Cantor Confusion
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From: mueckenh on 2 Feb 2007 02:14 On 1 Feb., 18:44, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 1, 11:53 am, mueck...(a)rz.fh-augsburg.de wrote: > > > Yes. That is not in question. What I dispute is that the difference > > depends on the cardinal number of the set. > > It may or may not. For instance the question > > "Is the cardinal number of the set a (finite) natural number > (if the set does not have a cardinal number then the > answer is no)?" > > certainly depends on the cardinal number of the set. Sets of type > I have a cardinal which is a (finite) natural number. Sets of type > II do not. My questions are: What numbers are subject to complete induction? What is the set the elements of which are subject to complete induction? > > Yes. That is not in question. What I dispute is that the difference > > (induction is possible or is not) depends on the cardinal number of > > the set. > > Induction on the elements of A > is always possible given any set A of > natural numbers. However, induction may not > be able to prove that something is true of A Why should it? But it can be able. For instance it can prove that every set of natural numbers is finite while the size of sets of natural numbers is unbounded from above. > > For example, every one of the elements > of A is finite. This can be shown by induction. > However, this does not show that the set > A is not (potentially) infinite. It does. Of course every set of natural numbers is finite, as proved by induction over all initial segments. (Potential infinity.) This simple truth has only been veiled by the arbitrary assumption that there is an actual infinity. As this assertion cannot be proved, some have argue that induction was not valid for the whole set. Regards, WM
From: Andy Smith on 2 Feb 2007 05:06 David Marcus <DavidMarcus(a)alumdotmit.edu> writes >Andy Smith wrote: (snip) > >> I am exercised by the idea of space filling curves, which is the same >> issue, covering something by something of a lower classical dimension, >> and so maybe the recursive fractal definition of reals does cover my >> concept of line. > >No. This is a different problem. Even Euclid treated lines as consisting >of points and planes as consisting of lines. A space-filling curve is a >is a continuous function from [0,1] to [0,1]^2 that is onto. The >surprising feature is that the function is continuous. > >> I asked some time before on the other thread, and didn't really get a >> straight answer from DM, > >If you don't get a "straight answer", maybe you didn't ask a straight >question. Your question was about changing the space-filling curve >construction, but it wasn't clear how you wanted to change it. > >> about a fractal scenario where case 1: we >> start of with the interval 0-1. On each iteration, each interval is >> divided in two, and we associate a real number with the mid-point of >> each section. After an infinite number of iterations, all the reals in >> [0,1] are covered, no problem. > >Do you mean that at stage n+1, we divide all the intervals from stage n >in half? So, at each stage we have twice as many intervals as before? If >we do this for all natural numbers n, we only get a countable number of >intervals (all together), so the mid-points of the intervals most >certainly do not include all real numbers in [0,1]. The only numbers you >get are numbers with a finite binary expansion. > OK, I understand that argument. You are saying that I can't generate a number with an infinite decimal expansion by a finite limit process, it just converges to some subset of Q. I can't just loosely say "an infinite number of iterations", it's not a number. But in that case, I misunderstand how the space filling curve iteration can cover all the points in the plane - if you take a cross-section of the plane along a line at e.g. 45 degrees, on iteration 1 it has 2 points covered, and then doubling each iteration. Doesn't the same argument then apply - you can't generate something infinite from a finite iteration? -- Andy Smith
From: Andy Smith on 2 Feb 2007 05:20 Dave Seaman <dseaman(a)no.such.host> writes >On Thu, 01 Feb 2007 21:52:26 GMT, Andy Smith wrote: >> Dave Seaman <dseaman(a)no.such.host> writes >> (snip) > >>>> I had previously imagined that the "Continuum Hypothesis" related to >>>> whether the set of reals (as 0 dimensional points, each with no >>>> neighbouring point) could cover the line, with dimension 1 (Which seemed >>>> like a good question). > >>>You are asking whether R and "the real line" are the same set. In order >>>for that question to have a nontrivial answer, you must have some meaning >>>in mind for "the real line" other than R. What is your definition? > >> Well until recently my naive view was that points and lines were >> different entities, and you might as well ask how many metres in an acre >> as how many points in a line. I would have said (in your terminology) >> that there is clearly an injection from the reals into the line, but >> that you can never cover it. But I doubt now if that is true, or if the >> question is a sensible one to start with. > >I don't follow. Is it your claim that: > > (1) A line does not have any points as members at all, or > (2) A line has points as members, but perhaps there aren't > enough points to fill it? > >Your metres-in-an-acre analogy seems to suggest (1), but otherwise you >seem to be saying (2). > >If (2) is your claim, then can you show me a point on the line that is >not represented by a real number? If you can do this, then evidently >"the line" to you must be something other than simply "the set of real >numbers", as it is to most people. Can you elaborate on this? > > Well I meant 2) - (and I can't find any location in the acre that is not covered by a metre line either). And I can't answer your question . Yes, of course, the moment that you specify a real number you have defined a location on the line but I am suffering from dimensional anxiety, particularly when you observe that any point on the line has no nearest neighbour. If I was a point sized entity wanting to walk from 0 to 1 without falling down a crack, how would I do it? (I know that is not a particularly sensible question, just illustrating my mis-perspective). > -- Andy Smith
From: mueckenh on 2 Feb 2007 05:36 On 2 Feb., 03:18, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <kAau6RggoiwFF...(a)phoenixsystems.demon.co.uk> Andy Smith <A...(a)phoenixsystems.co.uk> writes: > ... > > 1) Is there a set with cardinality greater than N but less than R ? > > The Continuum Hypothesis states that there isn't, but it is not provable. > And so some people use it as an axiom, others do not. > > > 2) I am (dimly aware) that mathematicians talk about an infinite range > > of cardinalities. Is that just an abstract concept, or can you point to > > some set and say e.g. that has a higher cardinality than the reals? > > That has already been shown by Cantor. In one of his proofs he did show > (in effect) that given a set K, the set of subsets of K has a cardinality > that is strictly larger than the cardinality of K itself. The most > succinct proof of this is due to Hessenberg. The proof is fairly simple. > Consider a set K and its powerset (i.e. set of subsets) P(k). Assume > that there is an injection f: K -> P(K) (those clearly do exist, consider > f(k) = {k}). Now we set out to prove that f (as given) can not be a > surjection. Clearly for each 'k' in K, f(k) is a subset of K, so we can > consider those 'k' where 'k' is an element of f(k) and those where 'k' is not > an element of f(k). That defines two subsets of K, call them Y (the 'k' > in that set map to subsets that contain 'k') and N. Now the question is, > is there a 'k' that maps to N? > (1) Assume 'k' in N, but by the definition of N, the elements do not map > to a subset that contains themselve, so 'k' not in N. > (2) Assume 'k' in Y, but by the definition of Y, the elements do map to > a subset that contains themselve, and N does not contain 'k', so > 'k' can not be in Y. > As 'k' is in neither Y nor N, there is no 'k' that maps to N and so the map > is not surjective. Alas, this proof has nothing to do with cardinality of sets. Regards, WM
From: mueckenh on 2 Feb 2007 05:48
On 2 Feb., 02:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170348341.624257.130...(a)p10g2000cwp.googlegroups.com> mueck....(a)rz.fh-augsburg.de writes: > > > On 1 Feb., 14:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1170330236.064219.62...(a)a75g2000cwd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > If you have a set and you add something to this set (in unary > > > > representation), then you have a super set of the former. The former > > > > is a subset of the latter. > > > > > > This is not defining what I asked for. What are the elements of the sets > > > you are talking about? > > > > The type of element is not restricted other than by the requirement > > that it be part of the real world. > > We were specifically talking about the sets I, II, III, etc. Or about 4 and > 5. I ask you how you define them as sets. But let me elaborate more on > this below: > > > > > > > > Bizarre. So 3 is the set of all existing (what does that mean) > > > > > > > sets with 3 elements, > > If 4 is the set of all existing (whatever that does mean) sets with 4 > elements and 5 is the set of all existing sets with 5 elements, we find > immediately that 4 is *not* a subset of 5. With this definition of > numbers as sets subsetting does not give what you wish. 4 is the set of all sets with 4 elements. 4 is also every set with 4 elements. In unary representation or by using peanuts or matchsticks 4 is a subset of 5. > > > > > It is circular. But please do not forget that I do not *define* a > > > > natural number! I leave that to Peano etc. I look for its *existence*. > > > > > > So it was not a definition at all. Note also that the Peano axioms to > > > *not* talk about representation. > > > > Therefore we have to investigate whether such a representation exists. > > Note that I use the definition of a number by Peano in order to look > > for the existence of that number. > > What does the latter sentence *mean*? But indeed, for most numbers some > representations do not exist, while other representations do exist. The > existence of a number is independent of the existence of a representation. This point of view has lead to the present mess-math. > For instance, for most rational numbers a decimal representation does not > exist. Correct, for instance for 1/7. > > > > > Of course it is. You must start at some point and eventually you will > > > > come back. Circular reasoning cannot be avoided. But that is not a > > > > problem when existence is concerned. The definition of a number is > > > > given on p. 3 and p 130 of my book. > > > > > > A definition that uses circular reasoning is not a definition. If a write: > > > a foo is a foo > > > > The definition of a number is given as non-circular as possible: p. 3: > > 1) 1 ist eine natürliche Zahl. > > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. > > in N is wrong here; N is not defined. N is already in (1), because (1) is identical to "1 is in N". > > A more proper version is: > 2) Jede natürliche Zahl a had einen bestimmten Nachfolger a', und > dass ist auch ein natürliche Zahl. What is every natural number if N is not defined? > > > 3) Es gibt keine Zahl mit dem Nachfolger 1. > > 4) Aus a' = b' folgt a = b. > > 5) Jede Menge M von natürlichen Zahlen, welche die Zahl 1 und zu jeder > > Zahl a in M auch den Nachfolger a' enthält, enthält alle natürlichen > > Zahlen. > > Yes, I know Peano pretty well. But this is *not* a circular definition. I > wonder why you think it is circular. I did not say it is circular. I said "as non-circular as possible". If it is not circular in your opinion, then be happy. (In fact every definition is circular, because every language is. You cannot explain anything without already using some unexplained wording. If you want to explaine the unexplained by the words already explained, you get circular. That is unavoidable. Cp. N being undefined but appearing in the Peano axioms. But that is not the point here.) > > > p. 130: > > > > 1) 1 in M. > > 2) If n in M then n + 1 in M. > I prefer the successor of n, rather than n + 1 here. I think that everybody able to read and understand these lines will know what "+ 1" means while the successor is not immediately clear. (The successor of n coul be n+2 or 2*n or 10*n or ....) > > 3) IN ist Durchschnitt aller Mengen M, die (1) und (2) erfüllen. > > That is an alternative definition, again not circular. It is better than Peano, because one does not use natural numbers, but is circular though as little as possible. You only need to know "+1" and the intersection of sets. (And what a set is, and what 1 is, and so on.) > > > To investigate the *existence* of number 3 we can use the definition > > of 3. > > And *in this context* the definition of 3 is: > succ(succ(1)) > no set at all. And the existence of 3 follows from the first set of > axioms you gave. The existence does not follow from axioms. Axioms state something in an arbitrary way. They can even state the famous pink elephant. Whether it exists remains to be investigated. But for this sake we need not define again what a pink elephant is. > Your definition is circular, this one is not. I do not give a definition but I look for the existence of the number already defined. Regards, WM |