™
Since „™¸¸¸ we have that i! œl„™ lœl lœl lœl l
The number of elements in each of these sets is the same: i!
Uncountable Sets
cÐÑ This set is clearly infinite, and ¸Î c ÐÑ ÐÑÑwe proved that for every set EE¸ÐE , Î c so the set is uncountable: lÐÑlÁic !
Note: the textbook also uses the term “denumerable” to mean “a countable set that is not finite.” Be aware that different books use the terms “denumerable,” “countable” (and the term “enumerable”) in slightly different ways.
We proved earlier that a set E0ÀÄE is infinite iff there exists a one-to-one function . Informally this means that an infinite set is “big enough” to have a one-to-one “copy” of put inside it Ð the “copy” is range Ð0ÑÑÞ In the same informal spirit, the next theorem says that a countable infinite set is “small enough” that a “copy” of the set can be put inside .
Theorem The set E0ÀEÄÞ is countable iff there exists a one-to-one function
Proof (É0ÀEÄE ) If there does exist a one-to-one function , then is equivalent to a subset of , namely, E ¸ range Ð0Ñ . If range Ð0Ñ is finite, then E is finite. But if range Ð0Ñ is an infinite subset of , then rangeÐ0Ñ ¸ (by the preceding Theorem) Þ By transitivity, then, E¸ Þ So E is countable.
(ÊE ) Suppose is countable. If lElœ!ß then Eœg and the empty function g ÀgÄ is one-to-one. If lEl œ 5 , then there is a bijection 0 À E Ä Ö"ß #ß ÞÞÞß 5× , and this same 0ÀEÄ is one-to-one If Eß in not finite, then by definition of countable ßE¸ , so there exists a bijection 0ÀEÄ , and, of course, this 0 is one-to-one. ñ
Corollary A subset of a countable set is countable.
Proof Suppose E0ÀEÄÞ is a countable set. Then there is a one-to-one function If F©E , then look at the restricted function 1œ0lFÞ Then 1ÀFÄ is one-to-one, so, by the theorem, Fñ is countable.
Corollary If FF©GG is an uncountable set and , then is uncountable. (More informally, a set that contains an uncountable set is uncountable.)
Proof If GF were a countable set, then its subset would have to be countable (by the preceding corollary. ñ
Next we will look at another example of a familiar set that is uncountable. Before doing so, we need to call attention to a fact about decimal representations of real numbers.
Two different decimal expansions can represent the same real number. For example, !Þ"! œ !Þ"!!!!!ÞÞÞ œ !Þ!****ÞÞÞ œ !Þ!* (why?). However, two different decimal expansions can represent the same real n umber only if one of the expansions ends in an infinite string of !* 's and the other ends in an infinite string of 's. Example The interval Ð!ß "Ñ is not equivalent to (and therefore, since the interval Ð!ß "Ñ is infinite, it is uncountable ).
Consider any 0 À Ä Ð!ß "Ñ . We will show 0 cannot be onto, so that no bijection can exist between and Ð!ß "Ñ .
Write decimal expansions of all the numbers in rangeÐ0Ñ À
0Ð"Ñœ<" œ!B . """#"$ B B ÞÞÞ B "8 ÞÞÞ
0Ð#Ñœ<# œ!B . #"###$ B B ÞÞÞ B #8 ÞÞÞ
0Ð$Ñœ<$ œ!B . $"$#$$ B B ÞÞÞ B $8 ÞÞÞ . . . 0Ð8Ñœ<8 œ!B . 8"8#8$ B B ÞÞÞB 88 ÞÞÞ ã Now define a real number Cœ!.CCC"#$ . . . C 8 . . . by setting
Cœ"if B Á 1 888 Cœ#888if B œ"
Then C−Ð!ß"Ñ Notice that
i) the decimal expansion of C is different from every decimal expansion in the th listC<8 specifically, the decimal expansions of and 8 disagree at the decimal place: CÁB888.
ii) the decimal expansion of C is not an alternate decimal representation for any of the
Therefore Ða8Ñ C Á <8 . In other words, C Â ran ( 0 ) and therefore 0 is not onto.
The argument in the example is referred to as “Cantor's diagonal argument.”
Example ‘ is uncountable.
i) One reason: Ð!ß "Ñ © ‘ and a set containing an uncountable set must be uncountable (see preceding corollary)
ii) Another reason: we saw earlier that Ð!ß "Ñ ¸ ‘ and a set equivalent to an uncountable set is uncountable (why ?)
iii) ‘‘¸ ‚ Ö!×ß so ‘ ‚ Ö!× is uncountable; and ‘ ‚ Ö!× © ‘## , so ‘ is uncountable. Similarly, ‘‘$% , , ... are each uncountable sets. Definition If a set E is equivalent to Ð!ß "Ñ , then we say that the cardinal number of the set E is - . More formally, if E ¸ Ð!ß "Ñ , then lEl œ -Þ
We think of - as an infinite cardinal number: it represents the “number of elements” in the set Ð!ß "Ñ, and in any set equivalent to Ð!ß "Ñ . For example,
- œ lÐ!ß "Ñl œ lÐ+ß ,Ñl œ l‘ l œ lÐ!ß ∞Ñl
In his development of set theory, Cantor chose the symbol “- ” for the cardinality of these sets because “c ontinuum” was an old name for the real number line.
At this point, we have not given a precise definition for “ ” between infinite cardinal numbers. However, since ‘©ß we intuitively expect that iŸ-à! and since ‘ ¸/ , that i-!! . When “ Ÿ ” and “ ” are officially defined ß “ i- ” turns out to be true.
Cantor conjectured, but was unable to prove , that there are no cardinal numbers of “intermediate size” between i-! and . This conjecture is called the “Continuum Hypothesis .” A complete understanding about the Continuum Hypothesis (CH) didn't come until another half-century passed.
Around 1930, a mathematician named Kurt Godel¨ proved that it is impossible to prove that CH is false (starting with the ZFC axioms for set theory). On the other hand, about 1960, a mathematician named Paul Cohen proved that is it impossible to prove that CH is true (starting with the ZFC axioms for set theory) that CH is true. These results, taken together, mean that CH is “undecidable” (in terms of the ZFC axioms).
One could go back to the ZFC Axioms and add “CH is true” as a new axiom; or add “CH is false” as a new axiom. The result would be two “different set theories” (these are analogous to Euclidean and non-Euclidean geometries that arise from modifying Euclid's parallel postulate.)
Fortunately , the issue of whether CH is true or false rarely comes up in doing mathematical research. Therefore there's usually no need to worry about making a choice between these “competing" versions of set theory.
Example Ò!ß"Ñ∪Ð#ß$ѸÐ!ß"Ñ
"" ## Bfor !ŸB" A bijection between the two sets is given by 0ÐBÑ œ "" ##Bfor #B$ ()Draw a reasonably careful picture!
Therefore lÒ!ß "Ñ ∪ Ð#ß $Ñl œ lÐ!ß "Ñl œ -Þ Theorem Suppose we have finitely many countable sets E"8 ß ÞÞÞß E . Then the product set E‚E‚ÞÞÞ‚E"# 8 is countable. In other words, the product of a finite number of countable sets is countable.
Proof Since each set E"8 ß ÞÞÞß E is countable, we know that there exist one-to-one functions:
0ÀEÄ"" 0ÀEÄ## ã 0ÀEÄ33 ã 0ÀEÄ88
Let :"8 ß ÞÞÞß : be the first 8 prime numbers and use these to define a function 0ÀE‚E‚ÞÞÞ‚EÄ"# 8 are follows:
0Ð+Ñ"" 0Ð+Ñ 33 0Ð+Ñ 88 0Ð Ð+"38 ß ÞÞÞß + ß ÞÞÞß + Ñ Ñ œ :" †... † :3 ... † :8 œ some 8 − Þ
This function 0 is one-to-one:
Suppose Ð+"38 ß ÞÞÞß + ß ÞÞÞß + Ñ Á Ð, "38"# ß ÞÞÞß , ß ÞÞÞß , Ñ − E ‚ E ‚ ÞÞÞ ‚ E 8
0Ð,Ñ"" 0Ð,Ñ 33 08Ð,Ñ 8 Then 0Ð Ð,"38 ß ÞÞÞß , ß ÞÞÞß , Ñ Ñ œ :" † ... † :3 ... † :8 œ some 7 − Þ
Since Ð+"38 ß ÞÞÞß + ß ÞÞÞß + Ñ Á Ð, "38 ß ÞÞÞß , ß ÞÞÞß , Ñ , we know that + 33 Á , for some 3 , and, because 00Ð+ÑÁ0Ð,Ñ33333 is one-to-one, this means that .
Therefore :783 occurs in the prime factorizations of both and , but a different number to times in each factorization). By the Fundamental Theorem of Arithmetic, we conclude that 7Á8Þ Therefore 0 is one-to-one.
Since we have produced a one-to-one function 0ÀE‚E‚ÞÞÞ‚EÄ"# 8 , a previous theorem tells us that E‚E‚ÞÞÞ‚E"# 8 is countable. ñ
Example Each of the following sets is countable:
™‚
™ ‚ Ö!ß "ß #× ‚ ‚
™=‚‚
™‚ g ‚ Ö"ß #× ‚ Theorem Suppose E"# ß E ß ÞÞÞß E 8 ß ÞÞÞ is a sequence of countable sets (one for each 8 in the index set ), and suppose that the 's are pairwise disjoint. Then ∞ is countable. EE888œ"
>2 Proof Let :8"#8 œ the 8 prime number. Then : ß : ß ÞÞÞß : ß ÞÞÞ is the infinite sequence of prime numbers.
Since each set E"8 ß ÞÞÞß E ß ÞÞÞ is countable, we know that there exist one-to-one functions
0ÀEÄ"" 0ÀEÄ## ã 0ÀEÄ33 ã 0ÀEÄ88 ã
Define a function ∞ as follows: 0À8œ" E8 Ä if ∞ then there is a unique for which . Then and we can B−8œ" E833 ß 3 B−E 0ÐBÑ− 0ÐBÑ3 let 0ÐBÑ œ :3 Þ
This is one-to-one: suppose ∞ 0BÁC−EÞ8œ" 8
i) If Bß C are in the same set E33 . Because 0 is one-to-one, 0 33 ÐBÑ Á 0 ÐCÑÞ 03ÐBÑ 03ÐCÑ Therefore 0ÐBÑœ:33 Á: œ0ÐCÑ , by the Fundamental Theorem of Arithmetic (because the exponents for :/3 are differ nt).
03ÐBÑ 08 ÐCÑ ii) If B−E38and C−E Ð where 3Á8Ñß then 0ÐBÑœ:3 and 0ÐCÑœ:8 Þ So 0ÐBÑ and 0ÐCÑ factor as powers of two different primes, :38 and : Þ By the Fundamental Theorem of Arithmetic, 0ÐBÑ Á 0ÐCÑÞ
Therefore 0ÐBÑ Á 0ÐCÑ , so 0 is one-to-one. Since we have produced a one-to-one function ∞∞, we conclude that is countable. 0À8œ" E88 Ä 8œ" E ñ
The next corollary says that we really do not need the hypothesis “pairwise disjoint” in the theorem. However it was notationally easier to first prove the theorem with the “pairwise disjoint” hypothesis, and (now) to worry about what happens if some of the sets have nonempty intersection. Corollary Suppose F"# ß F ß ÞÞÞß F 8 ß ÞÞÞ is a sequence of countable sets (one for each 8 in the indexing set ) Then ∞ is countable. ÞF8œ" 8
Proof Define EœF"" EœFF##" EœFÐF∪FÑ$$ "# ã E88 œF ÐF " ∪â∪F 8" Ñ ã
Since each E©F88 , each E 8 is countable. Also, the E 8 's are pairwise disjoint:
Consider E38 and E where 38Þ If B−E 33" œF ÐF ∪â∪F 3" Ñß then B−FÞ3"8"8 Since 3Ÿ8" , B−F ∪â∪F Þ Therefore BÂE ß so E∩EœgÞ38
Since ∞∞ We claim that these sets are actually equal. E©Fß888œ" E© 8 8œ" FÞ 8 Suppose B−∞ F Þ Pick the smallest 8œ8 for which B−F . Then BÂF for 8œ" 8!83! 38 and therefore B−E œF ÐF ∪â∪F Ñ , so B−∞ E . Therefore !88"8"8!! ! 8œ" ∞∞ 8œ"Eœ88 8œ" FÞ The theorem then applies to the 's, and we conclude that ∞∞ is countable. EFœEñ8888œ" 8œ"
Corollary If is a finite collection of countable sets, then 5 F"# ß F ß ÞÞÞß F 5 F " ∪ ÞÞÞ ∪ F 5 œ8œ" F 8 is countable.
(This seems completely plausible, since we have “less than a full sequence“ of countable sets. In order to apply the preceding corollary, we simply “pad out” the sets to form a sequence by adding empty sets to the list.)
Proof Define for each By the preceding corollary, ∞ is countable. Fœg8 85Þ8œ" F8 But ∞ since for each 8œ"FœF∪ÞÞÞ∪F8" 5 Fœg 8 85Þñ
In words we can summarize the preceding theorem and its corollaries as follows: the union of countably many countable sets is countable Å a finite number, or a sequence E"8 ß ÞÞÞß E ß ÞÞÞ
"# 8 8" Examples i) For 8 − ß let E8 œ Ö88 Þ ß ÞÞÞß 8 ß 8 ß ÞÞÞ × œ the set of all positive rationals that can be written with denominator . Then the set of all positive rationals ∞ 8œ œE8œ" 8 is countable. ii) The function 0ÐBÑ œ B is a bijection 0 À Ä œ the set of all negative rationals, so is countable. Since the set of all rationalsœ ∪ Ö!× ∪ œ the union of finitely many countable sets, is countable. Transcendental Numbers: An Extended Example
We have proved some not-too-complicated theorems about countable and uncountable sets. But these elementary facts are enough to let us prove an interesting result about real numbers.
: A rational number ; is the root of first-degree polynomial equation with integer coefficients, namely: ;B : œ ! (and, conversely, the root of such an equation is a rational number).
We can generalize the idea of a rational number by looking at the real numbers that are roots of higher degree polynomial equations with integer coefficients.
Definition A real number < is called an algebraic number if there is a polynomial
88" T ÐBÑ œ +88" B + B ÞÞÞ + "! B + with integer coefficients +!" ß + ß ÞÞÞß + 8 for which T Ð<Ñ œ !Þ
A little less formally: an algebraic real number is one that's a root of some polynomial equation with integer coefficients.
As remarked above: if TÐBÑ is a degree 1 polynomial with integer coefficients, then a root of TÐBÑœ ! is just a rational number. But, for example all of the following irrationals are also algebraic numbers.
# is an algebraic number, because it is a root of a quadratic equation with integer coefficients: B#œ!# .
""'* is an algebraic number because it is a root of a quadratic equation #' with integer coefficients: $B# $B & œ !Þ
( th #( is algebraic because it is a root of a degree polynomial equation with integer coefficients: B#œ!( .
A reasonable question would be: are there any real numbers that are not algebraic ?
Theorem The set ‘œÖ<− À < is algebraic × is countable.
8 Proof For each 8 " , let c™88 œ Ö+ B ÞÞÞ + "!8"! B + À + ß ÞÞÞß + ß + − × œŸ8 the set of all polynomials of degree with integer coefficients (the degree will be 8 if the coefficient +8 happens to be !). To name a polynomial in c8 , all we need to know is the list of its coefficients. The list of 8" coefficients is Ð8 "Ñ -tuple of integers À Ð+88" ß + ß ÞÞÞ ß + ! Ñ −™™™ ‚ ... ‚ œ . To put it 8" another way, the function 1Àc™8 Ä given by
8 1Ð+8"!88"! B ÞÞÞ + B + Ñ œ Ð+ ß + ß ÞÞÞ ß + Ñ is a bijection. 8" 8" Therefore c™8 ¸‚‚œ ... ™™ and ™ is a finite product of countable sets. So c8 is countable and therefore there is a bijection 0Àc Ä 8 .
Thus the polynomials in the set c8 can be listed in a sequence:
c8œ Ö T 8ß" ÐBÑß T 8ß# ÐBÑß ÞÞÞß T 8ß5 ÐBÑß ÞÞÞ ×
>2 ()Here,Tœ0Ð5Ñœ58ß5 the term of the sequence 0 .
Each polynomial T8ß" ÐBÑß T 8ß# ÐBÑß ÞÞÞß T 8ß5 ÐBÑß ÞÞÞ T 8ß5 in this list has degree Ÿ 8 , and therefore each equation TÐBÑœ!8ß5 has at most 8 roots . So the set
V8ß5 œ Ö< −‘ À T 8ß5 Ð<Ñ œ !× œ Ö< À < is a root of T 8ß5 ÐBÑ œ !× is finite
We now put all this together using our knowledge about countable sets.
∞ E8 œ5œ" V 8ß5 œ V 8ß" ∪ V 8ß# ∪ ÞÞÞ ∪ V 8ß5 ∪ ÞÞÞ œÖ<À < is a root of a polynomial of degree Ÿ8 with integer coefficients × is a countable union of countable (finite!) setsàE so 8 is countable.
Then ∞ is countable, since it is a countable union of countable sets. 8œ"E8 But ∞ is in one of the sets is a root of a polynomial of degree with <−8œ"88 Í< E Í < Ÿ8 integer coefficients is an algebraic number. So ∞ is countable. Í< œ8œ" E8 ñ
Definition A real number which is not algebraic is called transcendental . (Euler called these numbers “transcendental” because they “transcend the power of algebraic methods.” To be more politically correct, we might call the7 “polynomially challenged.” )
Corollary Transcendental numbers exist.
Proof Let “‘“ be the set of transcendental numbers. By definition, œ∪ . Since is countable and ‘“ is uncountable, cannot be empty. ì
In fact, this two-line argument actually proves much more: not only is ““ nonempty , but must be uncountable! In the sense of one-to-one correspondence, there are “many more” transcendental numbers than algebraic numbers on the real line ‘ .
This is an example of a “pure existence” proof it does tell us that any particular number is transcendental number, and the proof does not give us any computational hints about how to find a specific transcendental number. To do that is harder. Transcendental numbers were first shown to exist by Liouville in 1844. (Liouville used other (more difficult) methods. The ideas about countable and uncountable sets were not developed until the early 1870's by Georg Cantor. See the biography on the course website.) Two famous examples of transcendental numbers are / (proven to be transcendental by Hermite in 1873) and 1 (Lindemann, 1882).
One method for producing many transcendental numbers is contained in a theorem of the Russian # mathematician Gelfand (1934). It implies, for example, that # is transcendental. Gelfand's Theorem If αα" is an algebraic number, Á" 0 or , and is algebraic and not rational, then α" is transcendental.
The number /1 is also transcendental. This follows from Gelfand's Theorem (which allows complex algebraic numbers) if you know something about the arithmetic of complex numbers:
/œ/1113# = Ð/Ñ 3 3 and / 3 1 œ cos11 3 sin œ"Þ So /œÐ"Ñ1 3 , which is transcendental by Gelfand's Theorem Þ