An uncountable countable set
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From: Virgil on 17 Oct 2006 02:25 In article <45343e6f(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: > >>>>>>>>> How about this problem: Start with an empty vase. Add a ball to a > >>>>>>>>> vase > >>>>>>>>> at time 5. Remove it at time 6. How many balls are in the vase at > >>>>>>>>> time > >>>>>>>>> 10? > >>>>>>>>> > >>>>>>>>> Is this a nonsensical question? > >>>>>>>> Not if that's all that happens. However, that doesn't relate to the > >>>>>>>> ruse > >>>>>>>> in the vase problem under discussion. So, what's your point? > >>>>>>> Is this a reasonable translation into Mathematics of the above > >>>>>>> problem? > >>>>>> I gave you the translation, to the last iteration of which you did not > >>>>>> respond. > >>>>>>> "Let 1 signify that the ball is in the vase. Let 0 signify that it is > >>>>>>> not. Let A(t) signify the location of the ball at time t. The number > >>>>>>> of > >>>>>>> 'balls in the vase' at time t is A(t). Let > >>>>>>> > >>>>>>> A(t) = 1 if 5 < t < 6; 0 otherwise. > >>>>>>> > >>>>>>> What is A(10)?" > >>>>>> Think in terms on n, rather than t, and you'll slap yourself awake. > >>>>> Sorry, but perhaps I wasn't clear. I stated a problem above in English > >>>>> with one ball and you agreed it was a sensible problem. Then I asked if > >>>>> the translation above is a reasonable translation of the one-ball > >>>>> problem into Mathematics. If you gave your translation of the one-ball > >>>>> problem, I missed it. Regardless, my question is whether the > >>>>> translation > >>>>> above is acceptable. So, is the translation above for the one-ball > >>>>> problem reasonable/acceptable? > >>>> Yes, for that particular ball, you have described its state over time. > >>>> According to your rule, A(10)=0, since 10>6>5. Do go on. > >>> > >>> OK. Let's try one in reverse. First the Mathematics: > >>> > >>> > >>> Let B_1(t) = 1 if 5 < t < 7, > >>> 0 if t < 5 or t > 7, > >>> undefined otherwise. > >>> > >>> Let B_2(t) = 1 if 6 < t < 8, > >>> 0 if t < 6 or t > 8, > >>> undefined otherwise. > >>> > >>> Let V(t) = B_1(t) + B_2(t). What is V(9)? > >>> > >>> > >>> Now, how would you translate this into English ("balls", "vases", > >>> "time")? > >> That's not an infinite sequence, so it really has no bearing on the vase > >> problem as stated. I understand the simplistic logic with which you draw > >> your conclusions. Do you understand how it conflicts with other > >> simplistic logic? It's the difference between focusing on time vs. > >> iterations. Iteration-wise, it never can empty. There's something wrong > >> with your time-wise logic which has everything to do with the Zeno > >> machine and the indistinguishability of iterations outside of N. For > >> every n e N, iteration n results in 9n balls left over, a nonzero > >> number. If all steps are indexed by n in N, then this result holds for > >> the entire sequence. Within your experiment, with ball numbers all in N, > >> you never reach noon, and at every moment for the minute before it the > >> vase is nonempty. > > > > I didn't say it was an infinite sequence nor did I say it had a "bearing > > on the vase problem as stated". However, I asked you a question. I don't > > believe you answered my question. So, let me try again: > > > > How would you translate the mathematical problem I wrote above into > > English ("balls", "vases", "time")? > > > We could say we insert one ball, then another, then remove one, then the > other, and how many balls are in the vase after that? 0. That's the > sequence of events and the result. Now merely repeat similarly one insertion and one removal for each n in N to get the original vase problem.
From: imaginatorium on 17 Oct 2006 04:48 Tony Orlow wrote: > Virgil wrote: > > In article <45319846(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >>> Functions can exist at points at which their limits do not. > >>> There are even functions with domain R which are discontinuous at every > >>> rational argument but continuous at every irrational one. > >> That sounds vaguely interesting. Can you give an example? > > > > It is standard fare for anyone who knows any analysis. > > > > Let f: R --> R, be such that for each irrational x f(x) = 0, > > and for each rational x whose expression in lowest terms is p/q, > > let f(x) = 1/q. Then that function is provably continuous at each > > irrational and provably discontinuous at each rational except 0. It is > > provably the case that for each real a, lim_{x --> a} f(x) = 0, which > > establishes the claim. > > Can you give an example of two [distinct] irrational numbers, between which there > are no rational numbers? Of course he can't, because there aren't any. (My emendation in [ ], just in case the "two numbers" have the same value) But so what? What does this have to do with the claims about the function above? Do you understand the claims? Do you understand the mathematical definition of "continuous"? FWIW, I'm a bit slow at this stuff, and it took me a while to see why the function is continuous at each irrational. Of course, I had to think carefully about the definition of 'continuous', which wasn't terribly easy, but a lot less impossible than if I had no idea of the definition. Anyway, bumble on... Brian Chandler http://imaginatorium.org
From: Mike Kelly on 17 Oct 2006 06:48 Tony Orlow wrote: > cbrown(a)cbrownsystems.com wrote: > > Tony Orlow wrote: > >> cbrown(a)cbrownsystems.com wrote: > >>> Tony Orlow wrote: > >>>> cbrown(a)cbrownsystems.com wrote: > >>>>> Tony Orlow wrote: > >>>>>> cbrown(a)cbrownsystems.com wrote: > >>>>>>> Tony Orlow wrote: > >>>>>>>> cbrown(a)cbrownsystems.com wrote: > >>> <snip> > >>> > >>>>>>>>> Putting aside the question of /how/ (limit? sum of binary functions?) > >>>>>>>>> one determines the /number/ of balls in the vase at time t for a > >>>>>>>>> moment... > >>>>>>>>> > >>>>>>>>> Do you then agree that there is some explicit relationship described in > >>>>>>>>> the problem between what time it is, and whether any particular > >>>>>>>>> labelled ball, for example the ball labelled 15, is in the vase at that > >>>>>>>>> time? > >>>>>>>> For any finite time before noon, when iterations of the problem are > >>>>>>>> temporally distinguishable, yes, but at noon, no. > >>>>>>>> > >>>>>>> I don't understand why you think this would be the case. > >>>>>>> > >>>>>>> Why do you think the relationship holds for t < 0? > >>>>>>> > >>>>>>> Why you do think it does not hold for t >= 0? > >>>>>>> > >>>>>>> Cheers - Chas > >>>>>>> > >>>>>> Because for t>=0, n>=oo. > >>>>> Actually, for t>=0, there is /no/ natural number n such that t = -1/n. > >>>>> Similarly, for t = -1/pi, there is no natural number n such that t = > >>>>> -1/n. > >>>> Yeah, no idding. Who said oo was a natural number? > >>>> > >>> You just implied it; when you claimed that at t>=0, n>=oo; where I > >>> presume that by "n", you refer to the statement in the problem: > >>> > >>> "At time t = -1/n, where n is a natural number, we add balls labelled > >>> (10*(n-1)+1) through 10*n inclusive, and remove the ball labelled n". > >>> > >>> That statement obviously does not refer to removals or additions of > >>> balls at time t = 0, becuase there is no natural number n such that > >>> -1/n = 0. > >>> > >>> Do you agree with this conclusion? > >>> > >> Of course. That's what I was saying. > > > > Was it? When you said: > > > >>>>>> Because for t>=0, n>=oo. > > > > you seemed to be saying that we could use the rule: > > > >>> "At time t = -1/n, where n is a natural number, we add balls labelled > >>> (10*(n-1)+1) through 10*n inclusive, and remove the ball labelled n". > > > > and the fact that "for t>=0, n>=oo" to prove that we cannot conclude > > that ball 15 is not in the vase at t=0. Since "n>=oo" is never true, I > > don't see how your logic applies. > > > > In fact, from the problem statement, it follows logically that ball 15 > > is not in the vase at t=0. > > > > Now, I will grant this (alone) does not imply that you cannot /also/ > > prove that ball 15 /is/ in the vase at t=0. > > > > If you can /also/ prove that, then you have proven that the problem is > > contradictory - i.e., it contains assumptions that allow us to prove > > that something is both true and false. > > > > But even if you /did/ have such a proof, that would not change the fact > > that we can /also/ prove that ball 15 is not in the vase at t=0. > > > > Do you accept the above statements, or do you still claim that there is > > /no/ valid proof that ball 15 is not in the vase at t=0? > > > > 15 is a specific finite number for which we can state its times of entry > and exit. At its time of exit, balls 16 through 150 reside in the vase. > For every finite n in N, upon its removal, 9n balls remain. For every n > e N, there is a finite nonzero number of balls in the vase. Every > iteration in the sequence is indexed with an n in N. Therefore, nowhere > in the sequence is there anything other than a finite nonzero number of > balls in the vase. > > Now, where, specifically, in the fallacy in that argument? The fallacy is in your unjustified assumption that the above tells you how many balls are in the vase at noon. Noon is not anywhere in the sequence. -- mike.
From: Han de Bruijn on 17 Oct 2006 08:04 MoeBlee wrote: > Han.deBruijn(a)DTO.TUDelft.NL wrote: > >>MoeBlee schreef: >> >>>Han de Bruijn wrote: >>> >>>>Virgil wrote: >>>> >>>>>Axiom of infinity: There exists a set x such that the empty set is a >>>>>member of x and whenever y is in x, so is S(y). >>>> >>>>Which is actually the construction of the ordinals. Right? >>> >>>Wrong. >> >>Don't understand why that's wrong. Please explain. > > It's not the definition of 'ordinal' and there are ordinals that are > not "constructed" or proven to exist by the axiom of infinity. Huh? http://www.jboden.demon.co.uk/SetTheory/ordinals.html What I see there is that the empty set is a member of the (finite) ordinals, because 0 = { } . Right? http://en.wikipedia.org/wiki/Axiom_of_infinity We define the successor S(y) of y as y u { y } . Right? Now let { } be a member of the finite ordinals, then also { } u {{ }} is a member of the finite ordinals, hence the set {{},{{}}, .. }. Right? Why then does the axiom of infinity not define the (finite) ordinals? > The axiom of infinity is just not the construction of the ordinals. > Why don't you just read a set theory textbook rather than remain > ignorant about that which you are so opinionated. No matter how much textbooks about set theory I read, it all remains abacedabra for me. Han de Bruijn
From: Han de Bruijn on 17 Oct 2006 08:06
MoeBlee wrote: > Han de Bruijn wrote: > >>Virgil wrote: >> >>>In article <1160935613.121858.178420(a)m7g2000cwm.googlegroups.com>, >>> Han.deBruijn(a)DTO.TUDelft.NL wrote: >>> >>>>I find the idea absurd that natural numbers can be built >>>>by putting curly braces around the empty set. >>> >>>It appears as if much of useful mathematics is ultimately based on what >>>HdB finds absurd. >> >>The natural numbers can be defined without employing set theory. > > Yeah, so? > > And, by the way, natural numbers are not defined in set theory by a > method of curly braces. What then is the subtle difference between naturals and finite ordinals? Han de Bruijn |