Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Virgil on 29 Dec 2006 16:05 In article <1167394001.555223.74290(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1167256678.283824.142350(a)42g2000cwt.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > Yes. So it is essentially different from a finite path. The same > > > > with > > > > your trees, Each finite tree has a natural number as maximum level > > > > and > > > > has no infinite paths. The completed infinite tree has no maximum > > > > level > > > > and has infinite paths. So any conclusion you can make from the > > > > finite > > > > trees does not necessarily hold for the infinite tree. > > > > > > We conclude only that one can follow a path infinitely, and that always > > > there is an edge which continues it. There is no continuation without > > > an edge. ok? That is unavoidable and it is enough to prove that there > > > are not less edges than paths. > > > > Well, in that case, pray, show a proof. > > Follow a path. Look whether it splits somewhere without needing an > edge. Wm cannot throw the labor of proof on others for his own claims, but must complete such proofs himself, if he wishes to maintain those claims. So far, he has conspicuously failed to do so. As several people have posted proofs that there are "more" paths than edges in infinite binary trees, and WM has not been able to fault any of those proofs, > > > > > > For each two paths you can identify an edge where they separate. But > > > > there > > > > is *no* edge that separates (for instance) the path leading to 1/3 > > > > from > > > > all other paths. And through each and every edge run infinitely many > > > > paths. > > > > > > Yes. That shows that there are no infinite sequences of edges (or > > > digits) which individually represent nunmbers like 1/3 or ideas like > > > irratinal numbers. > > > > Why? > > Because non-distinguishable sets are not different sets. You just > proved that there is no element of a path which distinguished it from > every other path. Then WM is asserting the nonexistence in an infinite binary tree of anything but binary rationals (whose denominators are powers of 2). But 1/3 has binary representation as 1/3 = sum_{n =1..oo} 1/2^(2*n) = 0.01010101... which corresponds to a path in an infinite binary tree. > > > > > > You are extending conclusions valid for finite trees to the infinite > > > > tree > > > > without proof, it is just your intuition. > > > > > > Not at all. Forget about all finite trees. In the infinite tree there > > > is always an edge which continues a path. There is no continuation > > > without an edge. That is unavoidable and it is enough to prove that > > > there are not less edges than paths. > > > > In that case: prove it. As I have already shown, all edges can be made > > to represent the rational numbers where the denominator is a power of 2. > > That is because all edges are at a finite distance from the root. > > What distinguishes this union of rational trees from the complete tree? The existence of paths in the complete tree which are not possible in the union. For example the path which branches left if and only if the branch level is a perfect square. > Nothing. There is no node and no edge of the complete tree which is not > in the union of all rational trees. The path which branches left if and only if the branch level is a perfect square is not in any of WM's finite trees, so cannot be in the union. > > But if we subtract the rational paths from the complete tree, Then one has uncountably many paths left.
From: mueckenh on 29 Dec 2006 16:49 Franziska Neugebauer schrieb: > 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? What is your problem? Common irrational numbers are identical with my "ideas". You don't know or avoid to confess that the number of digits of their rational approximations is limited? But that doesn't change that fact. Regards, WM
From: mueckenh on 29 Dec 2006 16:51 William Hughes schrieb: > > 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. So it is. > However, there is never a maximum of the numbers > that it is possible to show exist. It is possible to show: When these numbers will exist, the due maximum will exist. That's enough. > 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. Why? How should we know which can possibly exist? Regards, WM
From: cbrown on 29 Dec 2006 16:52 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... So you agree that that /is/ what you call a "correct mathematical argument" - simply claiming something is valid or not valid. Of course it will be difficult to refute statements made with this justification. In the words of South Park's Towelie: "No man, /you're/ a towel!" > 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. If you specify what you mean by "250 times more", then your statement will have meaning; until then, it is nonsense. > > > > > > Thus "|N|/|R| < 1" is nonsense; as I previously stated > > quite clearly. > > I did not calculate |N|/|R| but edges per path. Wonderful. But what relevance does the calculation of "edges per path" have to do with the /actual question/: does there exist a surjective function from the set N to the set R? That is what is meant by "|N| >= |R|". The use of ">=" here has a /different meaning/ than "1 + 1/2 + ... >= 1", a difference which seems to elude you. In the latter case, we are not asking "is there a surjective function from 1 + 1/2 + ... onto 1?", because that would be simply nonsense. Similarly, to try to show that |N| >= |R| by calculating the limit of edges per path is equally nonsense - it requires misreading the meaning of ">=". > 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 therefore... the union of these calculations is a countable set of real numbers? And that shows a surjection from N to R how? > > > > > The number of edges accumulated by one path up to level n is a function > > > like that above: > > > f(n) = 2 - 1/2^n. Very easy. > > > > Yes; when you write it that way, it "looks" just like a valid statement > > about the real numbers. So, you "feel" that it is a both a sensible and > > a valid statement about things which are not real numbers; when instead > > it is nonsense. > > What is your definition of an infinite path? There is a countable set of edges. For each natural number n, there is a distinct edge which is the nth edge in the path p. If an edge e is the nth edge in the path p, then there is no other natural number m such that e is also the mth edge in the path p. An edge e is part of a path p iff there exists a natural number n such that e is the nth edge of p. More concisely: |edges| = |N|. A path p : N -> edges is an injection. An edge e is part of path p iff e is in the image of p. > You feel that it is not > the union of all finite paths... Where did I state that? Trivially, each edge in a path is a finite set with one element (the edge). The union of all singleton sets of edges is of course the same set as the set of all edges (i.e., the infinite path). The set of all singletons edges is a subset of the set of all finite paths; thus the union of all finite subsets of a path is the infinite path. Thus it's not a matter of what I "feel"; I just /proved/ that the infinite path is the union of all finite subsets of that path. But what is your point? An infinite path is also the union of all infinite proper subsets of that path. So what? > but something much much greater and > longer, more exciting and most mysterious? There really is nothing at all mysterious about it. It simply follows from the definitions of N and "an infinite path". What is mysterious is why you continue to claim to find a surjection from N to R. > Do you have a definition how > or any reason why the infinite path is not the union of all finite > paths? > No. Have you given up on your "rational relation" proof that |N| = |R|? Cheers - Chas
From: mueckenh on 29 Dec 2006 16:54
Newberry schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Dik T. Winter schrieb: > > > > Therefore: If you follow some path you will see that whenever it > > > > separates itself from another one, this happens by two edges .- one > > > > edge for the path, and the other edge for the other path. This process > > > > repeats and repeats without end. Nothing else happens. From the > > > > unavoidable and "inseparable" connection of separation and appearance > > > > of another edge we can conclude that every path which can be > > > > distinguished from another path runs through an edge which does not > > > > belong to the other path; call it "personalization of an edge. > > > > Therefore distinguishability of paths and personalization of edges are > > > > unavoidably connected. > > > > > > For each two paths you can identify an edge where they separate. But there > > > is *no* edge that separates (for instance) the path leading to 1/3 from > > > all other paths. And through each and every edge run infinitely many > > > paths. > > > > Yes. That shows that there are no infinite sequences of edges (or > > digits) which individually represent nunmbers like 1/3 or ideas like > > irratinal numbers. > > But there are algorithms that tell at EACH level whether to go left or > right. And there are paths of other numbers which, up to EACH level, behave exactly as the path of the binary representation of 1/3 does. Further, the algorithm for 0.010101... works only as long as you can determine whether the number n of the level is even or odd. For certain natural numbers like the first 10^100 digits of pi this is impossible. Regrads, WM |