The End of an Era - That of The Natural Numbers
in [Logic]
|
Prev: What are deliberately flawed & fallacious Arguments? Sophistry!
Next: sci.lang is not meant for advertising
From: Aatu Koskensilta on 24 May 2010 09:39 Marshall <marshall.spight(a)gmail.com> writes, to Nam: > You know, you often complain about how your "opponents" don't address > technical definitions (although of course they do.) Well, I'm for the moment done with Nam's technical definitions. But let me just comment on something you wrote a while ago, about whether "empty model" means anything at all: I'm disinclined to think so. I am under the impression that in ordinary usage, for something to qualify as a "model" its domain needs to have at least two elements. Certainly we can meet all the other criteria for a model with only a 1-element domain, but that's quite the degenerate case. In the cardinality-1 model, there is only one possible table for each arity; the constants are all the same; every possible such model of every possible signature can be fully axiomatized by the same degenerate axiom, "x=y". First, a correction: there are 2^n non-isomorphic one-element models for a language with n relation symbols -- each relation can either hold or fail to hold of the single element in the model. Second, an observation: in group theory, to pick a random example, it is often convenient to count the zero-element and one-element structures satisfying the group axioms as groups, and indeed not doing so makes certain definitions and statements of results come out quite needlessly convoluted. This conceptual simplification that results from the admittance of "degenerate" examples is a common phenomenon in mathematics. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Daryl McCullough on 24 May 2010 10:19 Nam Nguyen says... > >Daryl McCullough wrote: >> For example, suppose I tell you that all the coins in my pocket >> are quarters. > >So, per Tarski's concept of truth, _factually_, the set of coins in >your pocket is non-empty No, that's not true. "All coins in my pocket are quarters" is logically equivalent to "There are no coins in my pocket that are not quarters", which is consistent with (and implied by) "There are no coins in my pocket". >(iv) A universal statement "for all x A(x)" is true if and only if each >object satisfies "A(x)" Which is true if and only if there is no object that satisfies not A(x). >Note that "each object" would presuppose or require that there exist >objects. Nope. -- Daryl McCullough Ithaca, NY
From: Marshall on 24 May 2010 10:23 On May 24, 6:39 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Marshall <marshall.spi...(a)gmail.com> writes, to Nam: > > > You know, you often complain about how your "opponents" don't address > > technical definitions (although of course they do.) > > Well, I'm for the moment done with Nam's technical definitions. But let > me just comment on something you wrote a while ago, about whether "empty > model" means anything at all: > > I'm disinclined to think so. I am under the impression that in ordinary > usage, for something to qualify as a "model" its domain needs to have > at least two elements. Certainly we can meet all the other criteria for > a model with only a 1-element domain, but that's quite the degenerate > case. In the cardinality-1 model, there is only one possible table for > each arity; the constants are all the same; every possible such model > of every possible signature can be fully axiomatized by the same > degenerate axiom, "x=y". > > First, a correction: there are 2^n non-isomorphic one-element models for > a language with n relation symbols -- each relation can either hold or > fail to hold of the single element in the model. For predicates, but not for functions, right? Sure. If we restrict consideration to only those signatures that omit predicates, would my description hold? > Second, an observation: > in group theory, to pick a random example, it is often convenient to > count the zero-element and one-element structures satisfying the group > axioms as groups, and indeed not doing so makes certain definitions and > statements of results come out quite needlessly convoluted. This > conceptual simplification that results from the admittance of > "degenerate" examples is a common phenomenon in mathematics. Okay. I recall reading somewhere that a model has to have at least two elements in the domain. I thought it weird, so it stuck in my head. But doesn't a group require an identity element? How do you have an identity element with an empty domain? I am entirely down with the idea of simplification via the admittance of degenerate cases, but in this particular case, it appears to me that requiring the model to be nonempty is what results in the simplification. However this may be something that is only true in the context of universal algebra and not the wider context of model theory. Then again it could just be a misunderstanding on my part; I will defer to your superior knowledge here. Marshall
From: Daryl McCullough on 24 May 2010 10:31 In article <87pr0lwh9f.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta says... > >Marshall <marshall.spight(a)gmail.com> writes, to Nam: > >> You know, you often complain about how your "opponents" don't address >> technical definitions (although of course they do.) > >Well, I'm for the moment done with Nam's technical definitions. But let >me just comment on something you wrote a while ago, about whether "empty >model" means anything at all: > > I'm disinclined to think so. I am under the impression that in ordinary > usage, for something to qualify as a "model" its domain needs to have > at least two elements. Certainly we can meet all the other criteria for > a model with only a 1-element domain, but that's quite the degenerate > case. In the cardinality-1 model, there is only one possible table for > each arity; the constants are all the same; every possible such model > of every possible signature can be fully axiomatized by the same > degenerate axiom, "x=y". > >First, a correction: there are 2^n non-isomorphic one-element models for >a language with n relation symbols -- each relation can either hold or >fail to hold of the single element in the model. Second, an observation: >in group theory, to pick a random example, it is often convenient to >count the zero-element and one-element structures satisfying the group >axioms as groups, and indeed not doing so makes certain definitions and >statements of results come out quite needlessly convoluted. This >conceptual simplification that results from the admittance of >"degenerate" examples is a common phenomenon in mathematics. Here's the way I think about it. Suppose that we have a structure S for some first-order language L. The structure includes a domain of discourse D (plus denotations for constants, function symbols and relation symbols). Now, let D' be some subset of D. We can make sense of the complete theory for D' in the following sense: Let Th(D') = the set of all formulas Phi in the language L such that Phi *relativized* to D' is true in S. Then to me it makes sense to say that D' is a model of Th(D'), even in the case where D' is the empty set. -- Daryl McCullough Ithaca, NY
From: Aatu Koskensilta on 24 May 2010 10:38
Marshall <marshall.spight(a)gmail.com> writes: > If we restrict consideration to only those signatures that omit > predicates, would my description hold? Yes. > I recall reading somewhere that a model has to have at least two > elements in the domain. I thought it weird, so it stuck in my head. This does sound weird. I don't think there's any standard treatment that requires a model to have at least two elements. The standard requirement is that models be non-empty. > But doesn't a group require an identity element? How do you have an > identity element with an empty domain? My apologies -- I was actually thinking of semigroups here. > I am entirely down with the idea of simplification via the admittance > of degenerate cases, but in this particular case, it appears to me > that requiring the model to be nonempty is what results in the > simplification. In the model theory of first-order logic, sure. I was just pointing out that in other contexts where we consider mathematical structures that are essentially just models of these and those axioms allowing empty models is the more convenient convention. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |