Equivalent Sets Definition Let Eß F be sets. We say that E is equivalent to F iff there exists a bijection 0ÀEÄF. If Eis equivalent to F, we write E FÐor E For EµFor something similar: the notation varies from book to bookñ. It is intuitively clear that for finite sets E F iff Eand Fhave 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 E, Fß G be sets. Examples a) The identity mapping 0ÐBÑœBis 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 ÀFÄEis also a bijection, so F EÞTherefore 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 E to F or that there is a bijection from F to E.) c ÑSuppose E Fand F G. Then there are bijections 0ÀEÄFand 1ÀFÄG. Then 1 0 ÀEÄGis a bijection ( check! ) so E GÞ Therefore the relation is transitive. 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 0 will also include the point Ð!ß+Ñ, and the modified function 0 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 Examples 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. 1 1 1 1 5) ( ß Ñ because tan À ( ß Ñ Ä 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 / ß is a bijection. 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, Öß %ß 'ß )ß ÞÞÞ and Ö"&ß $!ß %&ß '!ß ÞÞÞ Þ By symmetry and transitivity, Öß %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 B 0Ð8Ñ œ 8 "8 if 8 is even if 8 is odd. (For example, 0ÐÑ 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 7is 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.
7 8 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 Ê 7 œ 8. 0 is onto : Let D À if D!, let 8œD À then 0Ð8Ñœ œdþ "Ð"DÑ if D Ÿ!ß let 8 œ " D Ðwhy? ) À then 0Ð8Ñ œ œ DÞ D 5) We saw earlier (near the end of the notes on functions) that there is a bijection 0À Ä œthe 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). Theorem Suppose that E G F H and Then i) E F G H ii) If E Fœg and G Hœgß then E F G HÞ Proof Since E Gand F H, we know that there are bijections 0ÀEÄGand 1ÀFÄHÞ i) Define 2ÀE FÄG H by the formula 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 - Gso since 0is 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 À Suppose Ð+ " ß, " Ñand Ð+ ß, Ñ E F 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 " " " " À 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 0 and 1 disagree so there is no problem with if 0 1 being a function--see Homework 9). We can write 2ÐBÑ œ 0ÐBÑ B E 1ÐBÑ if B FÞ 2is onto G HÀ if C G H, then C Gor C H.
If C G, then Ðsince 0 is onto) there is an + Efor 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 2is onto. 2 is one-to-one À Suppose BßC E F and 2ÐBÑ œ 2ÐCÑÞ If B Eand C F( or vice versa) ßthen 2ÐBÑœ0ÐBÑ Gand 2ÐBÑœ2ÐCÑœ1ÐCÑ HÞ This would mean G HÁg. So either Band C are in E, or B 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 1) Ð!ß "Ñ Ð!ß "Ñ and 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 E, E Î cðeñþ Proof For a finite set E this is clear because if E contains 8 elements, then cðeñ œ the set of all subsets of E contains 8 elements. Suppose, in general, that 0 is any function 0 À E Ä cðeñþ We claim that 0 cannot be onto and therefore 0 cannot be a bijection. This will show that E Î cðeñ. If + E, then 0Ð+Ñ cðeñ, 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 $, then cðeñ œ Ö g ß Ö" ß Ö ß Ö$ ß Ö"ß ß Ö"ß $ ß Öß $ ß Ö"ß ß $ Þ Suppose (say) 0ÀEÄcÐEÑ is the function given by 0Ð"Ñ œ Ö"ß ß 0ÐÑ œ Ö"ß $, 0Ð$Ñ œ Ö" Then " 0Ð"Ñß Â0ÐÑ and $Â0Ð$Ñ Let FœÖ+ EÀ+Â0Ð+Ñ. Then F E, so F cðeñþ We claim that Fis not in the range of 0. In the illustration, F œ Öß $, and, as promised, F is not in the range of 0. Suppose there were an B Efor which 0ÐBÑœFÞThen either B For BÂFÞ If B F, then B must satisfy the membership requirement for F: that B  0ÐBÑÞ But then BÂ0ÐBÑœF, so this is impossible. If B  F, then B fails to meet the membership requirement for F: so B 0ÐBÑÞ But then B 0ÐBÑœFwhich is impossible. Either way, the assumption that there is an B such that 0ÐBÑ œ F leads to a contradiction, so no such B can exist. Therefore 0 is not onto. ñ Example By the theorem, / cð). Notice that there is a one-to-one ( not onto) map 0À ÄcÐÑÐfor example, let 0Ð8Ñ œ Ö8 cñ Ð Ñ. If 0 À E Ä cðeñ is such a one-to-one function, then E 0ÒEÓ œ range Ð0Ñ cðeñ. So E is always equivalent to a subset of cðeñ but never equivalent to the whole set cðeñ. Intuitively, this means that cðeñ is a bigger infinite set than E. We can continue to apply the power set operation over and over to get / c Ð ), c Ð Ñ / cc Ð Ð ) Ñ, cc Ð Ð Ñ) / ccc ( Ð Ð ) Ñ),... and thus create a sequence of larger and larger nonequivalent infinite sets.
We proved that for every set E, E Î cðeñ. In particular, / cðñþ However, we can show that c Ð Ñ is equivalent to another familiar set: Ö!ß ". Recall that Ö!ß " denotes the set of all functions 0 À Ä Ö!ß " Þ Such an 0 is a sequence for which every term has value!, or ": if 0 Ö!ß ", we have values 0Ð"Ñß 0ÐÑß 0Ð$Ñß ÞÞÞß 0Ð8Ñß ÞÞÞ where each 0Ð8Ñ œ! or " In other words, Ö!ß " is the set of all binary sequences. Theorem c Ð Ñ Ö!ß " Proof Define a function F À Ö!ß " Ä cðñ as follows: if 0 Ö!ß ", then FÐ0Ñ œ Ö8 À 0Ð8Ñ œ " cðñ If 0 is a binary sequence, then FÐ0Ñis the subset of consisting of all those 8 for which >2 the 8 term of the sequence is ". For example suppose 0 has values then FÐ0Ñ œ Öß $ß 'ß ÞÞÞ Þ 0Ð"Ñß 0ÐÑß 0Ð$Ñß 0Ð%Ñß 0Ð&Ñß 0Ð'Ñß ÞÞÞ ll ll ll ll ll ll!ß "ß "ß!ß!ß " ß ÞÞÞ 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 FÐ0Ñ but 8  FÐ1Ñ, so FÐ0Ñ Á FÐ1Ñ. Similarly if 0Ð8Ñ œ! and 1Ð8Ñ œ ", then 8  FÐ0Ñ but 8 FÐ1Ñ, so FÐ0Ñ Á FÐ1Ñ. F is onto: Suppose E c Ð Ñ, that is, suppose E Þ Define a binary sequence 0as follows: if 0Ð8Ñ œ " 8 E! if 8  E Then 0 Ö!ß " and, by definition, FÐ0Ñ 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 let 5 œ Ö"ß ß ÞÞÞß 5. A set E is called finite if Eœgor if Ðb5 Ñ E 5 Þ Eis called infinite if Eis not finite (that is, if EÁgand Ða5 ÑE / 5 Þ Definition For a finite set E À
i) if Eœgß we say that! is the cardinal number of E Ðor that E has cardinality 0Ñ, and write lel œ!þ ii) if E 5 ßwe say that 5is the cardinal number of E Ðor that E 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 E is finite and E F, then Fis 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 E and Fare finite: a) le Fl œ lel lfl le Fl b) If E F œ gß then le Fl œ lel lflþ lel 6. If E is finite, then cðeñ is finite, and cðeñl œ Þ 7. If 8 and E" ß ÞÞÞß E8 are finite sets, then E" E ÞÞÞ E8 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 À 8 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< ÞWe want to prove Ða8 ) TÐ8Ñßwhere TÐ8Ñis the statement: if 8 <ß then 0 cannot be one-to-one. Since < represents a natural number so < " and so 8 Þ Therefore our induction starts with 8œÞ Base case If 8 œ, then < œ " and 0 À œ Ö"ß Ä Ö" œ " Þ Therefore 0 must be constant: 0Ð"Ñœ"œ0ÐÑÞSo 0is not one-to-one. Assume that TÐ8Ñis true for some particular value 8 œ 5Þ That is, our induction hypothesis is (*) If 0À Ä (where < is any natural number 5Ñ, then 0is not one-to-one. 5 < We want to prove: If 0ÀR Ä (where < is any natural number 5"Ñßthen 0is not one-to-one 5" < We prove this by contradiction. Suppose there is a one-to-one function 0ÀR Ä, where 5"<. (**) 5" < 0 À Ö"ß ß ÞÞÞß 5ß 5 " Ä Ö"ß ß ÞÞÞß < 0Ð5 "Ñ is one of the numbers "ß ß ÞÞÞß <Þ Delete 5 " from the first set Ðleaving 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 0l5 À 5 Ä < Ö0Ð5 "Ñ The set on the right Ðœthe codomain of 1Ñhas one less element than <, so there is a bijection 2À Ö0Ð5"Ñ Ä Þ < <" Composing functions, we get a one-to- one function 2 1À Ä ß 5 <" Since 5"<ßwe 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 which says that TÐ5 "Ñis true. By PMI, TÐ8Ñis 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 Î : ;Þ Proof One of : or ; is larger than the other say : ;Þ If 0 À : Ä ; is any function, the, by the Pigeonhole Principle, 0 cannot be one-to-one, so 0 cannot be a bijection. ñ The corollary tells us that a set E cannot be equivalent to both : and ; if : Á ; À otherwise, by transitivity, we would have : ; Þ This means that the cardinal number of E is welldefined: 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 0cannot 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 W is a set of R 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 Öß "*ß $$ ß then R œ $ and the subset Ö$$ works; so does the subset Öß "*ß $$ Þ Proof Let W œ Ö+ " ß + ß ÞÞÞß + where each + 3 Þ Consider the R subsets R Ö+ " ß Ö+ " ß + ß ÞÞÞß Ö+ " ß + ß ÞÞÞß + 3 ß ÞÞÞß Ö+ " ß + ß ÞÞÞß + R 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 + ÞÞÞ + + ÞÞÞ + ÞÞÞ + " 3 R " 3 5 Subtracting + ÞÞÞ + " 3 from both sides, we get that! + ÞÞÞ + R 3" 5 so RlÐ+ 3" ÞÞÞ + 5ÑÞ Therefore the subset Ö+ 3" ß ÞÞÞß + 5 works. Here, in Case ii), the pigeons are the R sets Ö+ " ß ÞÞÞ ß Ö+ " ß + ß ÞÞÞß + R 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,ßand there are 5" 's in +, then,+ consists of 651's followed 5,+ by 5!'s. We can cancel out all the!'s in,+ by dividing by "! À so is a natural "! 5 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 for some "! 5 œ 5& 5 œ ( "" "!" 4 œ (((( 4 4 Ðwe know is a "! 5 natural number; the 's and &'s in the denominator must cancel out with 's and &'s in the prime factorization of 7À œ4 ÑÞ 5 5 7 &,+ So is a natural number with only "'s for digits and divisible by ((((. "! 5,+
Example ( taken from the Spring 2008 Missouri MAA Mathematics Competition ) Consider the set of points œ ÖÐDßAÑ À D and A 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 ÖÐß!Ñß Ðß "Ñß Ðß Ñß Ðß $Ñß Ðß %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 Column 4 œ Ö Ð4ß!Ñß Ð4ß "Ñß Ð4ß Ñß Ð4ß $Ñß Ð4ß %Ñ Column 5 œ Ö Ð5ß!Ñß Ð5ß "Ñß Ð5ß Ñß Ð5ß $Ñß Ð5ß %Ñ Our first observation 1) tells us that two points in Column 4 have the same color: suppose they are Ð4ß 7Ñ and Ð4ß 8Ñ Ðwhere! Ÿ78Ÿ%ÑÞ Since Column 5 has the same color pattern, Ð5ß7Ñand Ð5ß8Ñare 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 Ðlike œ Öß %ß 'ß ÞÞÞ ÑÞ So we have to be careful not to jump to conclusions when making statements about infinite sets. Theorem The set E is infinite if and only if there exists a one-to-one function 0 À Ä EÞ 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 + 8's by induction. Once + 8 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 + Á + ß + ß ÞÞÞß + ) 8" " 8 In this way we have inductively defined + 8 for all 8 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 0ÀÄEby 0Ð8Ñœ+Þ 8 Since the + 8's are all different, 0is one-toone. For ( É ), we use the contrapositive: suppose E is finite and show that no such one-to-one function 0 can exist. So, if \ If Eœg, then there are no functions 0À ÄE ( see earlier discussion of ] Ñ If E œ Ö"ß ß ÞÞÞß 5} ß then there is a bijection 2 À E Ä 5 5 If there were a one-to-one function 0ÀÄEßthen we could restrict the domain of 0 to get a new one-to-one function 1 œ 0l5" À 5" Ä EÞ Then 2 1À5" Ä5 is one-to-one which contradicts the Pigeonhole PrincipleÞ E is finite, there is no one-to-one function 0 À Ä EÞ ñ
Note: A function 0ÀÄEis a sequence in EÞThe terms of the sequence are + " œ 0Ð"Ñß + œ 0ÐÑß 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 + ß ÞÞÞ in E. " 8 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 0Ð8Ñœ8is one-to-one. By the preceding theorem, is infinite. 2) The function 0 À Ä Ð!ß "Ñ given by 0Ð8Ñ œ 8 is one-to-one, so the interval Ð!ß "Ñ is infinite. " 3) Suppose E is infinite, so that there is a one-to-one function 0 À Ä E. If E F, then we can view 0 also as being a one-to-one function 0 À Ä FÞ So F is also infinite. The following theorem tells us that if a subset E is too big to be finite, then Eis equivalent to the whole set. For example, consider the set œ Öß %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 0Ð8Ñœ8is a bijection. But we want to prove it for an arbitrary infinite subset of. Theorem If Eis infinite and E, then E Þ Proof EÁgÞ By the Well-Ordering Principle (WOP), there is a smallest element in EÀcall it +Þ " Since E is infinite, there is a smallest element, call it +ß in EÖ+ ß " and + Á+ "Þ If we have chosen distinct points + " ß ÞÞÞß + 8 E, then (since E is infinite) E Ö+ " ß ÞÞÞß + 8 Á gß so we let + 8" be the smallest element in E Ö+ " ß ÞÞÞß + 8 Ð and + 8" will be different from all the preceding + 3's Ñ. In this way, we have defined (inductively) + 8 for all 8, and all the + 8's are different. Then define a function 0À ÄEby 0Ð8Ñœ+à this 0is a bijection. ñ 8 Definition The set E is called countable if i) E is finite, or ii) E is infinite and E Þ If E, we use the symbol i! to denote the cardinal number of EÀ llœiþ! ( i is aleph, the first letter of the Hebrew alphabet. i! is read as aleph-zero or aleph null or aleph naught (more British). This notation was first used by George Cantor th in his groundbreaking work of the theory of infinite sets (in the later part of the 19 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 g lgl œ! Ö+ß,ß- lö+ß,ß- lœ$ Infinite œ Öß %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 c Ð Ñ This set is clearly infinite, and Î c Ð Ñ Ðwe proved that for every set E, E Î cðeññ so the set is uncountable: lc Ð Ñl Á i! 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 E is infinite iff there exists a one-to-one function 0 À Ä E. Informally this means that an infinite set is big enough to have a one-to-one copy of put inside it Ðthe copy is range Ð0 ÑÑÞ 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 E is countable iff there exists a one-to-one function 0 À E Ä Þ Proof ( É ) If there does exist a one-to-one function 0 À E Ä, then E 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 range Ð0Ñ (by the preceding Theorem) Þ By transitivity, then, E Þ So Eis countable. ( Ê ) Suppose E is countable.
If lelœ!ßthen Eœgand the empty function g ÀgÄis one-to-one. If lel œ 5, then there is a bijection 0 À E Ä Ö"ß ß ÞÞÞß 5, and this same 0ÀEÄis one-to-one If E in not finite, then ß by definition of countable ß E, so there exists a bijection 0ÀEÄ, and, of course, this 0is one-to-one. ñ Corollary A subset of a countable set is countable. Proof Suppose E is a countable set. Then there is a one-to-one function 0 À E Ä Þ If F E, then look at the restricted function 1œ0lFÞ Then 1ÀFÄis one-to-one, so, by the theorem, Fis countable. ñ Corollary If F is an uncountable set and F G, then G is uncountable. ( More informally, a set that contains an uncountable set is uncountable.) Proof If G were a countable set, then its subset F 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 number 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"8ÞÞÞ 0ÐÑœ< œ!b. " B B$ ÞÞÞ B8ÞÞÞ 0Ð$Ñœ< $ œ!b. $" B$ B$$ ÞÞÞ B$8ÞÞÞ... 0Ð8Ñœ< 8 œ!b. 8" B8 B8$ ÞÞÞB88ÞÞÞ ã Now define a real number Cœ!. CCC... C... by setting " $ 8 if 1 C 8 œ " B 88 Á C œ if B œ " 8 88 Then C Ð!ß"Ñ Notice that i) the decimal expansion of C is different from every decimal expansion in the th list specifically, the decimal expansions of C and < 8 disagree at the 8 decimal place: C Á B. 8 88 ii) the decimal expansion of C is not an alternate decimal representation for any of the < 8 's because!. CCC " $... C8... does not end in either an infinite string of!'s or *'s. Therefore Ða8Ñ C Á < 8. In other words, C  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 continuum 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 i! Ÿ -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 Ðß$Ñ Ð!ß"Ñ A bijection between the two sets is given by 0ÐBÑ œ ( Draw a reasonably careful picture! ) " " B for!ÿb" " " B for B$ Therefore lò!ß "Ñ Ðß $Ñl œ lð!ß "Ñl œ -Þ
Theorem Suppose we have finitely many countable sets E" ß ÞÞÞß E8. Then the product set E " E ÞÞÞ E8 is countable. In other words, the product of a finite number of countable sets is countable. Proof Since each set E ß ÞÞÞß E " 8 is countable, we know that there exist one-to-one functions: 0" À E" Ä 0 À E Ä ã 03 À E3 Ä ã 08 À E8 Ä Let :" ß ÞÞÞß : 8 be the first 8 prime numbers and use these to define a function 0 À E " E ÞÞÞ E8 Ä are follows: 0Ð+Ñ 0Ð+Ñ 0Ð+Ñ " 3 8 " 3 8 " " 3 3 8 8 0Ð Ð+ ß ÞÞÞß + ß ÞÞÞß + Ñ Ñ œ :... :... : œ some 8 Þ This function 0 is one-to-one: Suppose Ð+ ß ÞÞÞß + ß ÞÞÞß + Ñ Á Ð, ß ÞÞÞß, ß ÞÞÞß, Ñ E E ÞÞÞ E " 3 8 " 3 8 " 8 0Ð,Ñ 0Ð,Ñ 08Ð,Ñ " 3 8 " 3 8 " " 3 3 8 Then 0Ð Ð, ß ÞÞÞß, ß ÞÞÞß, Ñ Ñ œ :... :... : œ some 7 Þ Since Ð+ " ß ÞÞÞß + 3ß ÞÞÞß + 8 Ñ Á Ð, " ß ÞÞÞß, 3ß ÞÞÞß, 8Ñ, we know that + 3 Á, 3 for some 3, and, because 0 is one-to-one, this means that 0 Ð+ Ñ Á 0 Ð, Ñ. 3 3 3 3 3 Therefore : 3 occurs in the prime factorizations of both 7 and 8, but a different number to times in each factorization). By the Fundamental Theorem of Arithmetic, we conclude that 7Á8Þ Therefore 0is one-to-one. Since we have produced a one-to-one function 0 À E E ÞÞÞ E Ä tells us that E E ÞÞÞ E is countable. ñ " 8 " 8, a previous theorem Example Each of the following sets is countable: Ö!ß "ß = g Ö"ß
Theorem Suppose E" ß E ß ÞÞÞß E8ß ÞÞÞ is a sequence of countable sets (one for each 8 in the index set ), and suppose that the E 's are pairwise disjoint. Then E is countable. 8 8œ" 8 >2 Proof Let : 8 œ the 8 prime number. Then :" ß : ß ÞÞÞß : 8ß ÞÞÞ is the infinite sequence of prime numbers. Since each set E ß ÞÞÞß E ß ÞÞÞ " 8 0 À E Ä 0 À E Ä ã 0 À E Ä ã 0 À E Ä ã " " 3 3 8 8 is countable, we know that there exist one-to-one functions Define a function 0À 8œ" 8œ" E Ä 8 as follows: if B E8ßthen there is a unique 3for which B E3. Then 0ÐBÑ 3 and we can 0ÐBÑ 3 let 0ÐBÑ œ : Þ 3 8œ" This 0 is one-to-one: suppose B Á C E Þ 8 i) If Bß C are in the same set E3. Because 03 is one-to-one, 03ÐBÑ Á 03ÐCÑÞ 03ÐBÑ 03ÐCÑ Therefore 0ÐBÑœ: 3 Á: 3 œ0ðcñ, by the Fundamental Theorem of Arithmetic (because the exponents for : are differ/ nt). ii) If B E3and C E8Ðwhere 3Á8Ñßthen 0ÐBÑœ: 3 and 0ÐCÑœ: 8 8 Þ So 0ÐBÑ and 0ÐCÑ factor as powers of two different primes, : 3 and : 8Þ 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 0À E Ä, we conclude that E is countable. ñ 8œ" 8 8œ" 8 3 03ÐBÑ 0 ÐCÑ 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 F8ß ÞÞÞ is a sequence of countable sets (one for each 8 in the indexing set ) Þ Then F is countable. Proof Define E œ F " " 8œ" 8 E œ F F " E$ œ F$ ÐF" F Ñ ã E œf ÐF â F 8 8 " 8" ã Since each E 8 F8, each E8 is countable. Also, the E8's are pairwise disjoint: Ñ Consider E3 and E8 where 38Þ If B E3 œf3 ÐF" â F3" Ñßthen B FÞ 3 Since 3Ÿ8", B F" â F8" Þ Therefore BÂE8ßso E3 E8 œ gþ Since E F ß E F Þ 8 8 8œ" 8 8œ" 8 We claim that these sets are actually equal. 8œ" Suppose B F8 Þ Pick the smallest 8œ8! for which B F8. Then BÂF3 for! 38! and therefore B E8 œf8 ÐF" â F8 " Ñ, so B!!! 8œ" E8. Therefore E œ F Þ 8œ" 8 8œ" 8 The theorem then applies to the E 's, and we conclude that F œ E is countable. ñ 8 8œ" 8 8œ" 8 Corollary If F ß F ß ÞÞÞß F is a finite collection of countable sets, then F ÞÞÞ F œ 5 8œ" F is countable. " 5 " 5 8 (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 F8 œ g for each 8 5Þ By the preceding corollary, 8œ" F8 is countable. But F œ F ÞÞÞ F since F œ g for each 8 5Þ ñ 8œ" 8 " 5 8 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" ß ÞÞÞß E8ß ÞÞÞ " 8 8" Examples i) For 8 ß let E 8 œ Ö 8Þ 8ß ÞÞÞß 8ß 8 ß ÞÞÞ œ the set of all positive rationals that can be written with denominator 8. Then œ the set of all positive rationals œ 8œ" E8 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 rationals œ 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 8 8" T ÐBÑ œ + 8B + 8" 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 TÐBÑ œ! 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 ( is algebraic because it is a root of a ( th 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 c8 œ Ö+ 8 B ÞÞÞ + " B +! À + 8 ß ÞÞÞß + " ß +! œ the set of all polynomials of degree Ÿ 8with integer coefficients (the degree will be 8 if the coefficient + 8 happens to be!). To name a polynomial in c 8, all we need to know is the list of its coefficients. The list of coefficients is Ð8 "Ñ-tuple of integers À Ð+ 8ß + 8" ß ÞÞÞ ß +! Ñ... œ 8". To put it 8" another way, the function 1Àc Ä given by 8 is a bijection. 8 1Ð+ 8 B ÞÞÞ + " B +! Ñ œ Ð+ 8ß + 8" ß ÞÞÞ ß +! Ñ
8" 8" Therefore c8... œ and is a finite product of countable sets. So c8 is countable and therefore there is a bijection 0À Äc 8. Thus the polynomials in the set can be listed in a sequence: c 8 c 8 œ Ö T 8ß" ÐBÑß T 8ß ÐBÑß ÞÞÞß T 8ß5 ÐBÑß ÞÞÞ ( Here, T œ 0Ð5Ñ œ the 5 term of the sequence 0. ) 8ß5 >2 Each polynomial T8ß" ÐBÑß T8ß ÐBÑß ÞÞÞß T8ß5ÐBÑß ÞÞÞ T8ß5 in this list has degree Ÿ 8, and therefore each equation T8ß5ÐBÑ œ! has at most 8 roots. So the set V œ Ö< À T Ð<Ñ œ! œ Ö< À < is a root of T ÐBÑ œ! is finite 8ß5 8ß5 8ß5 We now put all this together using our knowledge about countable sets. 8 5œ" 8ß5 8ß" 8ß 8ß5 E œ V œ V V ÞÞÞ V ÞÞÞ œö<à < is a root of a polynomial of degree Ÿ8 with integer coefficients is a countable union of countable (finite!) setsà so E 8 is countable. Then 8œ" E 8 is countable, since it is a countable union of countable sets. 8œ" 8 8 But < Í< is in one of the sets E Í < is a root of a polynomial of degree Ÿ8with integer coefficients Í< is an algebraic number. So œ 8œ" E8 is countable. 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 the 7 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 α is an algebraic number, α Á 0 or ", and " is algebraic and not rational, then α " is transcendental. The number is also transcendental. This follows from Gelfand's Theorem (which allows complex algebraic numbers) if you know something about the arithmetic of complex numbers: / 1 1 3 1 3 1 3 3 1 / œ / = Ð/ Ñ and / œ cos 13sin 1 œ "Þ 1 3 So / œ Ð"Ñ, which is transcendental by Gelfand's TheoremÞ