Cantor Confusion
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From: Virgil on 20 Oct 2006 14:28 In article <7c62b$4538b99a$82a1e228$5533(a)news2.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Franziska Neugebauer wrote: > > > Han de Bruijn wrote: > > > >>Franziska Neugebauer wrote: > >> > >>>Han de Bruijn wrote: > > > > [...] > > > >>>>Good! Now given two such members x and y. What does x = y mean? > >>> > >>>x = y :<-> Az(z e x <-> z e y) > >> > >>Of course. Because all members are sets. But I think this is an > >>infinite recursion with the equality definition. Does it end > >>somewhere? > > > > Where do you spot recursion? > > Two sets are equal if they have the same members. Two members are equal > if they have the same members ... Right? By that argument, any statement that can be repeated is recursive. Which makes every statement recursive. Unless HdB can think up one which cannot be repeated.
From: MoeBlee on 20 Oct 2006 14:59 Han de Bruijn wrote: > Ha! Mathematicians can't even define their most frequently used symbol, > which is the equality ' = '. And that is a prerequisite for their "is". Wrong. Equality is definable in set theory. MoeBlee
From: MoeBlee on 20 Oct 2006 15:03 MoeBlee wrote: > Han de Bruijn wrote: > > Ha! Mathematicians can't even define their most frequently used symbol, > > which is the equality ' = '. And that is a prerequisite for their "is". > > Wrong. Equality is definable in set theory. To be exact, I should qualify, that this is within the inherent first order limitation that no theory can preclude equality being satisfied by a mere equivalence relation, unless the first order semantics have a "built in" provision that equality be interepreted as the actual identity relation. MoeBlee
From: MoeBlee on 20 Oct 2006 15:32 Han de Bruijn wrote: > > x = y :<-> Az(z e x <-> z e y) > > Of course. Because all members are sets. But I think this is an infinite > recursion with the equality definition. Does it end somewhere? "Infinite recursion with the equality definition". You're amazing in your sheer ignorance and arrogance to opine on what you are completely ignorant. Read chapter one of virtually any textbook in set theory. There are three common but different approaches, which end up amounting to the same thing: 1. Identity is a logical predicate with semantics stipulated and axioms for identity. Then the axiom of extensionality is added, so that there is one primitive logical predicate symbol and one primitive non-logical symbol (membership symbol). 2. Identity is taken as a primitive non-logical predicate symbol with an identity theory with axioms but no stipulated semantics. Then set theory is an extension of identity theory, and the axiom of extensionality is added, so that there are two primitive non-logical predicate symbols. 3. Identity is not taken as primitive, but is defined from the primitive membership symbol. So there is just one primitive. And by the way, regarding my previous comment about the first order limitation (equivalence classes satisfying equality), since set theory has a finite number of primitives, we still can enforce both directions of Leibniz's law without special identity semantics and even though one direction of Leibniz's law is not statable as even a first order schema. MoeBlee
From: MoeBlee on 20 Oct 2006 16:09
Han de Bruijn wrote: > Two sets are equal if they have the same members. Two members are equal > if they have the same members ... Right? No, you need to cut that confusion at the root. We quantify for ALL objects, whether sets, members, classes, or whatever. For ANY x and ANY y, Az(zex <-> zey) -> x = y And identity theory already gives us: For any x and any y, x=y -> Az(zex <-> zey) So we have: For any x and any y, x=y <-> Az(zex <-> zey) I mean, really, if you can't even come to grips with the axiom of extensionality, this is not a failure of set theory, but of your own self-imposed ignorance and unwillingness to give the material even a modicum of clear attention. MoeBlee |