An uncountable countable set
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From: Tony Orlow on 9 Oct 2006 11:30 Virgil wrote: > 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 It's nice that you like yours. I'd hate to see you treating it badly. I don't mean to be mean to it. It's just that it kind of smells and keeps trying to crawl on my lap and drool. If you could keep it off the furniture, I'd appreciate it. Tony
From: Tony Orlow on 9 Oct 2006 11:31 Han de Bruijn wrote: > 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 > Gee, I guess if Virgil can irrelevantly bring in relativity, it's an open game, eh? ;) Tony
From: Ross A. Finlayson on 9 Oct 2006 11:45 Tony Orlow wrote: > Virgil wrote: > > In article <4529afa4(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: .... > >> That is not "the reals strictly between 0 and 1" but a subset thereof. > > > > So there is still no element within either set which is its LUB. > > If the Finlayson reals are used, indeed the LUB is the maximal member of > the set of reals in [0,1). Ross, is that correct? > > Tony Hi Tony, It depends how you use them because they are also the real numbers. In your sense of use, yes, it's permissible. You just can't then do other things that would right contradict that statement, you know. To reestablish your discussion about the entire set of real numbers again, including all the numbers between zero and one where there are only and everywhere real numbers, the segment of the real number line's continuum, involves that claim in context. A fellow Spinoza from some time ago is known for introducing the notion that the set of natural numbers is itself a continuum. Did you know that the closer scientists look at subatomic (well, and atomic) particles, the smaller the particles appear to be? Are they infinitesimal? Why is there gauge invariance? Ross
From: David Marcus on 9 Oct 2006 12:43 Tony Orlow wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> 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). > >> Given an unfillable vase and an infinite set of balls, we are to insert > >> 10 balls in the vase, remove 1, and repeat indefinitely. In order to > >> have a definite conclusion to this experiment in infinity, we will > >> perform the first iteration at a minute before noon, the next at a half > >> minute before noon, etc, so that iteration n (starting at 0) occurs at > >> noon-1/2^n) minutes, and the infinite sequence is done at noon. The > >> question is, what will we find in the vase at noon? > > > > OK. That is the English version. Now, what is the translation into > > Mathematics? > > Can you only eat a crumb at a time? I gave you the infinite series > interpretation of the problem in that paragraph, right after you > snipped. Perhaps you should comment after each entire paragraph, or > after reading the entire post. I'm not much into answering the same > question multiple times per person. 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. 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. When discussing mathematics, communication works best one step at a time. -- David Marcus
From: David Marcus on 9 Oct 2006 12:47
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. -- David Marcus |