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 EÁgß and E© ß the Well-Ordering Principle says there is a smallest member +−E" .
Since E is 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 ÞÞÞ contains 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 E is 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 F into 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$$ "# ã F88 œ E ÐE " ÞÞÞ E 8" Ñ
ã
To see that the F8 's are pairwise disjoint:
Consider FF78 and where, say, 78Þ=−F If 7 œ E7" ÐE ÞÞÞ E 7" Ñ , then B − E 7 . But that implies B  F88 œ E ÐE " ÞÞÞ E 7 ÞÞÞ E 8"78 ÑÞ So F F œ gÞ
To see that __ : --8œ"Fœ88 8œ" E Since , certainly we have __ . F©E88--8œ" F© 8 8œ" E 8 Conversely, if _ , then is in at least one of the 's. Pick B−-8œ" E88 B E the smallest 8B−EB−FB!8Þ8 such that !! Then because is not in
E"8" ÞÞÞ E! , the set “discarded” in the definition of F 8! . Therefore ___ , so . B−---8œ" F888 8œ" F œ 8œ" E
2) Since each FE88 is a subset of a countable set , Corollary 3 tells us that each F 8 is countable. We want to show that _ is countable. -8œ"F8
There is a one-to-one function 0ÀFÄ88 , for each 8 (since F 8 is countable). List the prime numbers as :"#$ œ #ß : œ $ß : œ &ß ÞÞÞß : 8 ß ÞÞÞÞ
Define _ as follows:if , then 08ÐBÑ 0À-8œ" F88 Ä B−F 0ÐBÑœ:8 −
For example, if B−F$$$ , then 0 ÀF Ä and 0ÐBÑ $ is some natural 0ÐBÑ$ numberÞ Suppose, to illustrate, that 0$ ÐBÑ œ '%Þ Then 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 ¶ Ð using, 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 E must be uncountable. ÐIf E were countable, then Corollary 3 would say that FÑ must 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