An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Tony Orlow on 24 Oct 2006 00:43 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.
From: Tony Orlow on 24 Oct 2006 00:44 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
From: Tony Orlow on 24 Oct 2006 00:45 David Marcus wrote: > Tony Orlow wrote: >> David Marcus wrote: >>> Tony Orlow wrote: >>>> Your examples of the circle and rectangle are good. Neither has a height >>>> outside of its x range. The height of the circle is 0 at x=-1 and x=1, >>>> because the circle actually exists there. To ask about its height at x=9 >>>> is like asking how the air quality was on the 85th floor of the World >>>> Trade Center yesterday. Similarly, it makes little sense to ask what >>>> happens at noon. There is no vase at noon. >>> Do you really mean to say that there is no vase at noon or do you mean >>> to say that the vase is not empty at noon? >> If noon exists at all, the vase is not empty. All finite naturals will >> have been removed, but an infinite number of infinitely-numbered balls >> will remain. > > "If noon exists at all"? How do we decide? > We decide on the basis of whether 1/n=0. Is that possible for n in N? Hmmmm......nope.
From: Tony Orlow on 24 Oct 2006 00:47 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.
From: Tony Orlow on 24 Oct 2006 00:48
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. |