These are theorems that I proved in class that, I think, cover the material in Section 5.3 much more efficiently. These theorems immediately imply Theorem 5.17-5.29 in the textbook. Theorem 1 If EßE¶Þ is an infinite subset of then Proof Since EÁgßand E© ßthe Well-Ordering Principle says there is a smallest member +−E" . Since Eis not finite ß E Ö+"# × Á gÞ Let +be the smallest member of E Ö+ " × Since E is not finite, we can always continue this process another step. We inductively define +8" to be the smallest member of EÖ+ " ß ÞÞÞß + 8 ×. The sequence +"# ß + ß ÞÞÞß + 8 ß ÞÞÞcontains every member of E (why? ), and the mapping 0 À E Ä defined by 0Ð+8 Ñ œ 8 is one-to-one and onto . Therefore E ¶ Þ ñ Corollary 2 A set E0ÀEÄ is countable iff there exists a one-to-one function (not necessarily onto!) Proof If such a function exists, then E¶range Ð0Ñ© . Therefore Eis finite or, by Theorem 1, E¶. Therefore E is countable. If EE is countable, then is either finite or denumerable so either there is a one-to-one function 0ÀEÄ5 © or there is a bijection 0ÀEÄ Þ Either way, we have a one-to-one function from EE to so is countable. ñ Corollary 3 A subset of a countable set is countable. Proof Suppose F©E where E is countable. By Corollary 2, there is a one-to-one function 0ÀEÄ Þ Then 1œ0lF is a one-to-one function from Finto so F is countable. Theorem 4 If EF and are countable, then E‚F is countable. Proof By assumption, there exist one-to-one functions 0ÀEÄ and 1ÀFÄ . For each ordered pair Ð+ß ,Ñ − E ‚ F, define 2Ð+ß ,Ñ œ #0Ð+Ñ $ 1Ð,Ñ − Þ Then 2 À E ‚ F Ä Þ (Example: if, say, 0Ð+Ñ œ #( and 1Ð,Ñ œ ""$ , then 2ÐÐ+ß ,ÑÑ œ ##( $ ""$ ÞÑ We claim that 2 is one-to-one: Suppose Ð+ß,ÑÁÐ-ß.Ñ−E‚F. Then either +Á- or ,Á.Þ If +Á-, then 0Ð+ÑÁ0Ð-Ñ since 0 is one-to-one. So #0Ð+Ñ $ 1Ð,Ñ Á# 0Ð-Ñ $ 1Ð.Ñ by the Fundamental Theorem of Arithmetic. If ,Á., then 1Ð,ÑÁ1Ð.Ñ since 1 is one-to-one. So #0Ð+Ñ $ 1Ð,Ñ Á# 0Ð-Ñ $ 1Ð.Ñ Either way, 2Ð+ß ,Ñ Á 2Ð-ß .Ñ, so 2 is one-to-one. Therefore, by Corollary 2, E‚F is countable. ñ Note: If EFG, , are countable then E‚F is countable so, by Theorem 4, ÐE‚FÑ‚G is also countable. But ÐE‚FÑ‚G ¶E‚F‚G, so E‚F‚G is countable. Continuing in this way, we can prove by induction that a product of finitely many countable sets is countable.) Theorem 5 Suppose, for each , that is countable. Then _ is countable. 8− E88-8œ" E (The union of a countable collection ÖE"8 ß ÞÞÞß E ß ÞÞÞ× of countable sets is countable.) Proof To show _ is countable, it is sufficient, by to produce a one-to-one map -8œ"E8 : _ (by Corollary 2). 0EÄ-8œ" 8 1) We begin by showing that we can find pairwise disjoint countable sets F8 such that __ (Then we will just need to show that _ is countable.) --8œ"Fœ88 8œ" EÞ - 8œ" F 8 Define FœE"" FœEE##" FœEÐEEÑ$$ "# ã F88 œ E ÐE " ÞÞÞ E 8" Ñ ã To see that the F8's are pairwise disjoint: Consider FF78 and where, say, 78Þ=−F If 7 œ E7" ÐE ÞÞÞ E 7" Ñ, then B − E 7. But that implies B  F88 œ E ÐE " ÞÞÞ E 7 ÞÞÞ E 8"78 ÑÞ So F F œ gÞ To see that __ : --8œ"Fœ88 8œ" E Since , certainly we have __. F©E88--8œ" F© 8 8œ" E 8 Conversely, if _ , then is in at least one of the 's. Pick B−-8œ" E88 B E the smallest 8B−EB−FB!8Þ8 such that !! Then because is not in E"8" ÞÞÞ E! , the set “discarded” in the definition of F 8! . Therefore ___, so . B−---8œ" F888 8œ" F œ 8œ" E 2) Since each FE88 is a subset of a countable set , Corollary 3 tells us that each F 8 is countable. We want to show that _ is countable. -8œ"F8 There is a one-to-one function 0ÀFÄ88, for each 8 (since F 8is countable). List the prime numbers as :"#$ œ #ß : œ $ß : œ &ß ÞÞÞß : 8 ß ÞÞÞÞ Define _ as follows:if , then 08ÐBÑ 0À-8œ" F88 Ä B−F 0ÐBÑœ:8 − For example, if B−F$$$, then 0 ÀF Ä and 0ÐBÑ $ is some natural 0ÐBÑ$ numberÞ Suppose, to illustrate, that 0$ ÐBÑ œ '%ÞThen 0ÐBÑ œ :$ '% '% œ:$ œ& . Notice that for each _ , there is exactly one for which B−-8œ"F8 8 B−F88 because the F 's are pairwise disjoint. This means there's no ambiguity in how 0 is being defined. We claim that is one-to-one. Suppose _ and 0BßC−FBÁCÞ-8œ" 8 i) if B−F87 and C−F, where 7Á8, then 0ÐBÑ87 0 ÐCÑ 0ÐBÑœ:87 Á: œ0ÐCÑ because :87 Á: 0ÐBÑ8 Fundamental Theorem of Arithmetic: :8 can't be factored as a product of :7's. ii) if BC and are both in the same F88, then 0ÐBÑÁ0ÐCÑ 8 because the function 0ÀFÄ88 is one-to-one. Therefore 0ÐBÑ88 0ÐCÑ 0ÐBÑœ:88 Á: œ0ÐCÑ 0ÐBÑ88 0ÐCÑ Fundamental Theorem of Arithmetic again: ::88 and have different numbers of :8's in their prime factorizations, so they can't be equal. Therefore 2 is one-to-one. Therefore _ is countable. -8œ"Fñ8 Corollary 6 A union of a finite number of countable sets is countable. (In particular, the union of two countable sets is countable.) (This corollary is just a minor “fussy” step from Theorem 5. The way Theorem 5 is stated, it applies to an infinite collection of countable sets E"8 ß ÞÞÞß E ß ÞÞÞ If we have only finitely many, we artificially create the others using g. ) Proof Suppose E"# ß E ß ÞÞÞß E 5 are countable sets. In order to apply Theorem 5, we define EœEœEÞÞÞœgÞ5" 5# 5$ Then _ , which is countable. E"5"5 ÞÞÞE œE ÞÞÞE ggÞÞÞœ-8œ" E 8 ñ Examples i) We proved earlier that the set of positive rationals is countable. The set of negative rationals is countable since ¶ Ðusing, for example, the function 0ÐBÑ œ BÑÞ The finite set Ö!× is countable. Therefore œ Ö!× is countable, by Corollary 6. ii) If F©E, and F is uncountable, then Emust be uncountable. ÐIf E were countable, then Corollary 3 would say that FÑmust be countable. For example, ‘‘ is uncountable because Ð!ß "Ñ © (in fact, we already knew ‘ is uncountable we proved ‘ ¶ Ð!ß "ÑÞ ‘‘‚ Ö!× ‚ Ö!× is uncountable (it's clearly equivalent to ‘ ), and ‚ Ö!× ‚ Ö!× © ‘‘$$. Therefore is uncountable. A similar argument shows that ‘8 is uncountable for any integer 8"Þ iii) Let ‘ be the set of irrational numbers. If were countable, then œ would be countable (by Corollary 6). Therefore is uncountable so there are “more” irrational numbers than there are rational numbers !! iv) A harder example: suppose E œ Ö+"# ß + ß ÞÞÞ , + 8 ß ÞÞÞ × © ‘% and let be an arbitrary % positive number. Let M+88 denote the open interval centered at with length #8" , that is M œ Ð+ %% ß + Ñ . Then E ©_ M (because each + is in the corresponding M Ñà 88##8# 8 8# -8œ" 8 8 8 but the total length of all the intervals M is _ % %. 8 !8œ" #8" This says, informally, that any countable subset of ‘ can be “covered” by a countable collection of open intervals whose total length is %%, where the positive number can be chosen as small as you like. So for a dramatic and counterintuitive example you can cover the set of all rational numbers with a countable collection of open intervals whose total length is "!"! xx.