An uncountable countable set
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From: Virgil on 25 Jun 2006 14:11 In article <1151224516.615769.116980(a)u72g2000cwu.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > mueckenh(a)rz.fh-augsburg.de says... > > > > >Daryl McCullough schrieb: > > >> Yes. The reasoning is very similar: > > >> > > >> If f is any function from N to P(N), then > > >> K(f) is an element of P(N) that is not in > > >> the image of f. Therefore, f is not a surjection. > > >> > > >> If L is any enumeration of reals (that is, L is a > > >> function from N to R), then D(L) is an element of > > >> R that is not in the enumeration L. Therefore, L > > >> is not complete. > > > > > >On the other hand we know that the set of individualizable real numbers > > >is countable. > > > > "individualizable" is not a very mathematically precise notion, so > > I don't think you can prove anything about it. > > Any countable alphabet has only countably many finite words (cp. the > set of finite subsets of |N). Any alphabet that we can use is countable > (in fact it is finite). Hence we have only countably many symbols, > names, notations. We can distinguish only countably many numbers. > > Further the set of elementary cells of the whole universe is finite. Do you have some personal line from God's mouth to your ear which allows you to "know" things that no one else knows? We may strongly suspect that the universe is finite, but we do not know it. > But mathematicians in general do not like to be frustrated by physical > needs. We get hungry and thirsty like anyone else. > > > >> It shows that every list of reals, there is at least one > > >> real that is not on the list. From that, it follows that > > >> there is no list containing all real numbers. > > > > > >No. > > > > Yes. > > The list of all names contains all we can denote as a number. The list of all reals which can be represented by unending strings of digits. > > >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. > > > > I don't know what you are talking about. Look at some examples: > > > > Here's an example: f(x) = x/2. That's a function from N to R. > > It generates the decimal expansions > > > > 0.00 > > 0.50 > > 1.00 > > 1.50 > > 2.00 > > ... > > > > The diagonal procedure produces a new real by adding one to > > each diagonal digit: > > > > D(f) = 1.61111111.... > > > > D(f) is definitely *not* in the image of f. > > But it is definitely in the image of a suitable g. Of what relevance is the fact that one can show that certain numbers ARE listed in considering whether certain numbers are NOT listed? > This can be done > with all diagonal numbers D ever constructed. Hence all diagonal > numbers ever constructed are countable. There are rules which will produce for any list of reals a distinct real not in the list for every subset of the naturals , P(N). For example: given a list of reals represented by their decimal expansions. construct a new real between 0 and 1 by Rule 1: If the nth digit following the decimal point of the nth number in the list is 3 the new number has an nth digit of 4, other wise it has an nth digit of 3 Rule 2: If the nth digit following the decimal point of the nth number in the list is 6 the new number has an nth digit of 7, other wise it has an nth digit of 6 Now for every subset, S, of N (member of P(N)), the rule is to use rule 1 above if n is a member of S and rule 2 if N is not a member of S. It should be clear that each subset of S produces a non-member of a given list and that futhermore different subset produce different non-members. Thus for any list of reals, we have shown that there is 1 non-member of that list for each member of P(N). Thus removing any list of reals from the set of all reals leaves as many reals as before. > > So you think that there is some function f such that *every* > > real is in the image of f? Then show it to me. > > I'll show it to you after you have constructed all numbers you wish to > be contained in my mapping. Wrong is only your assumption that there > are more than can be constructed. Nothing in the standard definition of the set of reals numbers requires constructability. If you want to do only constructionist mathematics, you are in the wrong place. > > There is no such thing like "all reals". But if the diagonal proof > fails in one case (see my next posting), it cannot be trusted in > others. > > > >There is no mapping including K as the image of at least one source > > >element, even if surjectivity is not at all in question! > > > > I can't tell exactly what that sentence means, but I think you > > are just saying > > > > there is no mapping f that includes K(f) in its image. > > > > which is equivalent to the more straight-forward statement > > > > forall f from N to P(N), K(f) is not in the image of f > > > > How are you saying anything different from that? But the > > latter statement immediately implies the statement > > "forall f from N to P(N), f is not surjective". > > 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.
From: Virgil on 25 Jun 2006 14:32 In article <1151225177.422298.193350(a)r2g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > mueckenh(a)rz.fh-augsburg.de says... > > > > >Such a function exists from |N --> |R for all real numbers which can be > > >individualized, i.e. which are really real numbers. > > > > If such a function exists, then demonstrate it. Give us a function > > f from N to R such that every real number that is "individualized" > > is in the image of f. > > Cantor's general proof breaks down by one counter example like this > one: Construct the diagonal number of the following list by digit > replacement 0 --> 1 (only the digits behind the point are to be > considered): > > 0.0 > 0.1 > 0.11 > 0.111 > ... > > It is D = 0.111... > > Do you really believe that actually infinitely many ones (1) can be > contained in 0,111... without exactly as many appearing in the list > numbers? Yes > *All* sequences of ones are present in the aleph_0 numbers of > the list. (aleph_0 - 1 = aleph_0) Wrong, as usual. Only finite sequences of 1's are included in your list. You are making the common error of assuming that the limit of a sequence is necessarily a member of the sequence. > > On the other hand, the limit 1/9 = 0.111... is not contained in the > monotonic sequence of the list numbers. Therefore you may insist that > 0.111... differs from each number of the list. But then the > representation 0.111... of 1/9 is not well defined. Why? The list > > 0.1 > 0.11 > 0.111 > ... > > contains all digits in places which can be indexed and identified by > natural numbers. The third digit behind the point, for instance, can be > indexed by 3 or, equivalently, by 0.111. The seventh digit by 0,1111111 > and so on. Without the possibility of being indexed, a place cannot be > identified and does not exist. So, if you insist that 0.111... has more > digits than any one of the list numbers, then there must be some digits > (at least one) at places which cannot be indexed and hence have no > well-defined positions. 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. > > 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. One can, at some length, show that the delta-epsilon definition validates that 1/10 + 1/100 + 1/1000 + ... is the sum of a geometric sequence with first term a = 1/10 and common ratio r = 1/10 , and that because -1 < r < 1, the infinite series has a finite sum of a/(1-r) = (1/10)/(1-1/10) = 1/9. > > And don't talk about quantifiers. Either D exists actually with all its > digits and the list numbers do exist completely too, then my reasoning > holds, or they do not, then Cantor's proof fails anyhow. Your reasoning does not hold when one regards non-terminating decimals as infinite series.
From: Virgil on 25 Jun 2006 14:38 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. > I don't want to start a new discussion here. But in case you are > interested: Elements of a set must be distinguished from one another. > In the universe there are less than 10^100 bits which can be use to > distinguish two elements. That assumes that we live in a finite universe, which may be suspected but has not been proved. In any case, it does not disrupt the purely mathematical theory of infinite series.
From: Virgil on 25 Jun 2006 14:44 In article <1151226100.639717.220500(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > If an almighty God cold not arrange to have the sun appear to stand > > still in the sky without disastrous side effects, such a God would > > hardly be almighty. > > Of course not. He would not even be able to construct a stone so heavy > that he was unable to lift it. > > If such a God first says: It is not good for the man to stay alone, so > we will create a women for him. But then curses this very man for > having obeyed his wife, this God is hardly knowing everything, not even > much about the psychology of women. > > Cantor stated that infinit is in God, in mathematics, and in nature > including physics, chemistry, ..., even sociology. He was wrong in all > three points. Physicists have already noticed that. How long will t > last until mathematicians get to know it? Mathematicians are no more subject to the whims and facies of physicists that physicists are to the whims of mathematics. Physicists may be constrained to have their theories reflect some aspects of what they view as external reality, but mathematicians are only constrained by the limits of their imaginations. Physicists continually want to subjugate mathematics to only their own needs and desires, and mathematicians continually refuse that sort of slavery.
From: Virgil on 25 Jun 2006 15:22
In article <1151226405.178213.286780(a)r2g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1151153501.336865.131980(a)m73g2000cwd.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > In order to be unambiguous transpositions must be carried out > > > > sequentially, meaning that only countably many can be applied. > > > > > > Not more is required. > > > > > > > > Mueckenh's analysis is not unambiguously a sequence of transpositions. > > > > > > > > Also he does not show that any single transposition can re-order a set > > > > having no first member into one which does have a first member. > > > > > > Of course that is impossible. It is just showing a contradiction in set > > > theory. > > > > In "mueckenh"'s set theory , perhaps, but not in anyone else's. > > > > Until "mueckenh" can show that a SINGLE transposition can switch an > > ordered set with a first element with one not having a first element, > > his alleged proof fails. > > Until it can be shown that Cantor's diagonal proof does reach the last > digit of the last list number, his proof fails. Cator constructs a rule which clearly shows how to find real numbers not on any list. "Mueckenh" cannot show that this rule does not do what it is claimed to do. So Cantor wins and "Mueckenh" > > > > > > > > Similarly in the opposite direction, no single transposition will cause > > > > a set with a first element to become one without a first element. > > > > > > Of course that is impossible. It is just showing that there is nothing > > > like an actually infinite set. > > > > Wrong! It is just showing that "mueckenh"'s "proof" is invalid. > > It is sowing that it does not deliver the result preferred by you, Only in your own opinion. > Virgil. Such proofs must be wrong in general, and even the allmighty > could change this fact. It takes a lot of chutzpah to speak for the almighty. |