Cantor Confusion
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From: Virgil on 30 Dec 2006 18:21 In article <1167514029.295041.300970(a)k21g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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? > > > > L_D must contain the diagonal. > > Thus, L_D must contain any element that > > can be shown to exist in the diagonal. > > It does, after the existence has been shown. In ZFC and NBG, that existence is implicit in the axioms.
From: Virgil on 30 Dec 2006 18:37 In article <1167514245.044650.160580(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > There is no such thing as a single "separated path" > > > > > > > > > > But the cardinal number of all these not being no-things is 2^aleph_0? > > > > > > > > What nothings? There are paths, but in an infinite tree, no path can be > > > > separated from all other paths > > > > > > Every (existing) path is separated from all other paths but no path > > > can be > > > separated from all other paths. > > > > > > > by any one node or edge. > > > > > > Not by one node or edge? By what else? > > > > To separate one path from all others requires an infinite set of edges > > or an infinite set of nodes. No finite suffices, as, for any finite set > > of edges and/or nodes, if there are any paths through all of them then > > there are uncountably many paths through all of them. > > You said: > It is enough to show that for every finite tree, the given path is not > in it, in order to show that it is not in the union. > > I say: > It is enough to show that for every finite set of edges, the given path > is not > individualized, in order to show that it is not individualized in the > infinite union. I have no ideal what WM's "individualizing" of a path means. But in any finite tree every path has a unique edge and a unique node, its last of each, neither of which is in any other path, so that edge and node are sufficient to "individualize" that path as separate from every other path in that tree. > > You said: > It is enough to show that for every finite tree, the given path is not > in it, in order to show that it is not in the union. Since that exactly conforms to the definition of unions, yes. An object is IN a union if and only if it is in at least one of the sets of that union, and thus is not in the union unless it is in at least one of the sets in the union.. > > I say: > Is the union of all finite trees different from the complete infinite > tree? Yes! > If yes: Are the irrational numbers a subset of the rational numbers? No! > > I will but WM keeps going off into his matheology despite his promises. > > Is the union of all finite trees different from the complete infinite > tree? Yes! > If no: What is the difference in terms of nodes and edges? > > You said: > I see no reason to do either. WM persists in his false claim that a > single node or edge is sufficient to separate one infinite path from > all > others. The truth is that it takes infinitely many edges or nodes to > achieve such separations. > > I say: > Infnitely many edges which are shared by two paths or more are not > sufficient and are of no value with regards to identify a path. It is not what is shared by two sets of paths which separate them but what is not shared. To separate one path from all others requires an infinite set of its included edges and/or nodes. For any such infinite set, any other path will excude infinitely many of those. This can be seen by noting that any two infinite paths can have no more than a finite number of nodes and/or edges in common. So each contains infinitely many not in the other.
From: Virgil on 30 Dec 2006 18:40 In article <1167514395.524777.101510(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > 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? > > > > Read 2-3. If you don't understand explain what you did not understand. > > > > > Common irrational numbers are identical with my "ideas". > > > > If your "ideas" are _identical_ with common irrational numbers how can > > there be "a difference"? > > > > 1) Either your ideas are _identical_ with irrational numbers then there > > is no difference. > > No difference to the common irratonal numbers. What is an uncommon irrational number? > > > 2) Or they are different. > > Different from what you believe is behind these "numbers". > > > > You have to decide which version of "ideas" shall be valid. Without that > > decision you contradict yourself. > > Common irrational numbers are identical with my "ideas" But the common > interpretation of common irrational numbers is false, in particular the > assertion of unlimited facility of approimation. The "facility" to approximate any square root of a rational to any desired degree of accuracy is inherent.
From: Virgil on 30 Dec 2006 18:43 In article <1167514949.394560.91780(a)n51g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Do you think that the union of all finite subtrees is identical with > the complete infinite tree? NO! > If not, what is the difference (in terms of edges or nodes)? Any path in the union must contain only finitely many edges or nodes, but in a compete infinite tree, all paths contain infinitely many of both.
From: Franziska Neugebauer on 30 Dec 2006 18:54
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > Franziska Neugebauer schrieb: [...] >> >> >> 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? >> >> Read 2-3. If you don't understand explain what you did not >> understand. >> >> > Common irrational numbers are identical with my "ideas". >> >> If your "ideas" are _identical_ with common irrational numbers how >> can there be "a difference"? >> >> 1) Either your ideas are _identical_ with irrational numbers then >> there is no difference. > > No difference to the common irratonal numbers. > >> 2) Or they are different. > > Different from what you believe is behind these "numbers". The contradiction of your answers remains in effect because what "I believe" is not part of your contradiction. Do you want to withdraw your answer to question 3.? >> You have to decide which version of "ideas" shall be valid. Without >> that decision you contradict yourself. > > Common irrational numbers are identical with my "ideas" Repetition of acknowledgement of your answer to question 2. > But the common interpretation of common irrational numbers is false, > in particular the assertion of unlimited facility of approimation. The contradiction of your answers remains in effect because "common interpretations" are not part of your contradiction. Do you want to withdraw your answer to question 3.? F. N. -- xyz |