Cantor Confusion
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From: mueckenh on 27 Dec 2006 16:50 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. And the numerical approximation of an idea like sqrt(2), written in a list, has not enough digits to determine whether it is different from infinitely many other numbers or numerical approximations of ideas contained in that list. > > > One can approximate an idea by a number. > > I do not see that one can in general numerically approximate an idea. Certainly not in general, but in some cases. Regards, WM
From: mueckenh on 27 Dec 2006 16:54 Virgil schrieb: > As every "personalization" separates a set of infinitely many paths > which contain that edge from a set of infinitely many which do not, such > an edge is not "personal" to a single edge, but only to a tree which is > , except for a finite initial sub-path, tree-isomorphic the the entire > tree. If the path representing an irrational number does not exist as an individuum in an infinite tree, then there is no individual existence of an irrational number. That's what I have been preaching for some years meanwhile. > > > > To assert that there are more paths separated by edges than are edges > > existing at all, is an error which cannot be explained other than by > > the desastrous and corrupting influence of the "logic" of set teory. > > ZFC or NBG set theory might corrupt the peculiar faiths of those who > believe as WM does, but is perfectly consistent with standard logic. You see that your standard logic enforces the opinion that an existing individual entity does not exist as an individual entity. If you think that is really consistent, then your thinking may be consistent but useless. ===================== >The same " limit" argument will conclude the in a tree in which no path > has a terminal node, every path has a terminal node. No, there are meaningful limits and meaningless limits. In particular in case of actuyl infinity-nonsense, there are many meaningless notions. > If either limit argument is valid they both are, and if either is > invalid they both are. And your comparison of both different limts is false in any case. ========================= > One can give an exact continued fraction of sqrt(2). Who says that the > only way to "represent" a number is in terms of some base like decimal? It is the only way to put it in trichotomy with all other numbers and, above all, an n-ary representation is required to apply Cantor's diagonal argument. =========================== > 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? ============================= > These things are all easily seen if one notes that any infinite path is > an infinite sequence of left/right branchings. But no infinite path exists individually? > For any natural n, cutting off the first n branchings of any (infinite) > path leaves an (infinite) path, and for fixed n, the set of all such > truncated paths is identical to the original set of all paths. And that is so, although no original path does exist. 2^aleph_0 = 0? Regards, WM
From: Virgil on 27 Dec 2006 16:55 In article <1167255246.090261.62550(a)h40g2000cwb.googlegroups.com>, 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. But what happens for every finite path need not be true for an infinite path. For example, every finite path has a root node at one end and a leaf node at the other, but this is not true for the union of all of them. > What is your definition of an infinite path? You feel that it is not > the union of all finite paths but something much much greater and > longer, more exciting and most mysterious? Do you have a definition how > or any reason why the infinite path is not the union of all finite > paths? One can, indeed, view an infinite path as the union of a nested sequence of finite paths, just as one views the first limit ordinal as the union of all finite ordinals, but one must be careful about presuming that properties of the finite sets carry over to the infinite union so obtained, as not all properties carry over.
From: mueckenh on 27 Dec 2006 16:57 Dik T. Winter schrieb: > In article <1167082643.909029.231380(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1166921237.502878.48560(a)h40g2000cwb.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > > > > Dik T. Winter wrote: > > > > > In article <1166845904.426550.122020(a)48g2000cwx.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > > > > > > What is level oo? The levels are indexed by natural numbers. > > > > > > The completed infinite tree? As long as you are talking about finite > > > trees, you are welcome. But in that case 1/3 is not a path in the tree. > > > > What is an infinite path? > > > > Do you think it has some fairy tale character? Some very special > > properties, different from any thing we can observe in the universe? > > > > A path is an infinite path if you can follow it without ever reaching > > an end - and that's all. > > 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. > > > 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. > > > To assert that there are more paths separated by edges than are edges > > existing at all, is an error which cannot be explained other than by > > the desastrous and corrupting influence of the "logic" of set teory. > > 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. Regards, WM
From: mueckenh on 27 Dec 2006 16:58
stephen(a)nomail.com schrieb: > In what system can you prove that #edges = 2 * #paths? In the infinite binary tree! (using proven mathematics and the knowledge that infinity is nothing but the endless continuation of he finite). Regards, WM |