An uncountable countable set
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From: Ross A. Finlayson on 8 Oct 2006 15:59 Virgil wrote: > In article <4529434c(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > > Virgil wrote: > > > In article <45251bc5(a)news2.lightlink.com>, > > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > >> Uh, no, the very conclusion that the vase empties, when at most one ball > > >> is removed at a time, implies that there is a last ball removed > > > > > > That is TO's assumption, contrary to the facts required by the > > > experiment. > > > > > > Infinite processes can end in finite time or else Zeno's 'paradoxes' > > > would prevent all action. > > > > Zeno's paradoxes involve a continuous motion at finite speed over finite > > time. The error is in considering each successively smaller time slice > > to be equal and add up to an infinite time. The vase problem is similar, > > but it's a paradox, not a fact. It's resolved with infinite series. > > > Webster1s Concise Electronic Dictionary > > 1. par·a·dox > (noun) [par·a·dox·es] > statement that seems contrary to common sense yet is perhaps true > > In this case, the emptiness of the vase at noon may be a paradox and yet > is true, at least in the world of mathematics. > > And the situation cannot arise in any other world. No, that's not correct. That's not a specialist definition. Ross
From: Virgil on 8 Oct 2006 16:04 In article <1160332251.241188.301420(a)i3g2000cwc.googlegroups.com>, Han.deBruijn(a)DTO.TUDelft.NL wrote: > David Marcus wrote: > > > Consider this situation: At time 5 one ball is added to a vase. At time > > 6, the ball is removed. > > > > Is the following a valid translation into mathematics? > > > > Let the value 1 denote that that the ball is in the vase and the value 0 > > that the ball is not in the vase. Let A(t) be the location of the ball > > at time t. Let > > > > A(t) = { 1 if 5 < t < 6; 0 if t < 5 or t > 6 }. > > No. And IMHO you are introducing a different model here than > the one we all agreed upon: > > http://groups.google.nl/group/sci.math/msg/d2573fcb63cbf1f0?hl=en& Not at all. It just requires a separate function like A(t) for each ball, say A_n(t) for the ball labeled with n. And then B(t) = Sum_{n in N} A_n(t) is well defined (and finite) at every point except those at which a ball is being entered or removed, which means that it is defined (and finite) at t = 0. And B(0) = 0 necessarily, as every A_n(0) = 0. > > Let's give a different, but analogous example. > Let p(V) be the pressure in a closed vessel having variable volume V. > Let C > 0 and > > p(V) = { C/V if V > 0; 0 if V <= 0 } . > > You can "define" what you want, but this is _not_ the proper model > of an ideal gas for V <= 0 . > > Likewise for the balls in a vase. The function A(t) may be a valid > translation only for times t _before_ noon. > The time domain is limited to t < 0. And therefore (t) is not really > like physical "time", which would flow smoothly through t = 0. > > But ah, a picture says more than a thousand words: > > http://hdebruijn.soo.dto.tudelft.nl/jaar2006/ballen.jpg But the function one should be graphing is not continuous at any of the time of entry or exit of any ball, nor at noon (except it is right-continuous at noon).
From: Virgil on 8 Oct 2006 16:35 In article <452948bf(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <45287184(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Strings with only finite bit positions. > > > > Wrong!!! Strings with only finite bit positions can still have > > infinitely many bit positions as there are infinitely many finite > > naturals. Finite naturals always have a finite most significant bit > > position and only finitely many non-zero digits. > > Incorrect. If every bit is in finite position, then there is no location > in the string where it can be said to have an infinite value. But there can still be infinitely many bit positions. In order to prohibit infinitely many bit positions one must have a most significant bit position as well as a least significant bit postion. A string is only finite if it has two ends, and a digit string without a most significant bit position has, at most, one end, and is therefore, endless and infinitely long. > > > > >>> So obviously this rule, given a starting point of 0, a finite natural > >>> and a finite-length bitstring, can never produce anything but another > >>> finite-length bitstring as a successor. So you've proven that N > >>> can contain only finite naturals. > >>> > >>> Unless you think that your rule allows an infinite bitstring successor > >>> to be formed from some finite bitstring? > >> You will not produce 1 bits in infinite positions without an infinite > >> number of successions. > > > > Which the naturals forbid. > > Correct. Given the restriction of finiteness on the values of the > elements and the constant finite difference between successors, it is > impossible for N to have "an infinite number of successions", even if > the bound on the successions is undefinable. To say that each is finite does not require that there be only finitely many of them. That particular conflation of senses has been TO's stumbling block from the beginning. he cannot conceive of an endless sequence of naturals without having some member itself being endless, despite the many contrary examples such as {1/n: n in N} > >> 1) ....00000 is a number. > > > > It is a digit string, and might be a numeral, but without a suitable > > context it is not a number. > > > > That's the equivalent of Peano Axiom 1. I declared it a number, and a > member of the set. So, it's a number. Call it a TO-number, if you like, but you do not get to call things numbers for others. > > > > >> 2) If x is a number, then the successive number, formed by inverting the > >> rightmost 0 and all 1's to the right of it, is also a number. > > > > And if it is NaN then neither are any of those other things. > > I declared it a number. What makes Peano's '0' a number? Peano has considerably more clout than TO. At least among mathematicians.
From: David Marcus on 8 Oct 2006 16:45 Tony Orlow wrote: > David R Tribble wrote: > > Virgil wrote: > >>> Except for the first 10 balls, each insertion follow a removal and with > >>> no exceptions each removal follows an insertion. > > > > Tony Orlow wrote: > >>> Which is why you have to have -9 balls at some point, so you can add 10, > >>> remove 1, and have an empty vase. > > > > David R Tribble wrote: > >>> "At some point". Is that at the last moment before noon, when the > >>> last 10 balls are added to the vase? > >>> > > > > Tony Orlow wrote: > >> Yes, at the end of the previous iteration. If the vase is to become > >> empty, it must be according to the rules of the gedanken. > > > > The rules don't mention a last moment. > > > The conclusion you come to is that the vase empties. As balls are > removed one at a time, that implies there is a last ball removed, does > it not? Please state the problem in English ("vase", "balls", "time", "remove") and also state your translation of the problem into Mathematics (sets, functions, numbers). -- David Marcus
From: Virgil on 8 Oct 2006 17:00
In article <452949b7(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <45286ce5$1(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> David R Tribble wrote: > > > >>> What about: > >>> sum{n=0 to oo} (10n+1 + ... + 10n+10) - sum{n=1 to oo} (n) > >>> The left half specifies the number of balls added to the vase, and > >>> the right half specifies those that are removed. > >>> > >> Do you mean: > >> sum{n=0 to oo} (10) - sum{n=0 to oo} (1)? > >> That sounds like what you re describing, and termwise the difference is > >> sum(n=0 to oo) (9). That's infinite, eh? > > > > But the sums are not given termwise in the question, but sumwise, so > > cannot be calculated termwise in your answer, but must be done sumwise. > > > > And sumwise they are no different. > > That's bass-ackwards. The gedanken specifically states that the > insertion of 10 and the removal of 1 are coupled as an iteration in the > process under discussion. As we are considering the expression suggested by David Tribble, <quote> What about: sum{n=0 to oo} (10n+1 + ... + 10n+10) - sum{n=1 to oo} (n) The left half specifies the number of balls added to the vase, and the right half specifies those that are removed. <\quote> The original problem is, for the moment, irrelevant. >Between every removal of 1 and the successive > removal of 1 is an insertion of 10. So, each term '+10' must be coupled > with a term '-1', which together make a '+9' per iteration. Where you > violate the condition of the experiment that specifies this coupling of > events into an iteration, you create your "discontinuity" at noon, and > the monster under your bed. 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. 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.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. 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. 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 B(t) is clearly defined and finite at every non-transition point, as being, essentially, a finite sum at every such non-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 |