An uncountable countable set
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From: Daryl McCullough on 26 Jun 2006 11:30 mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: >> >forall f from N to any set, K(f) is not in the image of f . >> >> Yes, that's true. And that implies that "forall f from N >> to P(N), f is not a surjection". > >It is not a problem of surjectivity. But the fact that f is not a surjection from N to P(N) *follows* trivially from the fact that K(f) is not in the image of f. >This set K may be in P(|N), but it very *defintion* is nonsense. There is nothing wrong with the definition of K(f). For any f, there is a corresponding K(f). >Define a bijective mapping from {1, a} on P({1}) = {{}, {1}}. a is not >a natural number. There are two bijections possible. The set K cannot >be mapped by a number although K is in the image of both the possible >mappings. >f: 1 --> {1} and a --> { } with K_f = { }, >g: 1 --> { } and a --> {1} with K_g = {1}. >Here we have certainly no problem with lacking elements. The claim made by Cantor is that for any set A, there is no surjection from A to P(A). Of course, if you construct a new set B by adding extra elements to A, then there might be a surjection from B to P(A). But why is that relevant? In your case, A = {1}, B = {1,a}. So there is a surjection from B to P(A), but there is no surjection from A to P(A). There is also no surjection from B to P(B) Do you disagree? Do you think that there *is* a surjection from some set A to P(A)? Then give an example. Your above example is not a counterexample. -- Daryl McCullough Ithaca, NY
From: Virgil on 26 Jun 2006 13:48 In article <1151318171.298241.313770(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > The list of all reals which can be represented by unending strings of > > digits. > > We cannot write or read an unending string. The assumption, such > strings existed, is wrong. The assumption that for every rational number we can describe precisely the unending string representing it is right. So some endless strings exist, even if only in the imagination. But that is where all mathematics exists! > > > > > > >which can be remedied by switching the mapping from f to g and this > > > > >is possible for every real number constructed. Therefore there is no > > > > >uncountable set of reals. > > > > You are saying that because there is a function on any list of infinite > > strings of digits which will produce a string not in that list, that > > there is another rule that will always produce a string in that list. > > > > I am not at all sure that your claim is true, but I am quite certain > > that it is irrelevant, even if true. > > It is true (include that very string into the list) and it is very > relevant. It is irrelevant, since what IS relevant is that no list of reals can be complete. Every such list omits as many reals as there are subsets on the set of naturals, as I demonstrated in previous posts. > > > Of what relevance is the fact that one can show that certain numbers ARE > > listed in considering whether certain numbers are NOT listed? > > All numbers you will ever produce can be listed. That is enough. In what list? Give me any listing which you claim lists every real, and I will give you a rule producing one nonmember of that list for every subset of N. > > > forall f from N to any set, K(f) is not in the image of f . > > > > That is OUR point! Which you conceded only with great reluctance. > > That is exactly the same point as in Cantors diagonal argument. K(f) is > not in the image of f. <==> The diagonal number is not in the list > (image of f: |N --> |R). They are both valid arguments that certain bijections are impossible,
From: Virgil on 26 Jun 2006 14:05 In article <1151318949.802756.289710(a)r2g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > 0.0 > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > > > > It is D = 0.111... > I do not believe that 0.111... is in the list. But I do not judge at > all. If it were in the list, then Cantor's proof would break down. If > it is not in he list, which contain every sequence of ones, which can > be indexed by natural numbers, then there are digits not defined by > natural numbers, hence undefined. Which digits in 0.111... = sum_{n in N} 10 ^(-n) are not indexed by members of N? > > > > You are making TO's vulgar false assumption that in order for a set of > > naturals to be infinite, some of its members must be infinite. That is > > no more true of sets of naturals than it is for sets of rationals or > > sets of reals. > > Of course it is true, but we need not dicuss it here. What is false is true? that needs immediate discussion if we are to proceed. > Either 0.111 has more ones than any list number, then it is undefined, > or it has not, then Cantor's proof breaks down. There are only these > two alternatives. Perhaps they are the only two that "mueckenh" is capable of considering, but others are more capable than he. 0.111... has more 1's that can be counted by any natural number, and is perfectly well defined as the complete decimal expansion of 1/9. > > > > > > So, whatever is your choice: There is a contradiction: Either 0.111... > > > differs from any list number and, therefore, does not exist at all, or > > > it does not differ from any list number; then the diagonal proof fails. > > > > You left out the only valid case in which 0.111... = 1/9. > > > > Non-terminating decimals are defined as the limits of sequences of > > terminating decimals as the number of digits increases without bound. > > Cantor's proof requires hat the diagonal number is a number which > differs at least at one place which can be identified from each of he > list numbers. There is no room for cloudy epsilons. No one is bothering with cloudy epsilons except "mueckenh". In any listing of reals, the number in position n does not require an endless expansion, merely one accurate through the nth decimal digit, in order to construct a non-member of the list. > Cantor's proof requires hat the diagonal number is a number which > differs at least at one place which can be identified from each of he > list numbers. There is no room for cloudy epsilons. Then blow your nose and get rid of them.
From: Virgil on 26 Jun 2006 14:13 In article <1151319193.126558.262450(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1151225575.137720.193510(a)u72g2000cwu.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > So either "mueckenh" must be declaring the set of rationals uncountable > > > > or he must admit that they can be well-ordered. > > > > > > They are neither well-orderable nor "countable". Because they just do > > > not exist. > > > > They do in my world. > > You may believe it, but it is wrong. The number [pi*10^10^100] does not > exist in any wordl. You just wrote down a complete description of a number which you claim does not exist. While that numbers may not have a known or even knowable decimal expansion, the fact that you have named that one validates its existence. There are lots of numbers whose existence is known and which are basic both to math and to science whose complete decimal expansions are not known and not knowable. That is no bar to their existence The idea that a number cannot exist without having a known or knowable complete decimal expansion is ridiculous.
From: Virgil on 26 Jun 2006 14:16
In article <1151319403.823915.220910(a)c74g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb:> > > > Virgil. Such proofs must be wrong in general, and even the allmighty > > > could change this fact. > > Should read: could not change this fact. > > Regards, WM Your lack of faith in an Almighty to be almighty, is self-contradictory. |