An uncountable countable set
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From: Virgil on 24 Oct 2006 01:30 In article <453d9ad0(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> Tony Orlow wrote: > >>>> David Marcus wrote: > >>>>> Tony Orlow wrote: > >>>>>> Virgil wrote: > >>>>>>> In article <4533d315(a)news2.lightlink.com>, > >>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>>>>>> Then let us put all the balls in at once before the first is > >>>>>>>>> removed and > >>>>>>>>> then remove them according to the original time schedule. > >>>>>>>> Great! You changed the problem and got a different conclusion. How > >>>>>>>> very....like you. > >>>>>>> Does TO claim that putting balls in earlier but taking them out as in > >>>>>>> the original will result in fewer balls at the end? > >>>>>> If the two are separate events, sure. > >>>>> Not sure what you mean by "separate events". Suppose we put all the > >>>>> balls in at one minute before noon and take them out according to the > >>>>> original schedule. How many balls are in the vase at noon? > >>>>> > >>>> empty. > >>> Suppose we put ball n in at 1/n before noon and remove it at 1/(n+1) > >>> before noon. How many balls in the vase at noon? > >> At all times >=-1 there will be 1 ball in the vase. > > > > And, the ball that is in the vase at noon, what is the number on the > > bal? "Infinity"? > > > > It would be some infinite number, yes, either absolute oo, or some > specific uncountable infinity, if such a limit is placed on the iterations. Which "infinity" does TO claim qualifies as a natural number? We are only have naturally numbered balls to start with, so unless TO manufactures his own balls, that's where he has to get them.
From: Virgil on 24 Oct 2006 01:32 In article <453d9b13(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> Tony Orlow wrote: > >>>> David Marcus wrote: > >>>>> Tony Orlow wrote: > >>>>>> Virgil wrote: > >>>>>>> In article <4533d315(a)news2.lightlink.com>, > >>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>>>>>> Then let us put all the balls in at once before the first is > >>>>>>>>> removed and > >>>>>>>>> then remove them according to the original time schedule. > >>>>>>>> Great! You changed the problem and got a different conclusion. How > >>>>>>>> very....like you. > >>>>>>> Does TO claim that putting balls in earlier but taking them out as in > >>>>>>> the original will result in fewer balls at the end? > >>>>>> If the two are separate events, sure. > >>>>> Not sure what you mean by "separate events". Suppose we put all the > >>>>> balls in at one minute before noon and take them out according to the > >>>>> original schedule. How many balls are in the vase at noon? > >>>> empty. > >>> Why? > >> Because of the infinite rate of removal without insertions at noon. > > > > OK. Just to recall, this vase has all the balls put in at one minute > > before noon, then taken out on the usual schedule. How many balls are in > > this vase at times before noon? > > > > Some supposedly infinite number, as only a finite number have been removed. But the vase ends up empty at noon anyway. The only relevant question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer directly violates the conditions of the problem. How does TO answer?
From: Virgil on 24 Oct 2006 01:32 In article <453d9bb4(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > Tony Orlow wrote: > >> You have agreed with everything so far. At every point before noon balls > >> remain. You claim nothing changes at noon. Is there something between > >> noon and "before noon", when those balls disappeared? If not, then they > >> must still be in there. > > > > Of course there is a "something" between "before noon" and "noon" where > > each ball disappears. At step n, time 2^-n min before noon, ball n is > > removed. This happens for every ball, since there is a step n for > > every ball. The balls are removed, one by one, one at each step, > > before noon. > > > > As each ball n is removed, how many remain? Can any be removed and leave > an empty vase? The more relevant question is "According to the rules set up in the problem, is each ball which is inserted into the vase before noon also removed from the vase before noon?" An affirmative answer confirms that the vase is empty at noon. A negative answer violates the conditions of the problem.
From: David Marcus on 24 Oct 2006 02:09 Tony Orlow wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> David Marcus wrote: > >>> Tony Orlow wrote: > >>>> David Marcus wrote: > >>>>> Tony Orlow wrote: > >>>>>> Either something happens an noon, or it doesn't. Where do you stand on > >>>>>> the matter? > >>>>> What does "something happens" mean, please? I really don't know what you > >>>>> mean. > >>>> ??? Do you live in the universe, or in a static picture? When "something > >>>> happens" o an object, some property or condition of it "changes". That > >>>> occurs within some time period, which includes at least one moment. > >>>> There is no moment in this problem where the vase is emptying, > >>>> therefore, that never "occurs". If you are going to insist that time is > >>>> a crucial element of this problem, then you should at least be familiar > >>>> with the fact that it's a continuum, and that events occurs within > >>>> intervals of that continuum. > >>> Thanks. That explains what "something happens" means. Now, please > >>> explain what "emptying" means. > >> "Empty" means not having balls. To become empty means there is a change > >> of state in the vase ("something happens" to the vase), from having > >> balls to not having balls. > >> > >> Now, when does this moment, or interval, occur? > > > > A reasonable question. Before I answer it, let me ask you a question. > > Suppose I make the following definitions: > > > > For n = 1,2,..., define > > > > A_n = -1/floor((n+9)/10), > > R_n = -1/n. > > > > For n = 1,2,..., define a function B_n by > > > > B_n(t) = 1 if A_n < t < R_n, > > 0 if t < A_n or t > R_n, > > undefined if t = A_n or t = R_n. > > > > Let V(t) = sum{n=1}^infty B_n(t). > > > > Then V(-1) = 1 and V(0) = 0. If we consider V to be a function of time, > > at what time does it become zero? > > Just answer the question, and stop beating around the bush. To recap, you wrote, "When 'something happens' to an object, some property or condition of it 'changes'. That occurs within some time period, which includes at least one moment." You also wrote, "To become empty means there is a change of state in the vase ('something happens' to the vase), from having balls to not having balls." From the definition of V, if t equals A_n or R_n, then V(t) is not defined. For other 1 <= t < 0, V(t) is positive. And, V(0) = 0. You asked when does the vase change from having balls to having no balls? Since, V(t) is positive (or undefined) for 1 <= t < 0 and V is zero at time zero, it would seem that according to *your* definition of "become empty", the vase becomes empty at noon. The only reason I say "seem" is that I don't know whether your definition allows V to be undefined at the times of addition and removal. We could either agree that at the addition and removal instants, the ball is not in the vase (thus changing the definition of V) or we could agree that "becomes empty" requires V to be defined at all times (in which case the vase never becomes empty). Regardless, V(0) = 0, so there are no balls in the vase at noon. -- David Marcus
From: David Marcus on 24 Oct 2006 02:13
Tony Orlow wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> stephen(a)nomail.com wrote: > >>> Also, supposing for the sake of argument that there are "infinitely > >>> number balls", if a ball is added at time -1/(2^floor(n/10)), and removed > >>> at time -1/(2^n)), then the balls added at time t=0, are those > >>> where -1/(2^floor(n/10)) = 0. But if -1/(2^floor(n/10)) = 0 > >>> then -1/(2^n) = 0 (making some reasonable assumptions about how arithmetic > >>> on these infinite numbers works), so those balls are also removed at noon and > >>> never spend any time in the vase. > >> Yes, the insertion/removal schedule instantly becomes infinitely fast in > >> a truly uncountable way. The only way to get a handle on it is to > >> explicitly state the level of infinity the iterations are allowed to > >> achieve at noon. When the iterations are restricted to finite values, > >> noon is never reached, but approached as a limit. > > > > Suppose we only do an insertion or removal at t = 1/n for n a natural > > number. What do you mean by "noon is never reached"? > > 1/n>0 Sorry, I meant t = -1/n. So, I assume your answer is that -1/n < 0. But, I don't follow. Translating "-1/n < 0" back into words, I get "all insertions and removals are before noon". However, I asked you what "noon is never reached" means. Are you saying that "noon is never reached" means that "all insertions and removals are before noon"? -- David Marcus |