Talk:Russell's paradox/Archive01

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I have removed this:

What the Barber paradox means is that English allows us to make statements that lead to contradictions whether you answer them "yes" or "no". Inconsistencies and ambiguities are no problem for a natural language; however, they are intolerable in a formal mathematical system. The fact that naive set theory allows us to make such a statement as Russel's paradox means that it is untenable as theory.

The Barber paradox doesn't point to any inconsistency or ambiguity: one simply has to conclude that there is no village where the barber shaves everyone who doesn't shave themselves. Such a village cannot exist, as the paradox shows. AxelBoldt 02:07 Nov 27, 2002 (UTC)

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Note: I have brute-force appended another article on this topic which was originally named "Russell's Paradox" (with a captial P). The two articles should be merged and copyedited into a better article.

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Quoting from article: The whole point of Russell's paradox is that the answer "such a set does not exist" means that the definition of "set" within a given theory is unsatisfactory. Why?
Here's my paradox:
Let there be a set S that contains A and doesn't contain A.
Is this a paradox? No. It is a contradiction. Such a set does not exist. I think this point should be explained in the article -- what is the definition of "set" unsatisfactory for? --Evgeni Sergeev 08:20, 1 Dec 2004 (UTC)

You are wrong. Such a set exists and it is empty. Some mathematical things are confusing if expressed in plain language. Usual words very often have several meanings. Mikkalai 17:04, 1 Dec 2004 (UTC)

That is very clear. I have not studied this mathematical field in detail. Mathematics is often confusing like this because mathematicians agree on certain things between themselves, such as what 'set' means. While in English, saying 'set' is an unambiguous term, if very general, in mathematics it points to a very precise formulation, different from anything else. The mathematical 'set' is a particular instance of the general English term 'set'. Are the 'set' axioms useful for proving something that is not obvious?--Evgeni Sergeev 11:39, 4 Dec 2004 (UTC)

Remark on my 'paradox': if I put 'doesn't contain A' before 'contains A', would it now be {A}, not an empty set?--Evgeni Sergeev 11:39, 4 Dec 2004 (UTC)

Contents 1 M is well-defined? 2 Note: This article uses specialized mathematical symbols. 3 Removed passage that makes no sense to me. 4 Question 4.1 contain 5 Disputed

M is well-defined?

I'm not sure. What does "well-defined" mean in a non-axiomatic system? Maybe nothing defined. Maybe M seems to be well-defined, but it's not w.-d. If it would be, then Russell's paradox couldn't be formalized. I think the problem is that M is not w.-d., but it seems to be w.-d. Gubbubu 23:50, 23 Apr 2005 (UTC)

Note: This article uses specialized mathematical symbols.

Note: This article uses specialized mathematical symbols. Where did this come from? I have read dozens of math articles on wikipedia, all with math symbols, and have never seen this note. It seems rather silly to me, but maybe there is a reason for it: I will not delete it (yet). --Taejo 09:21, 5 Jun 2005 (UTC)

Removed passage that makes no sense to me.

I removed the following passage which makes no sense to me:

"One completely overlooked grammatical approach is the same as Prior's solution to the Liar paradox, completely avoiding types and meta-theories. If one insists that grammatically well-formed sets must contain themselves, then Russell's paradox is obviated. In much the same way the listing of the elements a,b,c of a set does not indicate a set. The usual grammatical way to indicate a set is to put brackets around the elements, e.g. the set containing a,b,c is denoted {a,b,c}. Of course the only objection to Prior's solution is that "but I don't want to have to have sets contain themselves!" to which one of an infinite number of replies is "but I don't want to have to put brackets around a list of elements to show that they are a set!"

Paul August ☎ 14:10, Jun 16, 2005 (UTC)

Question

I can't even begin to see this logic. The paradox states: Let M be the set of all sets which do not contain themselves. Is there any set which does not contain itself? Answer is no. So to continue the logic based on a false statement will naturally introduce contradictions and paradoxical outcomes. No such set M exists and there is no paradox. (68.238.103.89)

I'm not sure I understand what you're saying. Are you claiming that every set contains itself? If that were the case, then no set could be a member of M, so M would be the empty set. This is different from stating that M does not exist. Factitious 08:49, Jun 24, 2005 (UTC)

It says: "The set of all sets that do not contain themselves as members." So, for example, the set A = {1, 2}, has two members, 1 and 2, neither of which is equal to A, so A does not contain itself as a member. Paul August ☎ 02:14, Jun 24, 2005 (UTC)

Whichever way you look at this, Russell's Paradox is not a paradox but a contradiction derived from an incorrect statement. Russell arived at a *paradox* (in fact an incorrect conclusion) because he started with a mathematical statement which is false, i.e. A | A does not contain A. The answer is simple: "All sets contain themselves." Anon. 8:52pm August 21, 2005

I don't think it's true that all sets contain themselves. Consider the empty set, which has no members. If it contained itself, then it would have at least one member. But it doesn't. Factitious 06:49, August 22, 2005 (UTC)

A set does not have to contain any members to contain itself. The empty set *does* contain itself, i.e. it contains *zero* members. It is untrue that a set has to contain at least one member in order to contain itself. When we speak of a set containing itself, we think conceptually of the set as a *whole*, i.e. it is irrelevant whether the set contains any members or not. 10H20 August 22, 2005 (Anon)

Completely wrong. In mathematics, when you say "A contains B" it is the bijective equivalent of saying "B is a member of A". It does not mean "A includes B" (the bijective equivalent of "B is a subset of A") as you suppose in your pseudo-demonstration. If if was the case, then effectively you would start counting the elements that are part of A. But for the verbal expressions "contains" or its reflexive "is a member of", you must ont count the elements in any set. You have just to prove the equality with one of the elements of a set.

So the empty set does not contain itself: you can't find any element in the empty set that is equal to the empty set. Conclusion: not all sets contain themselves.

This is also evident for example with the simple set A={1, 2, 3} with only three elements. None of these elements are equal to A, so A does not contain A. In fact, a very vast majority of sets defined everyday in the common language do not contain themselves.

The set of empty sets also exists. Its cardinality is 1, it does not contain itself, as it contains only one element, the empty set (whose cardinality is zero), and so the empty set is then distinct from the set of empty sets (due to cardinality difference).

But the set of sets is such a set that contains itself, because you can find an element in it (the set of sets) that is equald to the set itself. This is a very strange object not used in common language, because its definition is recursive. But recursive objects do exist in mathematics and are very useful.

A self-contained set does not have to be infinite (i.e. have an infinite cardinality). It may just contain itself and a finite number of other distinct elements, and its cardinality will be the one plus the finite number of these elements.

So please don't confure the very different concepts of containment (element of) and inclusion (subset). Mathematics have distinct symbols to note these relations.

81.49.124.22 03:54, 26 October 2005 (UTC)

Stephen Webb is correct. The empty set is a subset of itself, but it is not a member of itself. If a set contains anything, then it is not the empty set. Factitious 10:17, August 31, 2005 (UTC)

 I think you may be confusing the concept of a subset with that of containment.   

A = { 1, 2, 3 } B = { 1, 2, 3 } C = { 4, 5, 6, { 1, 2, 3 } } A contains three elements, namely 1, 2 and 3. B contains the same three elements. C contains 4 elements, namely 4, 5, 6 and the set { 1, 2, 3 }

A is a subset of B. B is a subset of A. B does not contain A. C contains A. C contains B. A is not a subset of C. Stephen Webb 02:43, 29 August 2005 (UTC)

?? A subset implies containment. I fail to see what you are saying. Also, how can you say the empty set is a *subset* of itself? The empty set is the set which *contains* no members. If what you say is true, then the empty set is indeed not empty, for it contains a subset, i.e. itself. This is ridiculous! :-) 9-9-2005 17H45 (Anon)

You're confusing inclusion with containment. The empty set has no members; that does not mean it has no subsets. Michael Hardy 21:46, 20 November 2005 (UTC)

... and your failure to see is only your failure to see. Michael Hardy 21:46, 20 November 2005 (UTC)

Hardy: A professor of Statistics? Where did you teach? Poor students. If inclusion and containment are not the same thing, then what is the difference? If a set has subsets, then are the subsets not members of the set? Of course they are and you don't fail to see, you simply do not see. 71.248.131.252 02:11, 1 February 2006 (UTC)

In fact, by the above reasoning, an empty set may contain as many members/elements as one wishes: we can have subsets of subsets of subsets of subsets, etc. To say as Webb says: A is a subset of B implies that A is *contained* in B. By definition, a set *contains* itself. A set's features, characteristics and its *nature* is determined by it's elements or lack thereof. Russell confused himself horribly and succeeded in confusing many others by using this very *logic* you state is true. i.e. that an empty set is a subset of itself. Again, by the same reasoning, A is a subset of A since we can take 1,2,3 and form a set from it. This reasoning is in error. The simple *truth* is that a paradox only existed in Russell's interpretation. 9-10-2005 9H58 (Anon)

No, "A is a subset of B" means that everything contained in A is also contained in B. Read that definition carefully, then think about whether the empty set is a subset of the empty set. Factitious 17:09, September 12, 2005 (UTC)

Well, if you look at it this way, then *every* set is a subset of itself, e.g. everything in A is also contained in A, therefore A is a subset of itself. It appears you are okay with saying a set is a subset of itself if it contains no members but you are not okay stating the same thing if a set has members. If you think of it this way, then you can easily state that no set contains the empty set unless it is specified in each set as in the above example of Webb. In fact Webb is mistaken to state that if A is a subset of B and B is a subset of A that B does not contain A: they both contain each other. Containment is determined by *membership*. We assume that the empty set is a subset of all sets because we can form an empty set from the members of any other set. This is containment. The empty set plays the same role in set theory as zero plays in number systems. We can say that 0 < 3 and 1 < 3 and 2 < 3 and thus infer that 0,1 and 2 are *contained* in 3. In this case the 'less than' operator has a similar function to containment. However, it would make *absolutely* no sense to state that 0 < 0. Would it? Of course not! We cannot have that a set contains itself and is a subset of itself for then we have a self-referential (or circular) definition that makes no sense. So we state that a set contains itself (i.e. it has or it does not have members) but it is not a subset of itself. Finally, if we have sets A and B with the same members, then we write A=B (A contains B and B contains A), we do not *write* A<B or B<A. Russell *perceived* a self-referential definition and caught himself in a bad tangle. Thus, every set contains itself but no set can be a subset of itself and Russell's paradox is still not a *paradox*. 14H27 9-12-2005 (Anon)

contain

I have no intention of being dragged into this discussion; it's one of the classic ones where people simply refuse to be convinced. But FWIW I want to point out that some of the confusion may be due to the unfortunate choice of word, "contain". It would be better to say "the set of all sets that are not elements of themselves", which is unambiguous.

The defenders in the discussion are quite correct as to the logic, and are consistently using "contain" to mean "contain as an element". Part of the reason they may be having some difficulty getting their point across, though, is that they seem to be claiming that "contain" does not mean "contain as a subset", and the fact is that, in actual mathematical usage, sometimes it does mean that. Usually the distinction is clear from context, but in a discussion like this one, where the difference is critical and is not clear to everyone, it would be better just to avoid the word. --Trovatore 20:54, 12 September 2005 (UTC)

"The set of all sets that are not elements of themselves". This is even more obscure IMO. The *logic* of the defenders is anything but *correct*. Sets by definition are *not* elements but rather a collection of elements. Sets contain elements that can be anything, including other sets. If a set contains other sets, then in this *context* those sets are elements. In other *contexts* the sets are not elements. The problem arises when one states that 'a set can be a subset of itself'. The empty set is not a subset of itself. No set is a subset of itself. When we talk about a set containing itself, we think *conceptually* of the set as a *whole*. In this case its members/elements or lack thereof are irrelevant. Russell's statement is problematic whichever you look at it. There is no set which does not *contain* itself and there is no set which is a *subset* of itself. Russell reckoned the empty set is a subset of itself but if his logic were correct, then every set is a subset of itself. In the end you have self-referential (or circular) definitions which introduce a peculiar result that appears to be a paradox. This confused poor Russell and has been confusing the academia ever since. Suppose you have an empty room. Is there an empty room within this empty room? By the same token suppose you have a furnished room. Is there a furnished room within this furnished room? What defines an empty room and what defines a furnished room? Is it the furniture or lack thereof? Or is it the room itself? 17H54 9-12-2005 (Anon)

As I said, I'm not going to get into it. --Trovatore 23:27, 12 September 2005 (UTC)

Why did you comment then? Should everyone believe you just because you think it's true? Perhaps you should have allowed the others to answer for themeselves. In my opinion, it would have been better had you not commented at all. (Anon)

I commented on the use of the word "contain". --Trovatore 00:58, 13 September 2005 (UTC)

Noted. As I wrote earlier, I think it is incorrect to use your suggestion because it is fundamentally erroneous: to say that x is both an element of A and x is also not an element of A (as stated in the Wiki article) is a false starting point to argue from. Thus any conclusions from such 'logic' are not paradoxes but *nonsense*. 9H25 9-13-2005 (Anon)

Let S={1}, T={1, 2}, U={3, {1}}. S is a member of U. S is not a member of T. S is a subset of T. S is not a subset of U. 1 is not a member of U. If you like, you can make up your own definitions of "member," "subset," "set," and so on, in such a way that those statements aren't true. But this article is about mathematics, so it uses the definitions that are used in mathematics. If you still don't get it, let's discuss it further on my talk page, since I think this is getting increasingly tiresome for people here. Factitious 05:31, 14 September 2005 (UTC)

Have no intention of producing my own definitions. Naive set theory definitions are sufficient. As this is a 'talk' page, I would prefer to discuss it here. There is a larger audience and chances are that someone might be able to prove me wrong. :-)

I agree that: If S={1}, T={1, 2}, U={3, {1}} then S is a member of U. S is a subset of T.

I do not agree that: S is not a member of T. Proof: If S is not a member of T, then we would not be able to form S from T. Since we can form S from T, it is both a member and a subset of T. S is not a subset of U. Proof: Since S is a member of U, it is also a subset of U for we form new sets from old sets by selecting elements of the old set. Thus, we can form a subset by selecting the set {1} from U. 1 is not a member of U. Proof: 1 is a member of {1} and {1} is a member of U. By transitivity, 1 must therefore also be a member of U. More simply put: x has property y and y has property z means x has property z.

I have not used anything but the standard naive set theory definitions. You may be getting a little confused about the following:

"A subset may be a member and a member may be a subset."

Placing {} around a member is redundant because by definition we assume the empty set belongs to all sets, yet we do not write {} as a member of every set. And how do we form an empty set from a non-empty set? Why, we simply remove all the elements/members. Does the empty set have {} as a member? I think not.

This Wiki article serves no purpose except to make academics (and I make no secret of this, I have a passionate dislike for most academics even though I hold a degree myself but do not think of myself as an academic) wallow in 'intellectual vanity'. Russell's paradox has no place in an encyclopedia. Most Real Analysis and Group Theory professors do not understand it because it is non-sense. To avoid being outcasts, they go along with the flow for Sir Russell was an individual of noble birth and God knows, too many English might be horribly offended. :-) Okay, this is already far too long. I am certain you are smart enough to figure this all out for yourself. As a service to a poor student who reads this article and leaves more confused, you might present this view in addition to the current article perhaps as an addendum. I will not post anymore on this page or any other page in Wikipedia. 8H59 9-14-2005 (Anon)

Your claim that x∈A and A∈B entail x∈B is false. Factitious 04:37, 15 September 2005 (UTC)

Pedagogical example: The glass (element) is in the cupboard (subset) and the cupboard (subset) is in the kitchen (set), therefore the glass (element) is in the kicthen (set). sqrt(2) is in the interval (1,2) and (1,2) is a subset of the interval (0,3), therefore sqrt(2) is in the interval (0,3). The proposition Y is in interval (1,2) entails the proposition Y is an element of interval (0,3) because all elements of interval (1,2) are elements in interval (0,3). Seems claim is true eh?

The set {{1}} has only one member. That member is {1}. {1} and 1 are different things. Therefore, 1 is not a member of {{1}}. Factitious 20:30, 16 September 2005 (UTC)

Agreed: {1} and 1 are different but are 1 and 1 different? They both define a set/subset and a subset/element of a set? Or, {1} is a set and 1 is a subset. What elements does the subset 1 have? Why, the parts that make up 1 of course! I said I would stop and here I am two posts later. Okay, this is the final post. Unless you can come up with a far better argument, Russell's antimony remains *non-sense* in my opinion.

1 is not a set, so it is not a subset. Factitious 20:45, 25 September 2005 (UTC)

Consider the set {{{...{1}...}}}, where there are aleph null pairs of brackets. This set contains one element, the set {{...{1}...}}, where there are aleph null minus one pairs of brackets. But aleph null minus one is aleph null; hence, the set contains itself as an element.

"Russell's paradox has no place in an encyclopedia." That's a bit strong, surely? 134.36.112.67 14:14, 26 September 2005 (UTC)

1 is not a set? Then what is 1? The whole idea behind the invention of set theory was to explain the properties of numbers. Of course 1 is a set: it consists of the parts that make up 1. Is this too hard for you to accept? I don't think it necessary to explain what these parts are. In fact, it is irrelevant except to know that they exist as a *concept*. All human knowledge is conceptual and usually *verifiable* to some or other degree. After all, that's what sets are: a concept of the collection of elements. And yes, Russell's paradox has no place in an encyclopedia:

1) It is not a paradox. 2) It is a proposition based on non-sense. To write 'M is the set of all A such that A is not

  an element of A' is *illogical* and untrue. By definition, all sets *contain* themselves.
  A set *cannot* be an element of itself. The aleph example provided is *incorrect*.
  The initial assumption that there are aleph null pairs of brackets is false since one of
  the brackets contain {1}. The logic that follows is nonsense.
  I do not doubt Russell had exceptional talent in logic. However, anyone with similar 
  talent can begin with a false premise that *leads* to a *paradox*. There are no shortage
  of these.

3) Wiki's policy is to publish information that is unambiguous and generally accepted even if

  it's proven to be false at some later time. I think this article should be discarded or
  rewritten in such a way that those who read it understand how a false premise can lead
  to a paradox. It is a good example of this. However, it surprises me how this topic is
  given so much  attention when there are hundreds of far more important topics. In fact,
  by strict definition of the word encyclopedic, anything that is published in a book 
  qualifies. In this sense, Russell's paradox is encyclopedic. In another sense, encyclopedic
  implies *circular education*: Russell's paradox does not *educate*, rather it confuses
  and misleads a learner/student. Knowledge is supposed to *edify*. By the same token,
  pornography is also encyclopedic but does it edify? I think not. Please conduct a survey
  and see how many readers actually understand this nonsense.

Disputed

The article claims that "in Cantor's system" the Russell set is "a well-defined set". Is there any evidence for this? Quoting from "Contributions to the Founding of the Theory of Transfinite Numbers:

By an "aggregate" (Menge) we are to understand any collection
into a whole (Zusammenfassung zu einem Ganzen) M
of definite and separate objects m of our intuition
or thought.

Now, how can the Russell set be considered to be a collection into a whole of definite and separate objects, when those objects themselves are aggregates that have not yet been "aggregated"--i.e., in today's terms, when their rank is as high as that of the Russell set itself? It seems clear that Cantor had in mind starting with familiar mathematical objects (naturals, reals, points) and aggregating them, and then further aggregating the aggregates. I see no justification for the notion that he thought every property had to have an extension. Indeed, doesn't that run counter to the notion of "limitation of size"?

(For Frege, on the other hand, what's written is clearly a fair cop--as I understand it he considered sets more or less to be extensions of properties.)

But I'm certainly no historian; if I'm wrong about this then someone please set me straight. Otherwise I think the page should be changed; Zermelo's axiomatization was not a way to fix problems in Cantor's concept of sets, but simply a provision of precise points of reference in applying it. --Trovatore 29 June 2005 02:49 (UTC)

• Cantor, Georg (1955). Contributions to the Founding of the Theory of Transfinite Numbers. Dover. ISBN 0-486-60045-9. 

I'm no historian either, but it seems to me that you are implicitly giving Cantor's definition a predicative reading (aggregates that have not yet been "aggregated", as you put it), while the distinction between predicative and impredicative arose only as a response to the paradox. Leibniz 16:20, 26 August 2005 (UTC)

So the distinction I'm interested in here isn't really predicative/impredicative; that refers not to sets, but to definitions of sets. As I read it, what Cantor says has nothing to do with whether a particular set has a definition. The intuition is, you've got some things, and you gather some of them together into an aggregate, without regard to whether there's any sort of rule that tells you which ones to pick. So it's more about the extensional/intensional distinction than the predicative/impredicative one, and what I'm saying is that Cantor's description is purely extensional. --Trovatore 04:30, 30 August 2005 (UTC)

The reason this is important is that there are several articles in the Wikipedia, such as Category:Set theory and Naive set theory, that imply that Cantor's conception of set led to the paradoxes, and had to be revised by Zermelo to eliminate those paradoxes. I think that's just wrong. The Cantorian conception was and remains fine; it was Frege's misinterpretation that led to the paradoxes. The Zermelo-Fraenkel axiomatization was very important, of course--but not because of the antinomies. It was important because it gave a precise point of reference.

I hope at some point to clean these articles up, but it's a tricky job. --Trovatore 04:34, 30 August 2005 (UTC)

That's an awful lot of verbiage just to say you don't find the Russell paradox to be paradoxical. That's just fine; you don't have to. Nevertheless it did utterly refute a certain view of sets (Frege's) that seemed plausible at the time to a great intellect, and the paradox still has value in defining the limits of attempts to make modifications of that view work.

What I question is not the value of the paradox, but whether (as apparently asserted) it really works against Cantor's conception the way it undeniably does against Frege's. --Trovatore 06:09, 26 October 2005 (UTC)

That's not verbiage. I included a logic demonstration with the mathematic justification. The paradox only exists because its sentence is not well-formed and forgets to specify a necessary condition (and because the reader does not see that it is really missing). If you take any false condition, you can conclude anything, including creating paradoxes... fr:Utilisateur:Verdy_p 14:36, 26 October 2005 (UTC)

You are being very nice to Russellophiles. The fact is that Russell's paradox was never a paradox, is not a paradox and will never be a paradox whichever way you look at it. —Preceding unsigned comment added by 192.67.48.22 (talk • contribs) 20:26, 28 October 2005

I have slightly changed the formulation to: "In Cantor's system, M would be a well-defined set". Obviously naive set theory is not internally consistent, so the 'paradox' is: if Cantor's system was internally consistent, M would be well-defined, yet cause an antinomy. Rasmus (talk) 15:20, 16 November 2005 (UTC)

Please see the top of this "Disputed" section, before it got muddled up by people who don't see what the paradox is about in the first place. The whole reason I added the disputed tag is that I'm not convinced the paradox applies to the Cantorian conception (at least, the later one, once he added "limitation of size"). As to whether naive set theory is internally consistent, that depends on what you mean: Certainly Fregean naive set theory is internally inconsistent. It's not clear that Cantorian naive set theory is. (And by the way, there's a very serious structural flaw in the way Wikipedia makes the division between "naive" and "axiomatic" set theory. What set theorists actually do--I am one--would be "naive" set theory according to the definitions given here.) --Trovatore 16:05, 16 November 2005 (UTC)

Well, the disputed tag has been there for four months now. It is only meant as a temporary measure to mark sections that are currently being discussed, not as a permanent thing. It really comes down to the unrestricted Comprehension axiom, which you are probably correct in saying only was included in Fregean set theory. I will try to change the formulation to this. If you disagree, please replace it with your own formulation, rather than reinserting the disputed tag. Rasmus (talk) 17:31, 16 November 2005 (UTC)

This version is fine with me. I didn't do that myself because I thought there might be people who want to defend the claim that Russell's paradox works against the Cantorian conception. If there are, I guess they can now make their case here. --Trovatore 17:55, 16 November 2005 (UTC)