Cantor Confusion
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From: William Hughes on 29 Dec 2006 19:02 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. - William Hughes
From: Virgil on 29 Dec 2006 19:12 In article <1167429748.336224.51060(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > 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? > > > That one which distinguishes a path from all its co-paths. What sort of an "entity" is that? It cannot be a mere node or an edge, since in an infinite tree any node (other than the root) or edge can only separate the uncountably many paths which contain it from the uncountably many which don't. > > > > > > > >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. > > The limit n --> oo covering all natural numbers is the only meaningful > limit in mathematics. What does it mean? > > > > > 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. > > > > > > ============================= > > > > > > > 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! > > Either there is an edge which separates a path from all other paths > Or it is impossible to separate a path from all other paths. It is possible, but it requires infinitely many edges and/or nodes to achieve. For any finite set of edges and/or nodes, if there is any path containing all of them then there is a last edge or node of the set in that path from which infinitely many other paths spring. > > > > > > > > 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. > > By what is a set which only exists of edges separated from another set > which also only exists of edges? Any two distinct paths have a "last" node in common, and are thereafter sepatrate, but that does not separate either one of them from ALL other paths. > > =================== > > Virgil: > 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. > > WM: > So each irrational number is in fact an uncountable set of numbers? That may be WM's missinterpretation, but it is no one else's. Each number, ratinal or not, has a infinite path in that infinite binary tree, with a few rationals having two paths > > Virgil: > 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. > > WM: > Call them bunches of paths. So any irrational number is represented by > a bunch of paths? Call then sets of paths. Then every irrational is represented by exactly one path or a set of paths containing only one path if you wish. > > WM: > So any irrational number is represented by a real interval? Where did WM find that idiocy? > > Virgil: > For every real in [0,1], there is a path > > WM: > One path? How does the path of 1/sqrt(2) differ from every other path? By being a different infinite sequence of nodes and edges. > > Virgil: > 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. > > WM: > These proofs, if not invalid, would at most prove the inconsistency of > ZFC. They only prove that ZFC and NBG are in conflict with WM's system, whatever that may be. Since WM has no more way of proving his system "true" that ZFC does, the most WM can claim is incompatibility of systems. Wm may have abounding faith in his system, but he has no proof of it.
From: Virgil on 29 Dec 2006 19:25 In article <1167430589.769445.209010(a)h40g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1167392633.967611.252860(a)n51g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Dik T. Winter schrieb: > > > > > > > > > > > Apparently your limits work only for real numbers in lists but no for > > > > > real numbers in trees. > > > > > > > > Oh, they do. If the trees are properly defined. > > > > > > Imagine a tree which contains only paths of rational numbers in > > > infinite representation, i.e., ending with a period of 000... or > > > 111.... > > > > > > The first rational tree contains only the paths 0.000... and 0.111... > > > > > > 0. > > > / \ > > > 0 1 > > > / \ > > > 0 1 > > > ............. > > > > > > The second tree contains the paths 0.000..., 0.0111..., 0.1000..., and > > > 0.111... > > > > > > and so on until all rational numbers are represented. > > > > > > Take the union of all these rational trees. It contains all paths > > > representing rational numbers. > > > > > > What distinguishes this union of rational trees from the complete tree? > > > Nothing. > > > > Except that it is easy to construct a path in the complete tree not in > > any member of the union, and therefore not in the union itself. > > Fine. Tell me the node or edge which differs from the nodes or edges of > the union. > > > > x = 0.101001000100001000001.... cannot be in any member of that union. > > Alternately, x = sum{n=1..oo} 1/(2 ^ ((n^2+n)/1) > > I know about your fairy tails, but these trees are constructed by edges > and nodes, rigorously. x = sum{n=1..oo} 1/(2 ^ ((n^2+n)/1) is constructed rigourously. And taking each 0 as a left branch and 1 as a right branch from the decimal point on builds a path not in any of the finite trees, so nt in the union. > Something that is claimed to exist in the > complete tree but not in the union of all rational trees must be > distinguished by a node or edge which is not in the latter, but in the > former. 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. We will do strict mathematics of nodes and edges here and not > matheology of possibly imaginable mythes. I will but WM keeps going off into his matheology despite his promises. > Point to the node supporting > the difference or withdraw your dream. WM's dream that one node or one edge can separate one infinite path from all others is nonsense. > > > > > There is no node and no edge of the complete tree which is not > > > in the union of all rational trees. > > > > But there are lots of paths which are not in that union. > > Same edges, different paths. But you do not begin to doubt ZFC? You > would rather prefer to doubt your own existence. 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. > > > > E.g., given any f:N -> N which is a strictly increasing polynomial > > function of degree greater than 1, then y = sum 1/2^f(n) is not in your > > union. > > That means, it is not anywhere. It is a convergent series so has a real number limit. And it has an easily constructable path in an infinite binary tree. > > > > > > > As a path is an ordered set of > > > edegs (and nodes), two paths are not different unless they differ by an > > > edge (or node). Hence, there is no path in the complete tree which is > > > not in the union of all rational trees. > > > > WRONG! See above for specific examples of transcendental numbers > > corresponding to paths not in your tree. > > Examples of dreams are not paths in a real tree, not even in a > Christmas tree. That WM rejects what is clearly possible, make him the one rejecting reality.
From: Franziska Neugebauer on 29 Dec 2006 19:44 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. 2) Or they are different. You have to decide which version of "ideas" shall be valid. Without that decision you contradict yourself. F. N. -- xyz
From: Dik T. Winter on 29 Dec 2006 20:26
In article <1167392887.194994.3670(a)h40g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Where is the contradiction with what I wrote? ZFC is not sufficient > > to give a definable well-ordering, but a definable well-ordering is not > > inconsistent with ZFC. > > Why don't you simply give the formula for the definable well ordering? > The formula could be evaluated, and, step by step, we would get a list > of all reals. A list of all reals would not be inconsistent with ZFC??? You can look it up yourself. However, this does give us a well-ordering, *not* a list. A list is a special *kind* of well-ordering. Namely a mapping from the natural numbers to the elements of the list. > > > > it follows from ZFC+V=L that a particular formula > > > > well-orders the reals, or indeed any set." > > > > > > Instead of casting assertions: Why don't you simply give the formula > > > for the definable well ordering? The formula could be evaluated, and, > > > step by step, we would get a list of all reals. > > > > You are too lazy to look it up? If I understand it well enough, V=L > > immediately gives a well-ordering of the reals. > > > > But whatever, you stated that a definable well-ordering of the reals would > > lead to a contradiction in ZFC. But you have nowhere shown any proof of > > that, and it is false. It has been proven that a definable well-ordering > > of the reals is consistent with ZFC. > > Why don't you simply give the formula for the definable well ordering? > The formula could be evaluated, and, step by step, we would get a list > of all reals. Wrong. It would not give a list. It would give a well-ordering. Do you not know the difference? And I doubt whether you could apply the formula one by one to each real. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |