Talk:Countable set/Archive 1

message: I checked about a dozen random textbooks I have, and all but 2 defined "denumerable" as "countably infinite", although the 2 that defined it as "countable" are quite well-known, so I'll put a comment indicating it's not completely universal agreement. Revolver 19:12, 9 Jul 2004 (UTC)
thread_title:  Whether ''countable'' implies ''infinite'' (part 2) 

message: By definition of some books (i.e. Walter Rudin's Principles of Mathematical Analysis) a countable set must be infinite, and a finite set is not countable.  However, he uses the terminology (not mentioned in Wikipedia): 'at most countable'. Both countable (infinite) and finite sets are 'at most countable'.  An infinite sum of finite sets, for example, is at most countable.  To prove that this sum is strictly 'countable' one must prove that it is infinite.
Nerses Ohanyan
:It's a bit strange that the use of countable to mean countably infinite wasn't mentioned in the article. I've added it now. --Zundark 07:15, 17 Sep 2004 (UTC)
replies: text: ???? But it's in the previous sentence? "A countable set is a set which is either finite or countably infinite."  I've reverted.  Noisy | Talk 09:58, 17 Sep 2004 (UTC)
replies: text: No, that's a different definition. By that definition every finite set is countable, whereas by the definition Nerses Ohanyan was talking about, no finite set is countable. So I'm going to revert back, perhaps with some extra clarification. --Zundark 10:09, 17 Sep 2004 (UTC)

text: OK - I think I understood on the third or fourth reading. Boy, I'm glad I'm not a mathematician. ;-)  Noisy | Talk 10:31, 17 Sep 2004 (UTC)
thread_title:  Whether ''countable'' implies ''infinite'' (part 3) 

message: Does countable mean that an element in set B can be indexed? Or am I missing something
In other words set A and set B are bijective
if I can write array[n]=Bk, where n is a natural number, is this sufficient for B to be countable? 59.145.141.130 10:49, 19 January 2006 (UTC)
replies: text: Um, yes. If I understand you correctly, that's the definition of countability. —Keenan Pepper 17:46, 19 January 2006 (UTC)
thread_title:  Query 

message: "THEOREM: (Assuming the axiom of choice) The union of countably many countable sets is countable."
I think we don't need the axiom of choice here, as we can choose by order of counting. The Infidel 09:45, 22 January 2006 (UTC)
replies: text: You do need at least a weak form of the Axiom of Choice. The result can't be proved in ZF. --Zundark 12:48, 22 January 2006 (UTC)
replies: text: But where? There's scetch of proof in the article. I can't see anything that can not be done by the enumeration functions of these countable sets (and set of sets). 84.160.220.72 14:49, 22 January 2006 (UTC)
replies: text: But first you must choose those infinitely many enumeration functions. --Zundark 16:04, 22 January 2006 (UTC)
replies: text: I'm not sure about that. If a set is countable, there exists an enumeration. Any one will do.And I don't need a choice function that gives an enumeration for every subset of enumerations but only one enumeration out of the full set of all enumerations. The Infidel 19:06, 22 January 2006 (UTC)
replies: text: Let the original sets be An, for n in N. You need an enumeration for each An. For each n, let En be the set of all enumerations for An. Each En is non-empty (as you point out). You need to choose an element from each of the infinitely many non-empty sets En. It's the Axiom of Choice that says you can do this. --Zundark 20:28, 22 January 2006 (UTC)

text: You're implicitly assuming the Axiom of Choice in deciding that it's possible to choose an element out of the Cartesian product of infinitely many sets (since that's equivalent to choosing one element from each of the infinitely many sets). Ruakh 20:55, 23 January 2006 (UTC)
replies: text: It's not necessary to choose an element, it suffices to know it's in there. No need to choose an enumeration, as long as we can show there is one. The Infidel 21:47, 23 January 2006 (UTC)
replies: text: The point is that you don't know there's an element in the Cartesian product if you don't have the Axiom of Choice. Without the Axiom of Choice, a Cartesian product of non-empty sets can be empty. --Zundark 22:44, 23 January 2006 (UTC)
replies: text: That was realy a point. I'm pretty sure I could construct a cartesian product as a restriction of some power set, but I never checked my "tools" if they depend on the axiom of choice or not. Can you provide any source, evidence or hint for your statement? The Infidel 18:26, 24 January 2006 (UTC)
replies: text: If the Axiom of Choice were false, then there would be a set of non-empty sets without a choice function. The Cartesian product of these non-empty sets would be empty, because any element of the Cartesian product gives a choice function. --Zundark 19:26, 24 January 2006 (UTC)

text: Remember that the Axiom of Choice is known to be true for any finite collection of sets; the difficulty comes in when we're talking about an infinite collection of sets. So, 1*2 is well-defined and nonzero, but 1*2*3*... might not be. Now, obviously 1*2*3*... is going to be greater than zero, because obviously the Axiom of Choice is true, but that doesn't change the fact that we're relying on the Axiom of Choice (or some equally non-ZF assumption). Ruakh 20:56, 24 January 2006 (UTC)
replies: text: I see that for infinite cardinals this is not (provably) incorrect, but still very counterintuitive. After all, as we can't proof axioms, they are not the least choosen because they are intuitive. The Infidel 21:10, 24 January 2006 (UTC)
thread_title:  Axiom of choice 

message: Someone recently added:
:Note that the number of elements of the set of natural numbers is not itself a number in the sense that you can do the usual kind of addition and subtraction.
I couldn't tell what this meant exactly, so I removed it. You can add cardinalities of sets, simply by taking two disjoint sets with those cardinalities and taking the cardinality of their union. —Keenan Pepper 17:17, 22 January 2006 (UTC)
replies: text: After your reverting, the article states "In mathematics the term countable is used to describe the size of a set, i.e. the number of elements it contains.". For infinite sets, this "number" is not a number in the "usual" sense: you cannot add up, subtract or divide it in any way we are used to do with natural numbers or the elements of a field or ring. For the "number" of elements of a countably infinte set, the equation \aleph_0 + 1 =\aleph_0 holds. If you remove anything in wikipedia that you don't understand exactly and without asking, there'll probably be nothing left. The Infidel 18:56, 22 January 2006 (UTC)
replies: text: The new version by Ruakh is better all around. —Keenan Pepper 21:58, 22 January 2006 (UTC)
replies: text: Thanks! I try. :-) Ruakh 22:51, 22 January 2006 (UTC)
thread_title:  Numbers or not numbers 

message: This page was getting rather long and unwieldy, so I've moved old discussions (going from 10 December 2001 to 7 March 2006) to Talk:Countable set/Archive 1; I hope no one minds. Ruakh 20:58, 29 April 2006 (UTC)
thread_title:  Archive 1 

message: This section currently asserts that the set of all infinite strings in English is countable. Since there is a one-to-one correspondence between {decimal expansions of real numbers} and {infinite lists of digits in English}, surely this is false? —The preceding unsigned comment was added by 193.170.117.10 (talk • contribs) .
:I'm not sure I agree that the section asserts quite that; rather, it asserts that {numbers that can be completely characterized by finite or infinite strings of English} is countable. Your argument is sound, though, and that definitely needs to be fixed; I'll take care of it. Thank you! Ruakh 19:39, 29 April 2006 (UTC)
replies: text: Define Dm as "The real number between zero and one whose n-th digit after the decimal point is '7' if the n-th mentionable number is a real number whose n-th digit after the decimal point is '2', '3', or '4' and is '3' otherwise.".  Now suppose that this is the k-th mentionable number.  What happens when you try to compute the k-th digit of Dm?  You go into an infinite loop; and never get a value.  So you cannot make your list without solving the Halting problem which is impossible. JRSpriggs 09:21, 9 May 2006 (UTC)
replies: text: To me it looks like Dm above, if it appears in the k-th position of the list, can't have a k-th digit in a manner compatible with its definition, and therefore doesn't appear on the list at all; I don't quite understand where the halting problem comes in. The set of mentionable numbers has to include some uncomputable numbers, but I don't see why not being able to compute the digits of numbers on the list prevents the list's being definable. Invisible Capybara 12:57, 9 May 2006 (UTC)

text: That's a good point. It's quite easily proved that if the set of mentionable numbers is well defined, then it must be countable (assuming that mention means complete characterized by a finite English string), but Cantor's diagonal argument implies that if a set is countable, then you can mention a real number that's not in the set. Here, that implies that if the set of mentionable numbers is well defined, then you can mention a number that's not in the set of mentionable numbers, which is impossible. Ergo, the set of mentionable numbers cannot be well defined. (I'm not sure if this is appropriate to this article, though; it's an interesting example of a proof that makes use of countability, but it might constitute original research. Plus, there have got to be better examples of proofs that make use of countability.) Ruakh 18:14, 9 May 2006 (UTC)
replies: text: I think it's reasonable to doubt that mentionable numbers are well-defined. I'm inclined to say that the mentionable numbers section should be removed unless a reference can be found which addresses this issue satisfactorily. Invisible Capybara 13:57, 10 May 2006 (UTC)
replies: text: It's more than reasonable to doubt it, seeing as it's easily disproved (using what we've already said). Ruakh 04:02, 11 May 2006 (UTC)

text: I disagree with your assessment. Some numbers that are clearly mentionable, e.g. one and pi. True, English isn't perfect, and there exist ambiguous English strings; but I don't think that's the problem here: even if English were completely well defined, the set of mentionable numbers could not be, as shown above. Ruakh 04:02, 11 May 2006 (UTC)
replies: text: I did not mean that there are not some numbers which are clearly mentionable.  I said that the *SET* of *ALL* mentionable numbers does not exist.  As compensation for the loss of that section, I added a section on the minimal model of set theory which includes most numbers that people are likely to mention in practice (unless they are interested in large cardinals).  JRSpriggs 07:24, 11 May 2006 (UTC)
replies: text: By the way, I think that the paradox which we have been discussing here is a version of the Berry paradox. JRSpriggs 07:37, 13 May 2006 (UTC)
thread_title:  Mentionable numbers 

message: The OED only lists 'denumerable' as synonymous with 'countable', and I have never heard of it used to mean 'countably infinite'. I have swapped round the terms in the article so that it defines the terms as synonyms, but then mentions the alternative use in the following paragraph. Oliverkroll 13:13, 15 May 2006 (UTC)
replies: text: That looks fine. I think it's less confusing this way, anyway. Ruakh 15:05, 15 May 2006 (UTC)
replies: text: According to "Webster's Encyclopedic Unabridged Dictionary of the English Language (1989)" and to my memory of what I was taught in school and what I have used all my long life, "countable" means either finite or equinumerous with the natural numbers and "denumerable" means equinumerous with the natural numbers (not finite).  I am changing it accordingly.  JRSpriggs 06:30, 16 May 2006 (UTC)
replies: text: On closer reading the OED actually does have 'denumerable' = 'countably infinite' as the primary meaning. Sorry about that Oliverkroll 14:35, 16 May 2006 (UTC)
replies: text: Incidentally, it defines countable (in the mathematical sense) as simply denumerable, with no elaboration. Indeed, I can't find any source that defines countable and denumerable differently; rather, it seems to me that mathematicians tend to use countable almost exclusively, and define it the broader way, while dictionaries tend to define both countable and denumerable the narrower way, with the broader way given as an afterthought or not at all. Ruakh 14:47, 16 May 2006 (UTC)
replies: text: Rudin's Introduction to Real Analysis defined countable as countably infinite, and "at most countable" as finite or countably infinite. So yes, the term does vary by author. I think authors use countable or denumerable and not both though. Althai 07:09, 30 November 2006 (UTC)
replies: text: I have made significant changes to match Walter Rudin's book. Teply (talk) 00:01, 11 December 2007 (UTC)
replies: text: Reverted non-standard usage. JRSpriggs (talk) 04:09, 11 December 2007 (UTC)
thread_title: Countable/denumerable

message: I came here from a link to Denumerable (I suspected it was a synonym of countable) and was rather confused to find that the lead did not define the term I was looking for. So I'm moving it there (it's common to mention synonyms in the lead anyway). --Taejo|대조 09:38, 13 June 2007 (UTC)
thread_title:  Denumerable in lede 

message: In the explanatory text of the article this paragraph appears: 
"The resulting mapping is like this: 0 ↔ (0,0), 1 ↔ (0,1), 2 ↔ (1,0), 3 ↔ (2,0), 4 ↔ (1,1), 5 ↔ (0,2), … It is evident that this mapping will cover all such ordered pairs."
It beguilingly asks us to believe what is taken to be an obvious correlation between the set of natural numbers and the cartesian product NxN, but no mapping is given to bolster the veracity of the correlation. I believe I have worked out the mapping to be such that the pair (i,j) correlates to the number (i+j)(i+j+1)/2 + (i+1). Thus, (0,0) maps to 1.
My maths is sufficient to show that the mapping is unique in that for (i,j) and (i,k) then j=k to get the same number; and similarly for (i,j) and (k,j). However, my maths is not sufficient to show that if the mapping is the same for (i,j) and (p,q) then we must have i=p and j=q, though, beguilingly, I suspect it's so.
Presuming this to be correct then it answers an earlier question concerning the position of the quotient m/n in the sequence.
—The preceding unsigned comment was added by Jimalton (talk • contribs) .
:I think you're looking at it the wrong way. It's really not trivial to see what natural number gets assigned to each ordered pair — especially since the path goes back and forth, rather than going the same way in each diagonal — but I think it's fairly self-evident that the path covers each ordered pair exactly once, and hence that you can number the ordered pairs by going along the path and numbering them sequentially. I don't think giving a mathematical formula would make that any more clear; indeed, I think it would distract from the real argument. Ruakh 02:55, 5 November 2006 (UTC)
replies: text: I think it would be nice to at least see an inductive proof. —The preceding unsigned comment was added by 74.109.7.4 (talk) 17:19, 13 February 2007 (UTC)

text: Oh, question: you do see the image above that paragraph, right? Because if that image failed to load for you, then I can definitely understand your confusion. Ruakh 02:56, 5 November 2006 (UTC)
thread_title: Cartesian product

message: THEOREM: (Assuming the axiom of choice) The union of countably many countable sets is countable. —The preceding unsigned comment was added by Roadrunner (talk • contribs) .
:You're guaranteed that for each of the countable sets you're uniting, there's some bijection from that set to the natural numbers. You are not guaranteed, unless you have the Axiom of Choice, that you can choose one such bijection for each of them. Since the proof assumes that these bijections have already been chosen, the proof relies on the Axiom of Choice (or some other non-ZF axiom). For a fuller explanation, see Talk:Countable set/Archive 1#Axiom of choice. Ruakh 21:33, 28 November 2006 (UTC)
replies: text: perhaps one could clarify slightly: You are not guaranteed, unless you have the Axiom of Choice, that you can choose one such bijection for each of them "simultaneously", meaning the bijections you choose is an element of a Cartesian product. Mct mht 23:32, 22 December 2006 (UTC)
thread_title:  Explain why axiom of choice is needed 

message: I am interested to know if bijection is a necessary requirement for answering the question: "Are two sets of the same size?" For comparing finite sets, one-to-one correspondence appears to yield onto mapping as a natural consequence rather than require it to reach a determination. For countably infinite set comparisons, one-to-one correspondence again seems sufficient with onto mapping following as a consequence, not preceding as a prerequisite. For sets of greater orders of infinity beyond countability, there are power set inequalities.  Is bijection forced by considering paradoxical results of those set theories only requiring one-to-one correspondence or power set inequalities to determine the size relation between two sets, and if so, what paradoxical results are avoided? Hotfeba 22:11, 9 April 2007 (UTC)
replies: text: The definition of having the same cardinality is that there is a bijection.  If two sets have the same cardinality, then necessarily there is a bijection.  That does not mean it is necessary in all cases to explicitly construct such a bijection in order to prove they have the same cardinality.  The Cantor-Bernstein theorem says that if there's an injection in each of the two directions, then there is a bijection. Michael Hardy 22:37, 9 April 2007 (UTC)
replies: text: True, but a theory in which cardinality by bijection is taken as axiomatic or by definition tends to have less generality than a theory where bijection is proved to be merely a theorem under that theory. Your citation of the theorem leading to bijection is directly on point. Hotfeba 23:04, 9 April 2007 (UTC)
replies: text: "Bijection" and "one-to-one correspondence" mean the same thing. Your question is incoherent. JRSpriggs 10:43, 10 April 2007 (UTC)
replies: text: Ahh, I see. "Bijection" is merely shorter jargon for "one-to-one correspondence" and is not necessary for the derivation of theorems in the theory we all assume we are talking about, unless the equivalence you speak of in your first statement is, either by definition or by necessity for derivation, not symmetric. In fact, I now see that one-to-one corespondence may be excessive for all finite set comparisons when all that is needed is a simple one-to-one relationship between one set's elements and the elements of the other in a choose-without-replacement scheme: if there are unmatched leftovers in either set, then the sets are of different sizes; if not, then they must be the same size. Unless you have a counter-example, it appears we can agree that the concepts of cardinality and bijection are not required to determine if two finite sets are of the same size. Hotfeba 20:26, 11 April 2007 (UTC)
replies: text: Re: "[…] the concepts of cardinality and bijection are not required to determine if two finite sets are of the same size.": Well, "cardinality" means "size", but overall, that's a true statement. The reason we formally define "cardinality" in terms of bijections is so we can apply the term to infinite sets; if we restrict consideration to finite sets, then there's no need for that. —RuakhTALK 00:01, 12 April 2007 (UTC)
replies: text: Apologies, but I was considering sets that humans or their computers could compare in reasonable periods of time. For programming design/deadline/delivery purposes, it is helpful to know that implementing strict cardinality tests on finite sets is not necessary by proof or definition. Hotfeba 00:12, 12 April 2007 (UTC)

text: Don't trolls try to engender anger? I'm not feeling angry, or even frustrated. —RuakhTALK 13:51, 12 April 2007 (UTC)
replies: text: Well, if you think that he is being reasonable, you are free to keep talking to him. But I think that he is either intellectually dishonest (my definition of a troll) or so stupid/confused that it is a waste of time to talk to him any more. JRSpriggs 02:27, 14 April 2007 (UTC)
thread_title:  Is bijection necessary? 

message: Bablefish suggests that the linked Japanese article really is about countably infinite sets, rather than countable sets as we define it here.  Can any reliable Japanese-speaker comment on :ja:可算無限集合?  — Arthur Rubin |  (talk) 23:55, 17 June 2007 (UTC)
replies: text: Even so, might it not be the closest equivalent article on their wiki? In which case, it should be linked anyhow. JRSpriggs 07:25, 18 June 2007 (UTC)

text: I don't see the difference. "Countable set" and "countably infinite set" are not synonymous, but if I had to write an article about each, they'd be almost word-for-word the same. —RuakhTALK 16:14, 18 June 2007 (UTC)
replies: text: I hadn't really considered that.  I had assumed this might have been the result of User:WAREL on one side or the other.  I guess it should remain, then. — Arthur Rubin |  (talk) 18:13, 18 June 2007 (UTC)
thread_title:  Japanese wikipedia 

message: From the article: "Remember our example of the Scrabble words."  This is the first occurance of the word "scrabble" (case insensitive) in the article.  What is the reader supposed to be remembering?  --Thinboy00 @901, i.e. 20:36, 4 December 2007 (UTC)
thread_title:  scrabble example