Cantor Confusion
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From: William Hughes on 14 Oct 2006 10:16 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > Han de Bruijn wrote: > > > Set Theory is simply not very useful. The main problem being that finite > > > sets in your axiom system are STATIC. They can not grow. > > > > Set theory provides for capturing the notion of mathematical growth. > > Sets don't grow, but growth is expressible in set theory. If there is a > > mathematical notion that set theory cannot express, then please say > > what it is. > > Obviously the notion of "rational relation" as used in the binary tree > cannot be expressed by mathematical notion: > Consider the binary tree which has (no finite paths but only) infinite > paths representing the real numbers between 0 and 1. The edges (like a, > b, and c below) connect the nodes, i.e., the binary digits. The set of > edges is countable, because we can enumerate them > > 0. > /a \ > 0 1 > /b \c / \ > 0 1 0 1 > ............. > > Now we set up a relation between paths and edges. Relate edge a to all > paths which begin with 0.0. Relate edge b to all paths which begin with > 0.00 and relate edge c to all paths which begin with 0.01. Half of edge > a is inherited by all paths which begin with 0.00, the other half of > edge a is inherited by all paths which begin with 0.01. So each finite path of length N is related to 1 + 1/2 +1/4 + ... + 1/2^N edges > Continuing in this manner in infinity, we get a limit which may or may not be related to anything. > we see that every single infinite path is > related to 1 + 1/2 + 1/ 4 + ... = 2 edges, which are not related to any > other path. No, the statement that what holds for finite paths also holds for infinite paths needs proof. Your provide none. - William Hughes
From: William Hughes on 14 Oct 2006 11:07 mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > Han de Bruijn wrote: > > > David Marcus wrote: > > > > Is your claim only that set theory is not useful or is contrary to > > > > common sense? Or, are you claiming something more, e.g., that set theory > > > > is mathematically inconsistent? > > > > > > I said that set theory is not *very* useful. I have developed (limited) > > > set theoretic applications myself, so I don't say it is useless. > > > > > Yes, a great deal of set theory is contrary to common sense. Especially > > > the infinitary part of it (: say cardinals, ordinals, aleph_0). > > > > > > I'm not interested in the question whether set theory is mathematically > > > inconsistent. What bothers me is whether it is _physically_ inconsistent > > > and I think - worse: I know - that it is. > > > > What does "physically inconsistent" mean? Wouldn't your comments be > > better posted to sci.physics? Most people in sci.math are (or at least > > think they are) discussing mathematics. > > Even worse, most of them truly believe their ideas on mathematics and > the functions (of mathematics as well as of their brains) were > independent of physics > You are confusing levels. Brains are physically based. Concepts produced by those brains (beauty, mathematics, ...) need not be. - William Hughes
From: William Hughes on 14 Oct 2006 11:21 Han de Bruijn wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > > imaginatorium(a)despammed.com schrieb: > > > >>The cranks universally proceed from some intuitions about the real > >>world that - essentially - there are no discontinuous functions in > >>physics. > > I don't know which "crank" you are talking about, but one of these > "cranks" says that - essentially - there are no continuous functions > either. This particular crank also claims that there is a largest integer (or more precisely, that there are a finite number of integers and he doesn't care whether there is a largest). So, this crank's views are clearly not based on common sense. - William Hughes
From: Alan Morgan on 14 Oct 2006 12:31 In article <1160833920.096796.262180(a)h48g2000cwc.googlegroups.com>, <mueckenh(a)rz.fh-augsburg.de> wrote: >Alan Morgan schrieb: > >> In article <990aa$452e542e$82a1e228$16180(a)news1.tudelft.nl>, >> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: >> >Dik T. Winter wrote: >> > >> >> In article <1160646886.830639.308620(a)c28g2000cwb.googlegroups.com> >> >> mueckenh(a)rz.fh-augsburg.de writes: >> >> ... >> >> > If every digit position is well defined, then 0.111... is covered "up >> >> > to every position" by the list numbers, which are simply the natural >> >> > indizes. I claim that covering "up to every" implies covering "every". >> >> >> >> Yes, you claim. Without proof. You state it is true for each finite >> >> sequence, so it is also true for the infinite sequence. That conclusion >> >> is simply wrong. >> > >> >That conclusion is simply right. And yours is wrong. Completed infinity >> >does not exist. >> >> sqrt(-1) doesn't exist either. Frankly, I have a much harder time >> believing in "imaginary" numbers than I do believing in infinite >> sets. > >But sqrt(-1) does not yield contradictions, as far as I know. It violates the heck out of my intuition. The objections to set theory seem to arise from someone's dislike for the conclusions or an inability to do mathematics correctly. No one has actually shown a contradiction yet. Alan -- Defendit numerus
From: William Hughes on 14 Oct 2006 13:30
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > COLLECTED ANSWERS > > > Once you have chosen your set of 100 elements it will have > > a largest size. Questions as to the method of choice have > > no bearing on this. > > > Correct. Once you have chosen how to use all the bits of the universe, > then you will have a largest number. And crucial to this discussion, once you have chosen that you are going to use all the bits in the universe you know that you will have a largest number. If you get your jollies from saying that there will be a largest integer, rather than there is a largest integer, far be it from me to stop you (but how you can have all the integers exist now, but only have the largest integer exist later is a problem you will have to solve). > Infinty does only come into the > play as long as this choice has not yet been fixed. Actually no. As well as the largest integer that will be described there is also the largest integer that can be described. (you need to both give your representation *and describe your representation*. E.g. The string "1" in the base 1 billion, represents the integer one billion. However, the string "1" does not represent this integer, you need to add "in the base 1 billion". We could do the usual formailization in terms of Turing machines, but this would do little except free us from such paradoxes as "let K be the largest integer that can be described plus 1".(note the usual interpretation is that K has not been described, you, however, claim that K cannot exist.)) And while the largest integer that will be described is not yet known (even in theory) the largest integer that can be described is known (at least in theory). So the largest integer that will be described cannot be chosen in an arbitrary manner. It must be less than or equal to the largest integer that can be described. As soon as you say "the set of integers consists of only those integers that will be described during the lifetime of the universe" you lose both actual and potential infinity. > > > It is true that in some cases the quantifier exchange is > >possible. However, the fact that the quantifier exchange > >is impossible in general means that you cannot use > >quantifier exchange in a proof without explicit > >justification of the step. Despite you protestations, > >this is exactly what you do. > > Despite of your belief I have proved that exchange is possible in > linear sets with only finite elements. If you state a counter example, > I can reject it by disproving it. You cannot, but only claim that for > an "infinite set" your position was correct - of course without being > able to show any infinity other than by the three "...". > Let N be the set of finite integers. Either N has an upper bound or it does not. If you claim that N has an upper bound we will have to go our merry ways. Otherwise let K be subset of N that does not have an upper bound. There, I have produced an infinite set without using the three "...". [You can of course, claim that N does not have an upper bound and N does not exist as a complete set. However, you wish to do more. You want to show that claiming "N does not have an upper bound and N exists as a complete set" leads to a contradiction.] > > Your putative proof of this "fact" depends on a step > in which quantifiers are switched without justification. > > Give a proof of that by a counter example using any concrete natural > numbers. > Let 0.111ppp be a number that has a 1 in every digit place corresponding to an element of the set N. Then every digit place is a finite natural number and therefore for every digit place n, there is an m(n) such that m(n) covers 0.111ppp up to digit place n. However, we cannot reverse the quantifiers. There does not exist a single M which covers 0.111ppp up to every finite natural number n. M would have to be an upper bound for the set N and we have assumed that such an upper bound does not exist. > > This requires proof. Try to produce one without > any unjustified quantifier exchange. > > I did never employ unjustified quntifier exchange, but if you dislike > that example, then consider the binary tree which i just posted another > time. This is a (slightly) obfuscated version of "{1,2,3,...,n} is bounded for all n in N. Therefore N is bounded". - William Hughes |