An uncountable countable set
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From: MoeBlee on 31 Oct 2006 15:48 Lester Zick wrote: > On 30 Oct 2006 16:52:02 -0800, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: > > >Tony Orlow wrote: > >> stephen(a)nomail.com wrote: > > > >> > The point is, there are different types of numbers, and statements > >> > that are true of one type of number need not be true of other > >> > types of numbers. > > > >> Well, then, you must be of the opinion that set theory is NOT the > >> foundation for all mathematics, but only some particular system of > >> numbers and ideas: a theory. That's good. > > > >Whether he thinks set theory is or is not a foundation, it doesn't > >follow that he should not think it is a foundation simply because there > >are different kinds of numbers. > > Huh? Maybe you could run that by us again, Moe. That there are different kinds of numbers does not entail that set theory cannot be a foundation. MoeBlee
From: Virgil on 31 Oct 2006 15:50 In article <4547707d(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <454632a3(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > >>> In article <4543b0b3(a)news2.lightlink.com>, > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> > >>> > >>>> The experiment occurred in [-1,0). Talk of time outside that range is > >>>> irrelevant. Times before that are imaginary, and times after that are > >>>> infinite. Only finite times change anything, so if something changes, > >>>> it's at a finite, negative time. > >>> Then let us change the experiment to include the insertion into the vase > >>> of a cube at one minute after noon. > >>> > >>> The experiment now ranges over [-1,1]. > >>> > >>> What are the contents of the vase at times in [0,1), TO? > >>> > >> An uncountable number of balls, all infinitely numbered. > > > > As none of them exist in the original problem, where does TO get them > > from? > > > > And how does he manage to make them come into existence on his command? > > > > Such magic is no part of mathematics. > > Coming from you, that's rich! On the contrary, it is TO who pretends to have a magic wand which he can wave to make the impossible happen. The original problem has only 1 ball per 1 finite natural Each such ball has a time of insertion and a time of removal. These times are all before noon. TO then "waves his wand" to pretend to find infinitely unnumbered balls created out of nothingness and inserted into the vase precisely at noon, when the original problem has nothing happening at noon. So who is invoking magic here, those who follow the rules, or TO.
From: Virgil on 31 Oct 2006 15:54 In article <45477121(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: > >>>>>> I am beginning to realize just how much trouble the axiom of > >>>>>> extensionality is causing here. That is what you're using, here, no? > >>>>>> The > >>>>>> sets are "equal" because they contain the same elements. That gives no > >>>>>> measure of how the sets compare at any given point in their > >>>>>> production. > >>>>>> Sets as sets are considered static and complete. However, when talking > >>>>>> about processes of adding and removing elements, the sets are not > >>>>>> static, but changing with each event. When speaking about what is in > >>>>>> the > >>>>>> set at time t, use a function for that sum on t, assume t is > >>>>>> continuous, > >>>>>> and check the limit as t->0. Then you won't run into silly paradoxes > >>>>>> and > >>>>>> unicorns. > >>>>> There is a lot of stuff in there. Let's go one step at a time. I > >>>>> believe > >>>>> that one thing you are saying is this: > >>>>> > >>>>> |IN\OUT| = 0, but defining IN and OUT and looking at |IN\OUT| is not > >>>>> |the > >>>>> correct translation of the balls and vase problem into Mathematics. > >>>>> > >>>>> Do you agree with this statement? > >>>> Yes. > >>> OK. Since you don't like the |IN\OUT| translation, let's see if we can > >>> take what you wrote, translate it into Mathematics, and get a > >>> translation that you like. > >>> > >>> You say, "When speaking about what is in the set at time t, use a > >>> function for that sum on t, assume t is continuous, and check the limit > >>> as t->0." > >>> > >>> Taking this one step at a time, first we have "use a function for that > >>> sum on t". How about we use the function V defined as follows? > >>> > >>> 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 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). > >>> > >>> Next you say, "assume t is continuous". Not sure what you mean. Maybe > >>> you mean assume the function is continuous? However, it seems that > >>> either the function we defined (e.g., V) is continuous or it isn't, > >>> i.e., it should be something we deduce, not assume. Let's skip this for > >>> now. I don't think we actually need it. > >>> > >>> Finally, you write, "check the limit as t->0". I would interpret this as > >>> saying that we should evaluate the limit of V(t) as t approaches zero > >>> from the left, i.e., > >>> > >>> lim_{t -> 0-} V(t). > >>> > >>> Do you agree that you are saying that the number of balls in the vase at > >>> noon is lim_{t -> 0-} V(t)? > >>> > >> Find limits of formulas on numbers, not limits of sets. > > > > I have no clue what you mean. There are no "limits of sets" in what I > > wrote. > > > >> Here's what I said to Stephen: > >> > >> out(n) is the number of balls removed upon completion of iteration n, > >> and is equal to n. > >> > >> in(n) is the number of balls inserted upon completion of iteration n, > >> and is equal to 10n. > >> > >> contains(n) is the number of balls in the vase upon completion of > >> iteration n, and is equal to in(n)-out(n)=9n. > >> > >> n(t) is the number of iterations completed at time t, equal to > >> floor(-1/t). > >> > >> contains(t) is the number of balls in the vase at time t, and is equal > >> to contains(n(t))=contains(floor(-1/t))=9*floor(-1/t). > >> > >> Lim(t->-0: 9*floor(-1/t)))=oo. The sum diverges in the limit. > > > > You seem to be agreeing with what I wrote, i.e., that you say that the > > number of balls in the vase at noon is lim_{t -> 0-} V(t). Care to > > confirm this? > > > > No that's a bad formulation. I gave you the correct formulation, which > states the number of balls in the vase as a function of t. TO did no such thing, as his formulation requires infinitely many unnumbered balls to come into existence out of nowhere after all numbered balls are removed from the vase.
From: Virgil on 31 Oct 2006 15:56 In article <454771cc(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: > >>>>>> David Marcus wrote: > >>>>>>> Tony Orlow wrote: > >>>>>>>> David Marcus wrote: > >>>>>>>>> Tony Orlow wrote: > >>>>>>>>>> David Marcus wrote: > >>>>>>>>>>> Tony Orlow wrote: > >>>>>>>>>>>> David Marcus wrote: > >>>>>>>>>>>>> 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? > >>>>>>>>>>>> Sure, all finite balls are gone at noon. > >>>>>>>>>>> Please note that there are no balls or time in the above > >>>>>>>>>>> mathematics problem. However, I'll take your "Sure" as > >>>>>>>>>>> agreement that V(0) = 0. > >>>>>>>>>> Okay. > >>>>>>>>>>> Let me ask you a question about this mathematics problem. > >>>>>>>>>>> Please answer without using the words "balls", "vase", > >>>>>>>>>>> "time", or "noon" (since these words do not occur in the > >>>>>>>>>>> problem). > >>>>>>>>>> I'll try. > >>>>>>>>>>> First some discussion: For each n, B_n(0) = 0 and B_n is > >>>>>>>>>>> continuous at zero. > >>>>>>>>>> What??? How do you conclude that anything besides time is > >>>>>>>>>> continuous at 0, where yo have an ordinal discontinuity???? > >>>>>>>>>> Please explain. > >>>>>>>>> I thought we agreed above to not use the word "time" in > >>>>>>>>> discussing this mathematics problem? > >>>>>>>> If that's what you want, then why don't you remove 't' from all > >>>>>>>> of your equations? > >>>>>>> It is just a letter. It stands for a real number. Would you > >>>>>>> prefer "x"? I'll switch to "x". > >>>>>> It is still related to n in such a way that x<0. > >>>>>>>>> As for your question, let's look at B_2 (the argument is > >>>>>>>>> similar for the other B_n). > >>>>>>>>> > >>>>>>>>> B_2(t) = 1 if A_2 <= t < R_2, > >>>>>>>>> 0 if t < A_2 or t >= R_2. > >>>>>>>>> > >>>>>>>>> Now, A_2 = -1 and R_2 = -1/2. So, > >>>>>>>>> > >>>>>>>>> B_2(t) = 1 if -1 <= t < -1/2, > >>>>>>>>> 0 if t < -1 or t >= -1/2. > >>>>>>>>> > >>>>>>>>> In particular, B_2(t) = 0 for t >= -1/2. So, the value of B_2 > >>>>>>>>> at zero is zero and the limit as we approach zero is zero. So, > >>>>>>>>> B_2 is continuous at zero. > >>>>>>>> Oh. For each ball, nothing is happening at 0 and B_n(0)=0. > >>>>>>>> That's for each finite ball that one can specify. > >>>>>>> I thought we agreed to not use the word "ball" in discussing this > >>>>>>> mathematics problem? Do you want me to change the letter "B" to a > >>>>>>> different letter, too? > >>>>>> Call it an element or a ball. I don't care. It doesn't matter. > >>>>>>>> However, lim(t->0: sum(B_n| B_n(t)=1))=oo. Why do you > >>>>>>>> conveniently forget that fact? > >>>>>>> Your notation is nonstandard, so I'm not sure what you mean. Do > >>>>>>> you mean to write > >>>>>>> > >>>>>>> lim_{x -> 0-} sum_n B_n(x) = oo ? > >>>>>>> > >>>>>>> If so, I don't understand why you think I've forgotten this fact. > >>>>>>> If you look in my previous post (or below), you will see that I > >>>>>>> wrote, "Now, V is the sum of the B_n. As t approaches zero from > >>>>>>> the left, V(t) grows without bound. In fact, given any large > >>>>>>> number M, there is an e < 0 such that for e < t < 0, V(t) > M." > >>>>>> Then don't you see a contradiction in the limit at that point > >>>>>> being oo, the value being 0, and there being no event to cause > >>>>>> that change? I do. > >>>>>>>>>>> In fact, for a given n, there is an e < 0 such that B_n(t) = > >>>>>>>>>>> 0 for e < t <= 0. > >>>>>>>>>> There is no e<0 such that e<t and B_n(t)=0. That's simply false. > >>>>>>>>> Let's look at B_2 again. We can take e = -1/2. Then B_2(t) = 0 > >>>>>>>>> for e < t <= 0. Similarly, for any other given B_n, we can find > >>>>>>>>> an e that does what I wrote. > >>>>>>>> Yes, okay, I misread that. Sorry. For each ball B_n that's true. > >>>>>>>> For the sum of balls n such that B_n(t)=1, it diverges as t->0. > >>>>>>>>>>> In other words, B_n is not changing near zero. > >>>>>>>>>> Infinitely more quickly but not. That's logical. And wrong. > >>>>>>>>> Not sure what you mea
From: Virgil on 31 Oct 2006 16:04
In article <45478bb7(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> I am beginning to realize just how much trouble the axiom of > >> extensionality is causing here. > > > > Oh, now the axiom of extensionality. > > > > When you buy into Robinson's non-standard analysis you buy into the > > axiom of extensionality, and all the other axioms of set theory, and > > mathematical logic - the whole kit and kaboodle - including the axiom > > of choice, ordinals, and uncountable cardinals, and all the > > "transfinitology" (even if not with platonistic committments) you so > > strenuously disclaim. > > > > MoeBlee > > > > Dear Moe - > > When I say there are problems with the axiom of extensionality, I refer > to the application of the fact that two sets, when viewed statically, > contain the same elements. When you deny that, you cannot then insist on anything that relies on that, such as Robinson's non-standard analysis. Such things as ZF set theory or Robinson's theories come all of a piece, not as parts catalogues from which you can pick only what you want and then go to someone else's catalog for other bits and pieces. |