Cantor Confusion
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From: Newberry on 27 Dec 2006 23:48 David Marcus wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > cbrown(a)cbrownsystems.com schrieb: > > > > > > This bijection between sets (initial segments {1,2,3,...n} and > > > > {2,4,6,...2n}) is only valid for finite sets. > > > > > > Suppose I instead counter-claim that this bijection is obviously "not > > > valid" for finite sets; only for infinite sets. > > > > > > Is that what you call a "correct mathematical argument"? Simply > > > claiming something is valid or not valid? How is it different from your > > > own argument? > > > > It would be as impossible to contradict as your other claims or as the > > following: > > In every finite initial segment of natural numbers, there are as many > > even numbers as odd numbers (plus or minus one number). But in the > > infinite set N there are 250 times more even numbers than odd numbers. > > Are you saying this is true? If so, why/how? > > > > Thus "|N|/|R| < 1" is nonsense; as I previously stated > > > quite clearly. > > > > I did not calculate |N|/|R| but edges per path. This calculation yields > > a real number for any finite path - and the infinite path is nothing > > else but the union of all finite paths. > > And, as you've said many times, anything true for finite is also true > for infinite. Oh, wait. There is no theorem that says that. So, how does > the fact that the infinite path is the union of its finite segments tell > you anything about the edges per infinite path? Because there is a one-to-one correspondence between integeres and nodes of the infinite paths. And it can be proved that An(1 <= (2*2^n - 2)/2^n <= 2) That is for ALL levels in the infinite tree: #edges >= #paths. > > -- > David Marcus
From: William Hughes on 27 Dec 2006 23:51 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > > Please give a counter example for a finite number n. > > > > > > > > > > > > No such counterexample exists. The claim is that despite > > > > > > the fact that given any n we can find a single line, L(n), that works > > > > > > for this n, there is no single line, L_D, that will work for all n. > > > > > > > > > > This claim is obviously false. At least my counter claim cannot be > > > > > disproved by an example.> > > > > > > > > see the example that disproves your counter claim below. > > > > (by mistake you snipped it) > > > > > > You are in error. You gave no example showing that not all natural > > > numbers can be in one line. > > > > > > > > > > > > > > > > > Remember, even if there is not a last line, every n eps N is finite. > > > > > > > > > > > > Which means that for any n we can find a single line L(n) that works > > > > > > for this n. It does not mean that that we can find a single > > > > > > line L_D, that works for all n. > > > > > > > > > > That is impossible, beause there is no "all n". > > > > > > > > "It does not mean that that we can find a single > > > > line L_D, that works for all n." written out in full is > > > > > > > > It does not mean that there exists an L_D with the property that: > > > > if a natural number n can be shown to exist in the diagonal > > > > n must be an element of L_D. > > > > > > If all natural numbers would exist, then, of course, also this L_D > > > would exist. If L_D does not exist, then we have the proof that not all > > > natural numbers do exist. > > > > No. L_D does not exist whether or not we assume that > > all natural numbers exist. > > As not all natural numbers do exist, the set is potentially infinite, > i.e., it is finite. It has a maximum. L_D. Taking this maximum and > adding 1 or takin L_D ^ L_D or so yields another maximum. In any cae > there is a maximum. No. At any time there is a maximum of the numbers that have been shown to exist. However, there is never a maximum of the numbers that it is possible to show exist. L_D does not only have to include numbers that have been shown to exist, it must include all numbers that it is possible to show exist. - William Hughes
From: Franziska Neugebauer on 28 Dec 2006 00:58 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> [...] >> >> >> Linguistic Question: Is it meaningfull to speak of an >> >> >> "approximation" if one denies the existence of the thing which >> >> >> is approximated (the irrational number)? >> >> > >> >> > Not as long as not all deny its existence. >> >> >> >> To me this reads >> >> >> >> As long as all deny the existence of the entity to be >> >> approximated it is not meaningful to speak of an >> >> "approximation". >> >> >> >> Is this correct? >> >> >> >> I do not deny its existence. Do you? What meaning does have an >> >> "approximation" to you if the approximated entity (irrational >> >> number) is supposed to not exist? >> > >> > Ther number does not exist, but the idea does exist. >> >> 1. So you agree that there _exist_ two entities x1, x2 e R which >> solve the equation x^2 = 2? > > Yes. >> >> 2. Is it true that what "we" call an "irrational number" (x1, x2)) is >> _identical_ to your "idea"? > > Yes. >> >> 3. If so, I cannot see what a meaningful, substantial difference >> between your "irrational idea" and the common "irrational number" >> could be. What - besides pure terminology - is that difference? > > The difference is that an idea has no digits. 2-3. You contradict yourself answering my question 3. that way. As you have agreed to in your answer to question 2. common "irrational numbers" are _identical_ to your "ideas". There are two ways out: 1) Withdraw your answer to question 2. 2) Withdraw your answer to question 3. Which do you prefer? F. N. -- xyz
From: Virgil on 28 Dec 2006 02:33 In article <MPG.1ffd12d28da12887989a7a(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Virgil wrote: > > In article <1167256020.615407.109810(a)42g2000cwt.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > > > > Note that in each finite tree, the complete edge assigned to a > > > > > > path is > > > > > > the last edge on the path. As there is no last edge in an > > > > > > infinite > > > > > > path, this does not hold in the infinite tree. > > > > > > > > > > So there are different paths without different edges? > > > > > > > > No. I never did state that. > > > > > > So at least one path has its own edge? > > > > Deliberate obfuscation. > > Do you think it is deliberate? Or, is WM really as confused as he > appears? Both!
From: Virgil on 28 Dec 2006 02:45
In article <1167281308.269041.104540(a)48g2000cwx.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > David Marcus wrote: > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > I did not calculate |N|/|R| but edges per path. This calculation yields > > > a real number for any finite path - and the infinite path is nothing > > > else but the union of all finite paths. If all properties of the finite trees carry over to infinite trees that infinite trees must be finite, as that is one of those properties. So that an argument that something holds for all finite cases can never be sufficient to prove it in any infinite case. Any such proof would need far better arguments than any yet provided. > > > > And, as you've said many times, anything true for finite is also true > > for infinite. Oh, wait. There is no theorem that says that. So, how does > > the fact that the infinite path is the union of its finite segments tell > > you anything about the edges per infinite path? > > Because there is a one-to-one correspondence between integeres and > nodes of the infinite paths. And it can be proved that > An(1 <= (2*2^n - 2)/2^n <= 2) > That is for ALL levels in the infinite tree: #edges >= #paths. For NO "levels of an infinite tree" is it true, but it is true for finite trees. > > > > > -- > > David Marcus |