Talk:Russell's paradox/Archive 1

message: 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)
replies: text: 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)
replies: text: 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)

text: 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)

message: 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)
thread_title:  M is well-defined? 

message: 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)
thread_title:  Note: This article uses specialized mathematical symbols. 

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

text: 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)

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

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

text: 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)

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

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

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

text: Consider the set , 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)
thread_title:  Question 

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

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

text: 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)

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

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

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

text: Consider the set , 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)
thread_title: contain