An uncountable countable set
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From: MoeBlee on 10 Oct 2006 20:49 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > The point is that the quantifier 'for all', just as that quantifier is > > understood throughout mathematics, does appear in the axiom of > > infinity. > > > Fraenkel, Abraham A., Bar-Hillel, Yehoshua, Levy, Azriel: "Foundations > of Set Theory", 2nd edn., North Holland, Amsterdam (1984), p. 46: > > AXIOM OF INFINITY Vla There exists at least one set Z with the > following properties: > (i) O eps Z > (ii) if x eps Z, also {x} eps Z. x is not to be regarded as a free variable in such contexts; otherwise the axiom would not be a sentence. It is common, sometimes even customary, to leave universal quantifiers off axoms and theorems, with the understanding that the universal quantifier is to be regarded as tacitly affixed. This is a common informality; and is known by anyone who indeed understands such notation used for set theory. MoeBlee
From: David Marcus on 10 Oct 2006 21:02 Ross A. Finlayson wrote: > 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. So, you aren't saying that ZF is inconsistent. You are saying that you prefer to use a different system of axioms. Is that correct? -- David Marcus
From: Ross A. Finlayson on 10 Oct 2006 21:11 David Marcus wrote: .... > > the null axiom theory. > > So, you aren't saying that ZF is inconsistent. You are saying that you > prefer to use a different system of axioms. Is that correct? > > -- > David Marcus Hi David, As you said, it's good to take these things one at a time, so, I am happy to explain my opinion about set theory to you. Having been explaining set theory for some time, I was able to to take part in some of these discussions here for example with everybody. No, David, Dave, I say ZF is inconsistent. That doesn't mean all its results are false. It just lets me say whatever I want about a less inconsistent set of "axioms". Ross
From: Tony Orlow on 11 Oct 2006 11:32 David Marcus wrote: > 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 > Interesting. I grew up on 115th and Riverside. :)
From: Tony Orlow on 11 Oct 2006 11:41
Lester Zick wrote: > 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. > If the line is defined by trichotomy, and 3<pi<4, isn't pi on that 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. Yes, and every finitely-indexed subdivision results in finite segments, but that cannot be truly when the segment is infinitely subdivided. > >> 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. > But specific infinite values and specific infinitesimals do. >>>>>>>> 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 |