Cantor Confusion
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From: Virgil on 2 Feb 2007 15:01 In article <1170400473.318379.151540(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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: > > > > > 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. Not so. Induction can only prove every set of naturals which is bounded above by a natural is finite, but that does not, according to the axiom of infinity, exhaust the possible sets of naturals. > > > > 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. Then lets see this alleged proof. What Induction says is precisely: Given a set, S, of natural numbers such that (a) S contains the first (smallest) natural, and (b) whenever a particular natural is a member of S, the successor of that natural is also a member of S, then every natural is a member of S. And induction says nothing else. > Of course every set of natural numbers is finite, as proved > by induction over all initial segments. That only proves that such initial segments as have a last element are finite. It says nothing at all about any initial segment which does not have a last element, and there is one.
From: Virgil on 2 Feb 2007 15:12 In article <1iAKSznLBxwFFwaz(a)phoenixsystems.demon.co.uk>, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > 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). Where does one find a "crack" between points not already filled by a point? The least upper bound property says that if any non-empty set of reals is bounded above, then there is a particular real number as its least upper bound filling any "crack" between that set and the set of reals greater than all of the members of that set. Similarly for sets of reals bounded below. So where is there room for any "crack" between points, which is not filled in by a least upper bound or greatest lower bound?
From: Virgil on 2 Feb 2007 15:20 In article <1170412604.844757.297400(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Alas for WM, it does have something to do with cardinalities of sets. It proves that given any set, S, and its power set P(S), the set of all subsets of S, then, according to Cantor's definition of order for cardinalities, Card(S) < Card(P(S)). That WM doesn't like it is irrelevant and illogical.
From: Andy Smith on 2 Feb 2007 15:29 Andy Smith <Andy(a)phoenixsystems.co.uk> writes (snip everything else) At root I think my problem comes down to achieving a suitably Zen-like perspective on the following apparently incompatible statements: 1) The real line is made up of an ordered and infinite set of points, and is connected. 2) No point on the real line has an adjacent point. regards -- Andy Smith
From: Virgil on 2 Feb 2007 15:37
In article <1170413330.341310.269660(a)p10g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 2 Feb., 02:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1170348341.624257.130...(a)p10g2000cwp.googlegroups.com> > > 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. The set of all sets with exactly 4 elements is not a subset of the set of all sets with exactly 5 elements since the set of all sets with exactly 5 elements does not contain any sets with less that five elemeNts. WM must have completely lost touch with reality to think otherwise. > > > > 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. The "mess" is only in the minds of those who imagine it into existence. For those of us who perceive no mess there is no mess. > > > For instance, for most rational numbers a decimal representation does not > > exist. > > Correct, for instance for 1/7. Wrong! That is one for which a decimal representation does exist. > > 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. When one accepts some words as primitives and undefined, definitions based only on them are not circular at all. > 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 ....) Successor is a primitive, but the Peano statements require it to have certain properties. > > > > 3) IN ist Durchschnitt aller Mengen M, die (1) und (2) erf�llen. > > > > 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.) In Peano, you do not have to know what any of the things are, all you have to know is that they work according to the rules stated. > > 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. But without a definition there can be no testing of existence. How can you tell if a "floop" exists if you can't even tell whether something is or is not a "floop"? If you don't know at all what you're looking for, you can't ever tell whether you've found it or not. |