An uncountable countable set
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From: Tony Orlow on 29 Sep 2006 16:29 Randy Poe wrote: > Tony Orlow wrote: >> imaginatorium(a)despammed.com wrote: >> At this point I can hardly remember what we were talking about, but of >> course that's partly because you snipped the entire context when you >> began this diversion. Randy and I were haggling over whether the vase >> empties, and if so, at what exact time? > > No we weren't. I was making precise statements about > the condition of the vase at different times, and you > are trying to get me to make vague and incorrect > restatements of those precise statements. > > The vase *is* empty at noon. It *is not* empty at > any time before noon. I am refusing to use the > verb "empties" because that implies a transition > which does not occur. > > - Randy > If at all times before noon the vase is not empty, but it is empty at noon, then that is when this event occurs. You say there is no transition? Great, the vase does not become empty at noon. Tony
From: Tony Orlow on 29 Sep 2006 16:34 Randy Poe wrote: > Tony Orlow wrote: >> Randy Poe wrote: >>> Tony Orlow wrote: >>>> Han de Bruijn wrote: >>>>> Virgil wrote: >>>>> >>>>>> In article <d12a9$451b74ad$82a1e228$6053(a)news1.tudelft.nl>, >>>>>> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: >>>>>> >>>>>>> Randy Poe wrote, about the Balls in a Vase problem: >>>>>>> >>>>>>>> It definitely empties, since every ball you put in is >>>>>>>> later taken out. >>>>>>> And _that_ individual calls himself a physicist? >>>>>> Does Han claim that there is any ball put in that is not taken out? >>>>> Nonsense question. Noon doesn't exist in this problem. >>>>> >>>>> Han de Bruijn >>>>> >>>> That's the question I am trying to pin down. If noon exists, that's when >>>> the vase supposedly empties, >>> Why does the existence of noon imply there is a time >>> which is the last time before noon? >>> >>> It doesn't. >>> >>> - Randy >>> >> I never said it did. When did I say that? > > I was responding to Han, who said that "If noon exists, that's when > the vase empties". > > Noon exists. > > But in order for the vase to transition from not-empty > to empty, there would have to be a last non-empty > moment. That would be the last time before noon. Yes, and at that last moment the last ball would have to be removed, and yet, at the moment before 10 balls would have to have been added. Can the vase contain -9 balls? :) > >> I will offer this simple >> logical argument. If the vase ever became empty, it would be because one >> ball was removed, > > Hence my continued statement that the vase does not > "become empty". It is non-empty at certain times and > empty at others. How do you reconcile.... There is no transitional moment. > ....with... > Noon is the first moment at which the vase is empty. Does the vase not go from non-empty to empty at noon? You're making no sense. If you can't answer that simple question without getting into trouble, that should be an indication that you're on shaky ground. > > But noon is not the transitional moment. There's no > time just before noon where the transition happened. > > - Randy > So, it was non-empty at every time before noon and empty at noon, but there was no transition between those two states? Then it must still be non-empty at noon. This is ridiculous. Tony
From: Tony Orlow on 29 Sep 2006 16:36 Virgil wrote: > In article <8cabe$451ccd62$82a1e228$12622(a)news1.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >> Virgil wrote: >> >>> In article <9fd0$451b7e7b$82a1e228$8977(a)news1.tudelft.nl>, >>> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: >>> >>>> Virgil wrote: >>>> >>>>> In article <451a8f41(a)news2.lightlink.com>, >>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>> For purposes of measure on the finite scale, infinitesimals can be >>>>>> considered nilpotent. That's all. Do you disagree? >>>>> I disagree that scale changes can convert between zero and non-zero. >>>>> >>>>> There are approximation methods is which products of small quantities >>>>> are regarded as negligible in comparison to the quantities themselves, >>>>> but they are always just approximations. >>>> Crucial question: are those "approximation methods" part of mathematics? >>>> I'll take Yes or No as a sufficient answer. >>> They are a part of the applications of mathematics to things other than >>> mathematics, so they are marginal. >> That sounds like a smart answer, but I don't buy it. >> Again: are those "approximation methods" part of mathematics? Yes or No. >> > > They are attempts to bend the mathematics to accommodate the needs of > the sciences, so one would have to say "Yes and No". Would you like syrup with your waffle?
From: Randy Poe on 29 Sep 2006 16:48 Tony Orlow wrote: > Randy Poe wrote: > > Tony Orlow wrote: > >> Randy Poe wrote: > >>> Tony Orlow wrote: > >>>> Han de Bruijn wrote: > >>>>> Virgil wrote: > >>>>> > >>>>>> In article <d12a9$451b74ad$82a1e228$6053(a)news1.tudelft.nl>, > >>>>>> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>>>>> > >>>>>>> Randy Poe wrote, about the Balls in a Vase problem: > >>>>>>> > >>>>>>>> It definitely empties, since every ball you put in is > >>>>>>>> later taken out. > >>>>>>> And _that_ individual calls himself a physicist? > >>>>>> Does Han claim that there is any ball put in that is not taken out? > >>>>> Nonsense question. Noon doesn't exist in this problem. > >>>>> > >>>>> Han de Bruijn > >>>>> > >>>> That's the question I am trying to pin down. If noon exists, that's when > >>>> the vase supposedly empties, > >>> Why does the existence of noon imply there is a time > >>> which is the last time before noon? > >>> > >>> It doesn't. > >>> > >>> - Randy > >>> > >> I never said it did. When did I say that? > > > > I was responding to Han, who said that "If noon exists, that's when > > the vase empties". > > > > Noon exists. > > > > But in order for the vase to transition from not-empty > > to empty, there would have to be a last non-empty > > moment. That would be the last time before noon. But there is no "last moment before noon". > Yes, and at that last moment the last ball would have to be removed, There is no "last moment" and no "last ball" > and > yet, at the moment before 10 balls would have to have been added. Can > the vase contain -9 balls? :) There is no "moment before the last moment". Have you not yet figured out yet that given any two different times, there are times in between them? That there's no such thing as the "next moment"? > > > > >> I will offer this simple > >> logical argument. If the vase ever became empty, it would be because one > >> ball was removed, > > > > Hence my continued statement that the vase does not > > "become empty". It is non-empty at certain times and > > empty at others. > > How do you reconcile.... > > There is no transitional moment. > > > > ...with... > > > Noon is the first moment at which the vase is empty. > > Does the vase not go from non-empty to empty at noon? No. > You're making no > sense. If you can't answer that simple question I answered it. The answer is "no". Somebody is asking you to think about infinitely high strips, and the situation is analogous. Think about the graph of tan(x), which you may or may not know grows without bound as x approaches 90 degrees. For values of x just above 90 degrees, tan(x) is large negative. Here's the graph: http://mathworld.wolfram.com/Tangent.html If you increase x from 0 to a point just above 90 degrees, the following things are true: (a) For every value of x below 90 degrees, y = tan(x) is positive. (b) For every value of x above 90 degrees, y = tan(x) is negative. (c) There is no point where y transitioned from positive to negative. If you plot the number of balls in the vase vs. time, it behaves much like one of those curves. You have a curve rising asymptotically toward t = noon, and a flat line at t >= noon. But there is no "transition" from the rising curve to the flat line, any more than there is a "transition" from the curve of tan(x), x<90 deg to tan(x), x>90 deg. The fact that this bothers you does not constitute my "getting into trouble". - Randy
From: Tony Orlow on 29 Sep 2006 16:54
stephen(a)nomail.com wrote: > Tony Orlow <tony(a)lightlink.com> wrote: >> stephen(a)nomail.com wrote: >>> Tony Orlow <tony(a)lightlink.com> wrote: >>>> stephen(a)nomail.com wrote: >>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>> stephen(a)nomail.com wrote: >>>>>>> Randy Poe <poespam-trap(a)yahoo.com> wrote: >>>>>>> >>>>>>> <snip> >>>>>>> >>>>>>>> What is the number of the ball which, when removed, >>>>>>>> makes the vase empty? >>>>>>>> I know the kind of nonsense you will spout in answer to >>>>>>>> those questions, but the answers within our axiom system >>>>>>>> are: (1) there is no t<noon which is the moment just >>>>>>>> before noon. For any t<noon, there is t < t' < noon. >>>>>>>> (2) There is no such ball. >>>>>>>> Here are the Tony gobbledgook answers: >>>>>>>> (1) noon - 1/oo >>>>>>>> (2) Ball number omega >>>>>>>> In TO-matics, one can confidently give an answer like >>>>>>>> number 2 despite the fact that one can also agree >>>>>>>> that no ball numbered omega is ever put into the >>>>>>>> vase. >>>>>>> In TO-matics, it is also possible to end up with >>>>>>> an empty vase by simply adding balls. According to TO-matics >>>>>>> >>>>>>> ..1111111111 = 1 + 1 + 1 + 1 + ... >>>>>>> >>>>>>> and >>>>>>> ..1111111111 + 1 = 0 >>>>>>> >>>>>>> So if you just keep on adding balls one at a time, >>>>>>> at some point, the number of balls becomes zero. >>>>>>> You have to add just the right number of balls. It is not >>>>>>> clear what that number is, but it is clear that it >>>>>>> exists in TO-matics. >>>>>>> >>>>>>>> But in mathematics and logic, we don't get to >>>>>>>> keep a set of self-contradictory assumptions around, >>>>>>>> only using the ones we want as needed. >>>>>>>> - Randy >>>>>>> Where's the fun in that? :) >>>>>>> >>>>>>> Stephen >>>>>> You drew that from my suggestion of the number circle, and that ...11111 >>>>>> could be considered equal to -1. Since then, I looked it up. I'm not the >>>>>> first to think that. It's one of two perspectives on the number line. >>>>>> It's either really straight, or circular with infinite radius, making it >>>>>> infinitesimally straight. The latter describes the finite universe, and >>>>>> the former, the limit. But, you knew that, and are just trying to have fun. >>>>>> Tony >>>>> I am just pointing out that according to your mathematics >>>>> that if you keep adding balls to the vase, you can end up >>>>> with an empty vase. The fact that other people may have >>>>> considered a number circle does not change the fact that the >>>>> number circle implies that if you keep on adding balls, eventually >>>>> you will have zero balls. >>>> That's a bastardization of the concept. There are two ways to look at >>>> the number circle, and you are combining them in mutually contradictory >>>> ways. >>> How is it a bastardization of the concept? You claim that >>> 1+1+1+1+ ... = ..11111111 > >> I wouldn't put it that way. ...1111 can be interpreted as the largest >> binary natural, if you claim all bit positions are finite. > > I do claim that all bit positions are finite, but there is no > largest binary natural. That is something you have invented. > That is what that binary string would represent, but of course it's not any exact number, being a boundless string. > So this "largest binary natural" of yours is not equal to 1+1+1+... > for some number of 1's? In other words, it is not in the > chain of successors? In what sense then is it a "natural number"? You can think of it that way if you want, but since this number doesn't exist as any specific quantity, I wouldn't try to do any math with it. That's why the concept of omega leads to so much hocus pocus. > >>> and that >>> ..11111111 + 1 = 0 > >> ...11111 can be interpreted indeed as -1, as is done every millions of >> times per microsecond all over the world in computers. > > I know of now computer that can hold ...111111 let alone > intreprets it as -1. Remember, ...111111 is an unending > string of 1's. If you fill the 2's complement register with 1 bits, it's equal to -1. This is true for any size register, and can be assumed true for an infinite register in that context. > >> Those are two different interpretations of -1. They aren't really >> compatible, as far as I can see, although Ross used to like to point out >> that temperatures below 0 Kelvin were somehow hotter than any positive >> temperature. There could be a connection. > > The moment you mention physics or Ross it is clear you are > just rambling. > If you read what I wrote, it's on-topic. >>> Why does that not apply to balls in the vase? Each ball >>> is a 1. If a add balls, I add 1's. Do I not eventually get 0? >>> If it does not apply to balls, what does it apply to, >>> and how do you determine when it applies? >>> > >> They are two different interpretations of the string. Do you honestly >> think that I think sum(x=1->oo: 1)=0? For god's sake, how long have I >> been saying it's infinite, and that's why any infinite set of naturals >> would have to include infinite naturals? Oy. Like I said, you're >> combining concepts that are mutually contradictory, from two different >> number systems. And, people complaint hat I try to use normal operations >> normally on infinite numbers.... > > You are the one who is combining things. In "my" system ...111111 does > not exist. Yet it does in your system, and it apparently is > a natural number, and it apparently equals 1+1+1+.... (does it > not need to be a successor of a successor of a successor and > so on of 1), and it apparently sometimes equals -1. Can you clearly > and concisely state the rules that govern ...11111111? If not, > please never mention it again. > ....1111 in binary means a 1 in every bit position, the value being sum(x=0->n: 2^x). The question is, for which n does the value become infinite? It is finite for all finite bit positions. Unless n becomes infinite, the sum is finite. So, if you insist that all bit positions are finite and the string countable, then the value is countable too, and finite. It's not a specific value, really, because there is no specific n when the bit positions are defined as all the "finite" positions. >>>>> So why is it okay to end up with zero balls, when you never remove >>>>> any at all, but it |