Cantor Confusion
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From: William Hughes on 11 Oct 2006 07:59 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > My conclusion is: > > > > > Either > > > > > (S is covered up to every position <==> S is completely covered by at > > > > > least one element of the infinite set of finite unary numbers > > > > > > > > Straight quatifier dyslexia. The fact that "for every x there exists > > > > a y such that" does not imply "there exists a y such that for every x" > > > > > > A nonsense argument. > > > > Hardly. For every integer x there exists an integer y such that > > x+y = 0. This does not imply that there exists a y such that for > > every x, x+y=0. So in general the implication does not hold. > > Uninteresting. > > > > You claim it holds in a specific case. > > Correct. > > > > >Your assertion is wrong in a linear set. > > > Give an > > > example where the linear set covers a number which is not covered by > > > one member of the linear set. > > > > > > > > > [If I understand your definition of cover] This is not possible. A > > number > > cannot be covered by a set but only by a member of a set. > > Correct. But it is asserted that 0.111... is not covered by a member of > the list > 0.1 > 0.11 > 0.111 > ... > but that it is covered by the whole list. Twaddle. This has never been asserted. Let set S be 0.1 0.11 0.111 .... Note that every member of set S has a finite number of ones. Therefore, 0.111... is not a member of set S. Then it is asserted that every member of set S is covered by some member of set S, but that no single member of set S covers every member of set S. Since 0.111... is not a member of set S, it is not asserted that it is covered by a member of set S. > You are correct. That is > impossible. > > > But this is a red herring. What we want is > > > > Given two sets, a linear set A and another set B, such > > that for every x in B there is a y in A such that y covers x. > > Then there is a y in A such that for every x in B, y > > covers x. > > > > It is easy to show that this can be false, if and only if > > A is a linear set with no largest element. I.e. it is > > not true. > > What is implied? Not my assertion is false but the assertion s false > that there are finished infinte sets. No, what is implied is that an infinite set with a linear ordering might not have a largest element. Since there is no reason that a set must have a largest element to be "finished" we cannot conclude that there are no finished infinite sets. [It would be absurd to say that a set can only be "finished" if it has a largest element. Take any ordered set without a largest element. Define a new ordering on the set by taking any element and declaring it larger than any other element (otherwise use the old ordering. So we can change an unfinished set to a finished set without adding or removing elements.] - William Hughes
From: briggs on 11 Oct 2006 08:04 In article <1160546562.540946.205860(a)e3g2000cwe.googlegroups.com>, cbrown(a)cbrownsystems.com writes: > Dik T. Winter wrote: >> In article <virgil-372F10.17374709102006(a)comcast.dca.giganews.com> Virgil <virgil(a)comcast.net> writes: >> > In article <J6w6LC.9rL(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> >> > wrote: >> > > In article <virgil-9E9CC6.02103209102006(a)comcast.dca.giganews.com> Virgil >> > > <virgil(a)comcast.net> writes: >> > > > > Dik T. Winter wrote: >> > > ... >> > > > > > The balls in vase problem suffers because the problem is not >> > > > > > well-defined. Most people in the discussion assume some implicit >> > > > > > definitions, well that does not work as other people assume other >> > > > > > definitions. How do you *define* the number of balls at noon? >> > > ... >> > > > How about the following model: >> > > >> > > And you also start with definitions, or a model. I did *not* state that >> > > it was difficult to define, or to make a model. But without such a >> > > definition or model we are in limbo. I think other (consistent) definitions >> > > or models are possible, giving a different outcome. >> > >> > Can you suggest one? One that does not ignore the numbering on the balls >> > as some others have tried to do. >> >> That does not matter, nor is that the problem. You gave a model where you >> find 0 as answer. I only state that I think there are also models where >> that is not the answer. Why is a limit of the number of balls over time >> not an answer? >> >> Let's give a simpler problem. At step 1 you add ball 1. At step n you >> remove ball n-1 and add ball n (simultaneously, I presume). > > When you say "at step n", do you have some particular time t associated > with that step? That's somewhat irrelevant. What matters is not what numeric time t is associated with each step. What matters is the [partial] ordering on the steps. Associating a numeric time t with each step is a way to ensure a total ordering. But that's more than we need. In particular, arranging matters so that all the step times come before a particular finite time is irrelevant -- it's a trick designed to fool our intuitions into delusions of physicality and all the implicit assumptions that come with physicality. We need not invoke "time" at all. Let me flesh this out a bit... The relevant ordering is such that if there is a ball n and step x and step y both involve ball n then trichotomy must apply. For all x, y if there exists an n such that x affects n and y affects n then exactly one of the following must hold: a. Step x comes before step y b. Step x and step y are the same step c. Step y comes before step x If step x and step y have no balls in common then no definite ordering relation between those two steps need exist. (Though transitivity or reflexivity may impose one). This gives rise to a family of total orderings. For each ball n the set of steps affecting ball n must be totally ordered. If we sample the set of balls in the vase then we must have an ordering relationship between the sampling and each and every step that affects a ball we are interested in. In this case, dichotomy must apply. For any sampling s and any step y, if there exists an n such that step y affects n and n is of interest to s then exactly one of the following must apply: a. The sampling s occurs before step y b. The sampling s occurs after step y (i.e. we handwave away the possibility of sampling a ball we care about exactly at a step that affects that ball) Of course, there is an obvious ordering that satisfies these constraints in Dik's example: Step n comes after step m iff n > m. Sample s occurs after all steps n
From: William Hughes on 11 Oct 2006 08:05 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > My conclusion is: > > > > > Either > > > > > (S is covered up to every position <==> S is completely covered by at > > > > > least one element of the infinite set of finite unary numbers > > > > > > > > Straight quatifier dyslexia. The fact that "for every x there exists > > > > a y such that" does not imply "there exists a y such that for every x" > > > > > > A nonsense argument. > > > > Hardly. For every integer x there exists an integer y such that > > x+y = 0. This does not imply that there exists a y such that for > > every x, x+y=0. So in general the implication does not hold. > > Uninteresting. > > > > You claim it holds in a specific case. > > Correct. > > > > >Your assertion is wrong in a linear set. > > > Give an > > > example where the linear set covers a number which is not covered by > > > one member of the linear set. > > > > > > > > > [If I understand your definition of cover] This is not possible. A > > number > > cannot be covered by a set but only by a member of a set. > > Correct. But it is asserted that 0.111... is not covered by a member of > the list > 0.1 > 0.11 > 0.111 > ... > but that it is covered by the whole list. Twaddle. This has never been asserted. Let set S be 0.1 0.11 0.111 .... Note that every member of set S has a finite number of ones. Therefore, 0.111... is not a member of set S. Then it is asserted that every member of set S is covered by some member of set S, but that no single member of set S covers every member of set S. Since 0.111... is not a member of set S, it is not asserted that it is covered by a member of set S. > You are correct. That is > impossible. > > > But this is a red herring. What we want is > > > > Given two sets, a linear set A and another set B, such > > that for every x in B there is a y in A such that y covers x. > > Then there is a y in A such that for every x in B, y > > covers x. > > > > It is easy to show that this can be false, if and only if > > A is a linear set with no largest element. I.e. it is > > not true. > > What is implied? Not my assertion is false but the assertion s false > that there are finished infinte sets. No, what is implied is that an infinite set with a linear ordering might not have a largest element. Since there is no reason that a set must have a largest element to be "finished" we cannot conclude that there are no finished infinite sets. [It would be absurd to say that a set can only be "finished" if it has a largest element. Take any ordered set without a largest element. Define a new ordering on the set by taking any element and declaring it larger than any other element (otherwise use the old ordering. So we can change an unfinished set to a finished set without adding or removing elements.] - William Hughes
From: Randy Poe on 11 Oct 2006 08:33 mueckenh(a)rz.fh-augsburg.de wrote: > David R Tribble schrieb: > > > Writing 1,2,3,... is but cheating > > > > Which of your rules does it break? > > The assumption of the complete set is inconsistent as we have seen by > discussion the vase. Inconsistent with what? - Randy
From: Han de Bruijn on 11 Oct 2006 09:04
Dik T. Winter wrote: > In article <b7f51$452a1029$82a1e228$25909(a)news2.tudelft.nl> > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > > arguments why it is admissible to allow for infinite sets. > > Because we can think about them. You can also think about the existence of angels and devils. Han de Bruijn |