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 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 ÀEÄGis 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+Ñ0will 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 ∞Ñbecause 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 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. 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− Àthen 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 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). 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‚HÀSuppose 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−Gor C−H. IfC−G , then Ðsince 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 0ÐBÑœFÞThen either B−F or BÂFÞ If B − F, then B must satisfy the membership requirement for F: that B  0ÐBÑÞ But then BÂ0ÐBÑœF, so this is impossible.