An uncountable countable set
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From: Virgil on 21 Aug 2006 00:23 In article <ecb7ae$9qo$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > If, instead of talking about limits, we apply some unit infinity n, then > we can maintain that 1/n>1/2n, one being half the infinitesimal value of > the other. We might get in trouble, but I see none ahead, and no one's > convinced me there's any down this stream. No one can convince TO of anything that does not meet the peculiar criteria of his "intuition", how well proven it may be ever or obvious to everyone else. his attitude is always " if I don't like it, it isn't true." Such obstinacy would only be justified by a level of genius that TO is all too obviously lacking.
From: imaginatorium on 21 Aug 2006 01:30 Tony Orlow wrote: > Dik T. Winter wrote: <snip> > > Division by 0 can not be handled in a ring. > > According to MathWorld, indeed, there is an exception for 0 in the > optional Multiplicative Inverse condition. Only required due to the > aversion to infinitesimals. Alas! Why cannot we tie this all together? > > http://mathworld.wolfram.com/Ring.html Thanks for that link - it includes a motivation for the name 'ring'. As for "tying together", well if you take the axioms of a ring (even I think the absolute minimal set of axioms: group under addition, semigroup under multiplication, distributive law), you have (for any x, y, in the ring): x . 0 = x . ( y - y ) = x . y - x . y = 0 Note that we only rely on the *additive* properties of inverses, plus the distributive law to get this. Therefore there cannot be a nonzero p such that x . 0 = p, which is what we would need to have a multiplicative inverse of zero. Of course, you can append an object to a ring and call it Bigun (or anything else) and investigate the resulting structure (see javascript and my lens calculators for a practical example), but this structure will not be a ring. You may huff and puff at this if you like. Brian Chandler http://imaginatorium.org
From: Virgil on 21 Aug 2006 04:01 In article <ecb927$ce6$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Proofs depend on starting assumptions, rules > and facts. Since TO steadfastly refuses to lay his out, he can never prove anything.
From: mueckenh on 21 Aug 2006 04:22 Dik T. Winter schrieb: > In article <1155998763.083248.291940(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > Where can I find that quote? I have problems with reading utf-8. > > > > Works, p. 278, first paragraph (his famous article on the diagonal > > proof). > > I will look it up when I am back at work. > > > > Well, let me be clear. Constructable numbers is used in two senses in > > > mathematics. The sense I meant is "computable numbers". So let me > > > refrase with this word: > > > The set of computable numbers is countable, but the diagonal is > > > not computable. > > > > No diagonal is computable, except in simple cases like my list, because > > no infinite list is computable, again with the exception of closed > > formulas and simple cases like my list. But that is completely > > irrelevant. > > You apparently do not know what the word "computable" does mean. To > be precise: all algebraic numbers are computable in the mathematical > sense. To be precise: Each algebraic number is computable. But not all. Because then the list of all algebraic numbers was computable. Not even an infinite list of numbers which are independent of each other is computable. (Independent means: the list cannot be computed automatically by one or several given formulas). Regards, WM
From: mueckenh on 21 Aug 2006 04:26
Dik T. Winter schrieb: > In article <1155999150.167954.72660(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > > I do not only *claim* that 0.111... can be indexed by the natural numbers; > > > > > I can *prove* it, as I have done again and again. > > > > > > > > You supposed that all digits of 0.111... could be indexed, and than > > > > you proved that they can be indexed. > > > > > > No. I *defined* 0.111... such that all digits could be indexed. > > > > I *defined* that all digit positions which can be indexed belong to > > numbers of my list. The indexes are natural numbers and all natural > > numbers are in my list. > > > > And I showed that your number is not in my list. > > > > Hence one of our definitions must be wrong. > > No. *Both* definitions are correct, there is no contradiction between > them. It is your later assertion that is incorrect: > "If all digit positions can be indexed, the number is in the list". > It does not follow from your definition. You are dreaming. It does follow from the form of the numbers of my list. Try to come back to reality. Simply show how a digit position n can be indexed the complete number of which, i.e., the complete sequence of digit positions 1 to n is not in the list. > > > Either there are not all natural numbers in my list and nowhere else, > > or your number cannot be indexed completely. > > Nothing of that. Give me an index position in my number that is not in > the list. (By definition, there are none.) But all natural numbers > are in your list, as are all index positions of my number. But the > number itself is not in the list, because it is not a natural number. In my list there are all numbers which can index the position n+1 if they can index the position n. But they cannot index a position larger than all naturals. All positions which are not larger than all naturals are in my list. You assert that a set of small numbers can surpass a large number if there are enough small numbers. That is wrong. 0.111...1 < 0.111... such that any attempt to index digit positions of 0.111... leaves unindexed positions. It is the same with the staircase: If the total height is H, then there must be at least one stair of height H. If the total height is omega and omega is a number larger than any natural number, then there must be at least one stair of height omega. > All digit positions are indeed in the list, and it is itself not in the > list. Why must it be in the list itself? Consider the digit positions > as bricks. I state the number I have is built of bricks, but I do not > state that the number I have is a brick itelf. That is what *you* are > claiming. You forget that the numbers of my list leave no other outcome. > > > > > > Because the sum of infinitely many differences of 1 would make up an > > > > infinite number. > > > > > > That is not a proof. Indeed, the sum of infinitely many differences of > > > 1 make up an infinite number, > > > > are you sure? Indeed, even in set theory a sum of infinitely many 1's > > is infinite? > > I would think so, yes. We have infinitely many differences between natural numbers. Hence they sum up to a non-natural number. Contradiction. > Regards, WM |