An uncountable countable set
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From: Han de Bruijn on 9 Oct 2006 04:12 David Marcus wrote: > 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. > > Then what is the translation into Mathematics? The mathematical model for the number of balls in the vase is this. Let t_k = - 1/2^k for (k = 0,1,2, ... in N). Note that t < 0 . Then the number of balls is B(k) = 9 + 9.ln(-1/t_k)/ln(2) = 9.(k+1) for t_k < t < t_(k+1) : a staircase. Han de Bruijn
From: Virgil on 9 Oct 2006 04:31 In article <820e2$452a0460$82a1e228$23624(a)news2.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > David Marcus wrote: > > > 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. > > > > Then what is the translation into Mathematics? > > The mathematical model for the number of balls in the vase is this. > Let t_k = - 1/2^k for (k = 0,1,2, ... in N). Note that t < 0 . > Then the number of balls is B(k) = 9 + 9.ln(-1/t_k)/ln(2) = 9.(k+1) > for t_k < t < t_(k+1) : a staircase. > > Han de Bruijn I like mine better: 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
From: Han de Bruijn on 9 Oct 2006 05:32 Virgil wrote: > In article <e3ab0$4529ffd9$82a1e228$22026(a)news2.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>Virgil wrote: >> >>>In article <1160332251.241188.301420(a)i3g2000cwc.googlegroups.com>, >>> Han.deBruijn(a)DTO.TUDelft.NL wrote: >>> >>>>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). >> >>Correct. I have "continuized" the function. The true (discrete) function >>should be like a staircase with a stair at each red ball. But a discrete >>version likewise explodes (i.e. not implodes) at noon. > > Actually not precisely "at" noon. > > 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 Wrong. Because you cannot jump over noon. The vase has become so infinitely heavy at that time that a black hole is being formed. And clocks begin to tick slower and slower, sooo slow that it's forever impossible to reach noon. Look what modern GR can do! Han de Bruijn
From: Ross A. Finlayson on 9 Oct 2006 10:46 David Marcus wrote: > Ross A. Finlayson wrote: > > Have you heard of Burali-Forti, a "paradox"? There's no set of > > ordinals nor cardinals in ZF. > > > > There is a universe, and there's not in ZF. There are smaller sets > > than the universe that are sufficient for many statements, in terms of > > quantifying over their elements. There is quite regular use of the > > word universe in the practice of naive set theory. > > > > Basically it's an anti-foundation mindset, and then some. Quantify > > over sets: the result is not a set. There's nothing else over which > > to quantify. So, foundation is an exercise in vacuity. > > > > In set theory, where basically "pure" set theory means every item is a > > set, there are only sets, every thing is a set, and there's reason to > > consider why the transfer principle holds true, that everything is a > > set. Basically I see a contradiction in the existence of the universal > > quantifier without a universe. > > > > That gets into things along the lines of that infinite sets are > > irregular, and then some. > > 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? > > -- > David Marcus {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. Have a nice day, Ross
From: Tony Orlow on 9 Oct 2006 11:07
Virgil wrote: > In article <4529ac7e(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> 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. >>> >> What David wrote didn't even make sense to me, which is why I >> re-expressed it. What does {10n+1+...+10n+10} mean except 100n+55, and >> what does that signify? Why is he summing this expression from 0, and >> summing n from 1? Are we really summing n, anyway? We're not adding the >> numbers on the balls to each other, we're just counting them 10- or 1 or >> 9-at-a-time. So, it's not the original problem which is irrelevant, but >> this meaningless expression which is irrelevant to the original problem. I'll take your lack of response as assent. >> >>>> 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. >> Does time obey the law of trichotomy? Can something occur, not before, >> not after, and not at the same time as another event, given that the >> events happen instantaneously? > > In general relativity , two events cannot be said to occur at "the same > time"" unless they are also in the same place. ???? Where did anyone mention relativistic spacetime? I asked about time as a dimension. Can we consider it to be linear and continuous? Is there a fourth option besides the three of trichotomy? >>> 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. >> What is "in transition". Can't we consider the addition or removal of a >> ball to be instantaneous? > > Would TO's "occurring instantaneously" mean that it does not take place > at any point in the time stream? I don't think so. The function I > described is undefined at all such points in time at which there is a > transition in position from before that moment to after that moment. It means it takes place at a particular point in time, and that there is not a measurable "period" of transition. If you say it's "undefined" at that point, then define it. All you have to do is choose whether the event occurs at that point, or before, or after it. That is, in transition from f(before t)=x to f(after t)=y, either f(t)=x or f(t)=y. Either x<f(t)<=y or x<=f(t)<y. > >>> 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. >> Correct, a linearly increasing finite value. > > Not increasing between points of discontinuity nor over any interval of > time containing times past noon. Linearly increasing with each iteration, with a condensation point at noon where iterations occur infinitely fast. > >>> 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 >> I understand your logic, but the basis is incorrect. As WM >> understandably complains, using the "completed set of naturals" as a >> measure of anything simply does not work. > > It works in ZFC and NBG. That it does not work for TO or for "Mueckenh" > (though for different reasons) is their problem, not ours. > Not for different reasons at all. The difference is whether this problem causes one to reject infinite values entirely, or seek a better representation of them. > >> You're focused on the Twilight >> Zone. > > Even twilight beats the stygian darkness in which TO is operating. Says he in the Cave of Treasures. |