An uncountable countable set
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From: David Marcus on 10 Oct 2006 14:55 Tony Orlow wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> This part of Ross I don't quite follow, but perhaps we should discuss > >> it. What is your background, if you don't mind my asking? > > > > Is that relevant? Or, are you just curious? > > Curious, > and probably, > therefore, > relevant... > ...Tony http://www.davidmarcus.com/Personal.htm -- David Marcus
From: Virgil on 10 Oct 2006 15:12 Ross A. Finlayson wrote: > An axiomless theory still has logical axioms, it just doesn't have > proper, or illogical, axioms. Ross flubs again.
From: Ross A. Finlayson on 10 Oct 2006 15:21 David Marcus wrote: > Ross A. Finlayson wrote: > > David Marcus wrote: > > > Ross A. Finlayson wrote: > > > > David Marcus wrote: > > > > > Ross A. Finlayson wrote: > > > > > > David Marcus wrote: > > > > > > > I thought you said there was a contradiction in ZF. In the context of > > > > > > > ZF, the Burali-Forti argument shows that there is no set of all > > > > > > > ordinals, but does not lead to a contradiction. So, do you still say > > > > > > > there is a contradiction in ZF? If so, what is it? > > > > > > > > > > > > {For any x: x is a set} = emptyset <=/=> {For any x: x is a set} = U > > > > > > (V, L) > > > > > > > > > > > > That says, for any x, that's the empty set, and, for any x, that's the > > > > > > universal set, it seems sufficient to show the universe non-empty. > > > > > > > > > > > > There is no set of ordinals nor cardinals in ZF. Yet, because there's > > > > > > the axiom of infinity, those infinite ordinals/cardinals as there ever > > > > > > would be are claimed to exist, basically where they're all hereditarily > > > > > > finite, those ordinals of the cumulative hierarchy. > > > > > > > > > > Sorry, but I don't follow. Are you saying this is a contradiction within > > > > > ZF? By "within" I mean that ZF proves this contradiction. > > > > > > > You seem like you know what you're talking about, which is good. > > > > > > > > Basically, yes, I say the existence of the (universal) quantifier, > > > > where the word universal is in parentheses because there's the mutually > > > > implicit existential quantifier, that the existence of the universal > > > > quantifier in ZF lead to illustration of a contradiction derivable from > > > > ZF. > > > > > > OK. But the Burali-Forti argument does not (as far as I know) lead to a > > > contradictionn derivable from ZF. So, what is the contradiction that you > > > can derive from ZF? > > > > > > > I have some other arguments along those lines as well, of similar tack, > > > > where I advocate axiom-free natural deduction, as a return of sorts to > > > > a more "naive" set theory with post-Cantorian acknowledgement. > > > > > > > > That is to say, there are thousands of pages more of my opinion about > > > > these matters readily available. > > > > > > -- > > > David Marcus > > > > Hi David, > > > > I just mentioned Burali-Forti to see if you had heard of it. Basically > > the consideration is that because of the mechanistic structure of the > > ordinals, the set of ordinals, or counting numbers extended to the > > hyperfinite, because of the mechanistic structure of the ordinals the > > set of ordinals would also be an ordinal. > > > > That's an example of where the transfer principle holds and what is > > true for elements of the set is true for the set, here, in terms of set > > membership, definition. > > > > So, in ZF there is no set of ordinals. That's basically the very same > > argument as that there is no complete infinity, in for example the > > finite ordinals also known as the whole, counting, or natural numbers, > > that the set of those can not exist, for, the successor in natural > > generation of the mechanistic ordinal is yet another mechanistic > > ordinal. > > > > So, looking at the Burali-Forti "paradox" illustrates how unresolved > > "paradoxes" earlier in development of definition, in this case about > > infinity, reexhibit themselves under further analysis of the system. > > > > I hope that was not misguiding, about inconsistency in ZF. I think the > > existence of a universe is implicit with the existence of a quantifier. > > ZF claims a quantifier, with "restricted comprehension", as has been > > discussed here before. Yet, even the definition is couched in terms of > > "unrestricted comprehension." > > > > An axiomless theory still has logical axioms, it just doesn't have > > proper, or illogical, axioms. > > Sorry, but I don't follow. Are you still saying that ZF is inconsistent > or when you say you were "misguiding, about the inconsistency of ZF", do > you mean that you are not saying that ZF is inconsistent? > > -- > David Marcus Hi David, Burali-Forti is not directly about inconsistency in ZF, although I hope you would consider the above discussion and how the Axiom of Infinity states there is a completed infinity, and how it would carry over that the Burali-Forti set, or set of ordinals, exists. Similarly, where it seems to me that universal quantification sans the universe does not work, there is a universe, via ZF's Axiom of Foundation there is not a universe, and those are not mutually compatible inferences. So, I toss foundation. That obviously leads to a system concerning the Russell paradox, or Russell set. The Russell set ends up variously being the empty set or set of all finite ordinals. Can that not exist? I think the axioms besides regularity/foundation of ZFC are theorems of the null axiom theory. Ross
From: Ross A. Finlayson on 10 Oct 2006 15:23 Virgil wrote: > Ross A. Finlayson wrote: > > > An axiomless theory still has logical axioms, it just doesn't have > > proper, or illogical, axioms. > > Ross flubs again. Excuse me, "non-logical". What are you talking about? Ross
From: Lester Zick on 10 Oct 2006 19:41
On Tue, 10 Oct 2006 13:56:43 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >Lester Zick wrote: >> Tony, I'm going to strip as much as seems unessential. >> >> On Mon, 09 Oct 2006 15:56:32 -0400, Tony Orlow <tony(a)lightlink.com> >> wrote: >> >>> Lester Zick wrote: >>>> On Sun, 08 Oct 2006 22:07:23 -0400, Tony Orlow <tony(a)lightlink.com> >>>> wrote: >> >> [. . .] >> >>>> As far as transcendentals are concerned, Tony, the only thing that can >>>> lie on a real number line in common with rationals/irrationals are >>>> straight line segment approximations. That's the only linear order >>>> possible. So either you give up transcendentals or a real number line. >>> The trichotomy or real quantity itself defines a linear order. Each such >>> value is greater than or less than every different value. Pi is >>> transcendental - is it less than or greater than 3? Is there any doubt >>> about that? >> >> No but that only applies to linear approximations for pi, Tony. It has >> nothing to do with the actual value of pi which lies off to the side >> of any real number line. Imagine if you will the linear order you >> speak of superimposed on a real number plane instead of a straight >> line (which isn't even possible either). Then pi has to lie off to one >> side of the straight line. So the metric for the straight line becomes >> a non linear variable instead of linear. Thus the fact that the value >> of pi lies between 3 and 4 doesn't show any value for pi and never >> will. All you wind uyp with is a linear metric which approximates pi >> which doesn't provide us with any real number line running along the >> lines of 1, 2, e, 3, pi. So the real number line metric is variable. >> > >Pinpointing particular points on the line may require varying degrees of >computation/construction. A natural requires only a finite number of >iterations of increment, a rational may require a division operation, >and an irrational or a transcendental like pi may require an infinite >computation to exactly pinpoint the value. That doesn't mean that value >doesn't exist as a point on the line. It just mans specifying it >requires an infinite process. It requires an infinite process only because it isn't on the line. The only thing an infinite process determines is how close the curve comes to a straight line without ever getting there. Every other rational/ irrational on the line can be located with finite processes because they are on the line. >>>>> Perhaps not, but if, say, y=1/x is both oo and -oo at x=0, and oo=-oo, >>>>> then the function is continuous in every respect, which is what we might >>>>> desire in such a fundamental algebraic relation. >>>> But for the division operation x never becomes zero. Which indicates >>>> that there can be no plus or minus infinity and no continuity. >>> Is 0 part of the continuum, or just another arbitrary "limit" >>> discontinuity? When you ask yourself, "If I divide this finite space >>> into individual points, how many will I have?", what answer do you get? >>> How much of the space between 0 and 1 does each real in that interval >>> occupy, and how many are there? >> >> Well points don't occupy any space at all, Tony. > >Right, their measure is 0. > >Intersection ends in >> points but subdivision never ends in points. > >In the limit, subdivision results in infinitesimal segments, where the >values of the endpoints cannot be distinguished on any finite scale. Oh I don't know about that, Tony. Certainly the first bisection in a series is exactly half the length of the starting finite segment. >That's the difference >> between intersection and differentiation. Although not a natural zero >> certainly exists but the question is how it's used. Zero used as a >> limit is approachable but not reachable. Same with infinity. Zero used >> as a divisor is undefinable. Maybe you're trying to combine arithmetic >> and infinitesimal calculus operations in ways that are incompatible? >> > >It's true that absolute 0 has an indeterminate reciprocal of oo, and >that regular arithmetic won't work on those. But, a specific unit >infinity lends itself to arithmetic quite nicely. So does zero. It just doesn't have any meaning when used for division. Nor does infinity when used to multiply. >>>>>>> When it comes down to this argument, Wolfgang's argument, I agree with >>>>>>> his logic concerning the naturals and the identity function between >>>>>>> element count and value. >>>>>> To me "element count" and the number of commas are the same. >>>>> Sure, that sounds okay to me. In the naturals, the first is 1, the >>>>> second 2, etc. What is the aleph_0th? >>>> I don't know what you're asking here, Tony. If there is no real number >>>> line aleph there are no aleph ordinals either. There can be aleph >>>> infinitesimals but that represents a continual process of subdivision >>>> and not one of division in which case the ordinality would be one of >>>> relation between various curvatures where straight lines would be >>>> first or minimal and the ordinality of others judged in relation to >>>> it. I think the question you really need to be asking in this context >>>> is whether there can be such a thing as an open set. If not then the >>>> question becomes how to close open sets and whether there can be >>>> anything besides finites in closed sets. >>>> >>> Well, yes, that is the question, but I would pose it in a different way. >>> >>> Let's say there CAN be open sets, by which I assume you mean boundless, >>> without definite range, either element-wise, or value-wise given a >>> formula defining membership and any kind of measure (is that what you >>> mean?). Such a set can be declared as "existing" simply by defining the >>> rules of the object. I don't see that it *can't* exist. >> >> Well the question I have then is whether such a set can be infinite or >> contain only finites without limit? > >I think it can be either, depending on how it's defined. If only >finitely-indexed elements may exist in the set, then it can't be >infinite in my book. There must exist elements that require infinite >specification, such as pi in the reals. Well the problem here is that sqrt(2) also requires an infinite series in arithmetic terms even though it doesn't in geometric terms. Pi doesn't admit of finite specification in eith |