An uncountable countable set
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From: David Marcus on 27 Oct 2006 11:49 Tony Orlow wrote: > David Marcus wrote: > > Tony Orlow wrote: > >> Mike Kelly wrote: > >>> Now correct me if I'm wrong, but I think you agreed that every > >>> "specific" ball has been removed before noon. And indeed the problem > >>> statement doesn't mention any "non-specific" balls, so it seems that > >>> the vase must be empty. However, you believe that in order to "reach > >>> noon" one must have iterations where "non specific" balls without > >>> natural numbers are inserted into the vase and thus, if the problem > >>> makes sense and "noon" is meaningful, the vase is non-empty at noon. Is > >>> this a fair summary of your position? > >>> > >>> If so, I'd like to make clear that I have no idea in the world why you > >>> hold such a notion. It seems utterly illogical to me and it baffles me > >>> why you hold to it so doggedly. So, I'd like to try and understand why > >>> you think that it is the case. If you can explain it cogently, maybe > >>> I'll be convinced that you make sense. And maybe if you can't explain, > >>> you'll admit that you might be wrong? > >>> > >>> Let's start simply so there is less room for mutual incomprehension. > >>> Let's imagine a new experiment. In this experiment, we have the same > >>> infinite vase and the same infinite set of balls with natural numbers > >>> on them. Let's call the time one minute to noon -1 and noon 0. Note > >>> that time is a real-valued variable that can have any real value. At > >>> time -1/n we insert ball n into the vase. > >>> > >>> My question : what do you think is in the vase at noon? > >> A countable infinity of balls. > > > > So, "noon exists" in this case, even though nothing happens at noon. > > Not really, but there is a big difference between this and the original > experiment. If noon did exist here as the time of any event (insertion), > then you would have an UNcountably infinite set of balls. Presumably, > given only naturals, such that nothing is inserted at noon, by noon all > naturals have been inserted, for the countable infinity. Then insertions > stop, and the vase has what it has. The issue with the original problem > is that, if it empties, it has to have done it before noon, because > nothing happens at noon. You conclude there is a change of state when > nothing happens. I conclude there is not. So, noon doesn't exist in this case either? -- David Marcus
From: David Marcus on 27 Oct 2006 11:52 Tony Orlow wrote: > David Marcus wrote: > > You are mentioning balls and time and a vase. But, what I'm asking is > > completely separate from that. I'm just asking about a math problem. > > Please just consider the following mathematical definitions and > > completely ignore that they may or may not be relevant/related/similar > > to the vase and balls problem: > > > > -------------------------- > > For n = 1,2,..., let > > > > A_n = -1/floor((n+9)/10), > > R_n = -1/n. > > > > For n = 1,2,..., define a function B_n: R -> R by > > > > B_n(t) = 1 if A_n <= t < R_n, > > 0 if t < A_n or t >= R_n. > > > > Let V(t) = sum_n B_n(t). > > -------------------------- > > > > Just looking at these definitions of sequences and functions from R (the > > real numbers) to R, and assuming that the sum is defined as it would be > > in a Freshman Calculus class, are you saying that V(0) is not equal to > > 0? > > On the surface, you math appears correct, but that doesn't mend the > obvious contradiction in having an event occur in a time continuum > without occupying at least one moment. It doesn't explain how a > divergent sum converges to 0. Basically, what you prove, if V(0)=0, is > that all finite naturals are removed by noon. I never disagreed with > that. However, to actually reach noon requires infinite naturals. Sure, > if V is defined as the sum of all finite balls, V(0)=0. But, I've > already said that, several times, haven't I? Isn't that an answer to > your question? I think it is an answer. Just to be sure, please confirm that you agree that, with the definitions above, V(0) = 0. Is that correct? -- David Marcus
From: David Marcus on 27 Oct 2006 11:57 Tony Orlow wrote: > David Marcus wrote: > > Your question "Is there a smallest infinite number?" lacks context. You > > need to state what "numbers" you are considering. Lots of things can be > > constructed/defined that people refer to as "numbers". However, these > > "numbers" differ in many details. If you assume that all subjects that > > use the word "number" are talking about the same thing, then it is > > hardly surprising that you would become confused. > > I don't consider transfinite "numbers" to be real numbers at all. I'm > not interested in that nonsense, to be honest. I see it as a dead end. > > If there is a definition for "number" in general, and for "infinite", > then there cannot both be a smallest infinite number and not be. A moot point, since there is no definition for "'number' in general", as I just said. -- David Marcus
From: Tony Orlow on 27 Oct 2006 11:59 Ross A. Finlayson wrote: > Tony Orlow wrote: >> The Inverse Function Rule uses infinite-case induction to finely order >> infinite sets of reals mapped from a standard set, N. Where there is a >> bijection between N and a set S using f(n)=s, there is a mapping from S >> to N using g(s)=n, where g(f(x))=f(g(x)) (inverse functions for the >> bijection). The size of the set S over the interval [a,b] is given by >> floor(g(b)-g(a)+1). (I think I wrote that correctly). This works for all >> finite sets of reals. The number of square roots, for instance, between >> 1 and 100 is floor(100^2-1^2+1), 10000 square roots, from sqrt(1) to >> sqrt(10000). IFR can easily be used to show that the evens are half as >> numerous as the naturals, and other interesting "facts". >> >> EF is the special case of IFR mapping the naturals in [0,oo) to the >> reals in [0,1), using the mapping function f(n)=n/oo. Isn't that how you >> define the equivalency function? Given this mapping, we can say >> g(s)=s*oo, so that over the entire real line, we have oo^2 reals, oo in >> each unit interval, over oo unit intervals. Does that sound about right? >> >> Tony > > Isn't there symmetry about the origin thus it's 2 times oo^2? Oh sure, it could be. I generally think of N as the positive side of the real line, so that would be right, but one could use N as both sides. As long as one chooses a standard infinite length, IFR works over the real line. > > Obviously half of the integers are even. I think that's obvious. > > What are cases against use or validity of IFR? How do you address > those? > > Ross > Oh, there have been plenty of objections. Foremost, I think, is its dependence on infinite-case induction, hence Chas' diagonal staircase counterexample for such induction. But, I haven't seen a valid counterexample yet, though that was a good try on Chas' part. Of course, it leads to the notion of infinitesimals like EF does - that seems to be out of vogue this century, and gets plenty of objections. Virgil complained at one point that not every function has an inverse function, but any set of reals bijected with another does so with an inherently invertible function, or there's no bijection. Also it only applies to real values, since it depends on a value range, and the infinite range of the line is not considered real by standard mathematicians. I don't care - call it N, and express sets over the real range as formulas on N, or Big'un. It works for finite and infinite sets of real values mapped from the naturals. Do you see any problems with it? You might want to play with it. Maybe you'll extend it somehow. I certainly was happy to see it mesh with your EF. I had been trying to apply it to the relationship between the continuum and the naturals using f(x)=log_oo(x) or something, but that caused problems. EF is the correct mapping between the two, according to IFR. :) Rock on! Tony
From: David Marcus on 27 Oct 2006 12:01
Tony Orlow wrote: > stephen(a)nomail.com wrote: > > What are you talking about? I defined two sets. There are no > > balls or vases. There are simply the two sets > > > > IN = { n | -1/(2^floor(n/10)) < 0 } > > OUT = { n | -1/(2^n) < 0 } > > For each n e N, IN(n)=10*OUT(n). Stephen defined sets IN and OUT. He didn't define sets "IN(n)" and "OUT (n)". So, you seem to be answering a question he didn't ask. Given Stephen's definitions of IN and OUT, is IN = OUT? -- David Marcus |