An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 9 Oct 2006 19:17 In article <452ab8cd$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <452ab325(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> David Marcus wrote: > > > >>> I snipped it because it wasn't a statement of the problem, as far as I > >>> could see, but rather various conclusions that one might draw. > >> I drew those conclusions from the statement of the problem, with and > >> without the labels. > > > > But without the labels, it is a different problem. > > Define "different". "Different" means not the same. as in "2+ 2=?" being a different problem than "3 + 3 = ?". > Why do the labels matter. Suppose one rolls a pair of dice with no labels on the faces, or deals a hand of cards with no labels on the faces of those cards. Why do the labels matter? > >> While > >>> these may be correct, until the problem is stated mathematically, it is > >>> impossible to tell. Please give the statement of the problem in > >>> mathematics without also giving any conclusions or derived properties. > >> Yeah, I compared and contrasted the two approaches. Please reread. > > > > As TO was, as usual, wrong, please rewrite. > > As Vigil was, as usual, clueless, please rethink. Rethought and came up with: Let A_n(t) be equal to 0 at all times, t, when the nth ball is out of the vase, 1 at all times, t, when the nth ball is in the vase, and undefined at all times, t, when the nth ball is in transition (times at which a ball changes location). Note that noon is not a time of transition for any ball, though it is a cluster point of such times. let B(t) = Sum_{n in N} A_n(t) represent the number of balls in the vase at any non-transition time t. B(t) is clearly defined and finite at every non-transition point, as being, essentially, a finite sum at every such non-transition point, and is undefined at each transition point. Further, A_n(noon) = 0 for every n, so B(noon) = 0. Similarly when t > noon, every A_n(t) = 0, so B(t) = 0 there also.
From: Virgil on 9 Oct 2006 19:21 In article <452abf31(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <452ab540(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > >>> In article <452aa5c9(a)news2.lightlink.com>, > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> > >>> > >>>> If you increment a natural n times, you have added n to it. If successor > >>>> is increment, and there are an infinite !number! of such increments, you > >>>> have added this infinite number to your starting value. Adding an > >>>> infinite number to a finite yields an infinite. Therefore, the infinite > >>>> set includes infinite values. > >>> Without a limit ordinals in between or as the larger, one never can have > >>> two ordinals separated by infinitely many successors of the smaller. > >> Proof, please? > > > > Why? TO never proves any of his claims. > > "Why, to show me how it done properly And if I do not choose to cast my pearls before...?
From: Ross A. Finlayson on 9 Oct 2006 20:25 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. Ross
From: Tony Orlow on 10 Oct 2006 13:56 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. >>>> 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. 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. >>>>>> 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. > >> The question then becomes, can we tell anything about the "size" of the >> set, when the elements themselves are not associated with any measure, >> and there is no bound to the count of the set? The only way I can see to >> measure an unending set is to establish a standard unending set, and >> formulaically compare the two. Unfortunately, the "standard" unedning >> set is unending due to there being no largest finite, and not due to >> infinite measure persay. > > And I would agree. But the elements in the set never become non > finite. So "infinite" in that sense only means successive finite > elements "without limit". Right, that's the standard countably infinite aleph_0. It's really just "potentially" infinite. But there are different kinds of infinities > in this sense which unfortunately only make sense in relation to one > another in terms of the infinitesimal calculus used
From: David Marcus on 10 Oct 2006 14:53
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 |