Cantor Confusion
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From: mueckenh on 27 Dec 2006 17:02 Dik T. Winter schrieb: > > What is an infinite path? > > Why do you not answer my question? (Note: rood -> root.) The second left branch is assigned to half of all paths. If the number of all paths is a meaningful notion, then half of it is a meaningful notion too. > The point is that you think that also in the infinite tree you can assign > shares of edges as 1 + 1/2 + 1/4 + ... to a path. But that is false. > There is *no* edge that is shared by finitely many paths. So there is no irational number? ================================ > In the finite case all paths end at a terminal node. So when you compare > number of edges to number of paths at finite levels you are comparing > edges to paths with terminal nodes. But even without terminal nodes, the paths consist of edges. Where no edges are, there is no path. Even "in the infinite" that is true. The number of paths cannot "overtake" the number of edges "in the infinite". ================================= > Note that you conclude things about the infinite here based on the finite. We cannot conclude anything other than based on the finite. Regards, WM
From: Virgil on 27 Dec 2006 17:12 In article <1167256020.615407.109810(a)42g2000cwt.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > You should know that you can *not* reverse absolutely converging series. > > Proof? A series, at least in standard mathematics, is really a sequence of the finite partial sums of terms in another sequence. And one cannot take the terms of a sequence in reverse order as there cannot be a first term for such a reversed order. > > > And that is what you are doing. Because that would result in a series > > without start. > > No reordering of an absolutely converging series changes its sum. No permutation of terms does, but the result must still be a sequence of terms whose partial sums are to be a new but still convergent sequence. > > > > > =============================== > > > > 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. No path has its own edge in an infinite tree, as through each edge pass uncountably many paths. It is only in finite trees that each path can have its own private edge, the one ending in that path's own terminal node.
From: Virgil on 27 Dec 2006 17:17 In article <1167256258.863156.191060(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > 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. Maybe WM's "idea of sqrt(2)" contains no digits, but his ideas are notoriously inept. My "idea of sqrt(2)" can be carried out to enough digits to distinguish it from any given rational, which is all that is needed.
From: Virgil on 27 Dec 2006 17:40 In article <1167256495.807852.223380(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. But every irrational DOES have a path. The problem is that some rationals have two paths. > > > > > > 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. In standard infinite trees, an edge only separates the uncountable set of paths through it from the uncountable set of paths not through it. As WM has not explained how a path in an infinite tree can be "separated by an edge", no logic is needed to reject his thesis. And any corruption involved here is strictly within WM's version of logic. > > > > 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. Which "individual entity" is that? > >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. All the limits involved here are equally meaningless. In particular > > > 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. My comparison of the equivalence of their validity is spot on. > > ========================= > > > 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 False. One can do it for arbitrary rationals without any non-natural representation at all. > > =========================== > > > 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 by any one node or edge. > > ============================= > > > 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? If that is how you choose to misunderstand things! > > > 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. Who says no original path exists? They all exist but are not "separated" by the means you claim.
From: Virgil on 27 Dec 2006 17:58
In article <1167256678.283824.142350(a)42g2000cwt.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1167082643.909029.231380(a)f1g2000cwa.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > 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. It is not enough or WM would have proceeded to do the proof. What Wm continually ignores is that in an infinite tree, through each edge pass infinitely many paths, indeed, uncountably many, so that each new edge produces a new infinity of 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. It is a delusion that there are only two paths, unless those two edges are both terminal edges. If neither edge is a terminal edge, then there are at least 4 paths, 2 through each, for longer paths more edges, and for endless paths uncountably many edges. > > > 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. In an infinite tree one is merely "personalizing" an whole uncountably infinite set of paths, not a single path. > > > Therefore distinguishability of paths and personalization of edges are > > > unavoidably connected. One can, at best. pair up each edge with the uncountably infinite set of paths passing through it, but thus does not support WM's delusions. > > > > For each two paths you can identify an edge where they separate. But there is an infiniteness of paths through the edge and another not through that edge. It is not just the two edges splitting from some node, it is all the paths that continue on through further nodes and edges that must be accounted for, but which WM ignores. > > 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. False. For every real in [0,1], there is a path (and sometimes more than one), and for every two reals there is a node at which their paths ultimately separate. > > > > > 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. Except that WM is incapable of producing that alleged proof, while others are quite capable of providing proofs that his claim is false, at least in ZFC and NBG. |