Cantor Confusion
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From: Virgil on 23 Jan 2007 14:15 In article <1169545722.090162.16030(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > >> > nevertheless the union of {1,...,n} and > > >> > {1,..., m} and the infinite union of all segments are defined. > > >> > > >> We are writing about trees. > > > > > > The trees have levels. For them we have the same as for the initial > > > segments given above. > > > > Again: Your notations > > > > T(1) U T(2) U ... > > > > and > > > > U {T(i) | i e N } > > > > are undefined. > > You are in error. The union of the trees T(n) and T(n+1) is defined. The union of two trees, if union means what it usually means in set theory, is not a tree at all. What WM is trying, but failing, to say is that when one takes the union of the set of nodes of two such trees and the union of the set of edges of two such trees, the concatention of those two separate unions forms a tree, but that concatenation of unions is not, itself, a union according to any mathematical meaning of "union". > Therefore the union of all finite trees is > defined. No more so than the union of two trees. > > If you try to construct the tree with n levels, do you fail at some > number of levels? No. Therefore the union is defined for every n. More > is not feasible. A union of trees is not a tree, and what is defineable as a tree conjoined from separate trees is not a union of trees. In any event, the set of paths of the complete infinite binary tree is uncountable, and attempts by WM to show otherwise are futile on two levels: one, WM is too inept to find such a proof, were it to exist, and two, it does not exist, and a counterproof does exist. > > It is an equivocation (fallacy) to claim that your tree-union (which > > selects the deepest out of two trees as "union") is a union. It has > > been explained by Virgil, WH and me that a set-theoretical union of > > trees is hardly a tree. Hence you are writing on undefined notations. > > > > It has been claimed, but falsely. If two trees, T(n) and T(m) are > idential down to level n but T(m) contains some moere levels, then the > union of both is T(m). There is a "conjunction" of trees which can be T(m), but the /union/ of T(n) and T(m) is not a tree at all > > Further, both Virgil and William understand that the union defined by > me is identical to te complete tree T as far as nodes and edges and > levels are concerned. WM's imagined "conjunction" of trees may be the infinite tree he claims, but such "unions" are not trees at all. > They merely doubt the identity of path due to > some inexplicable religious belief in a death religion. It is ingrained in the the faith of such true believers as WM to attribute the doubts of others to their religious beliefs. We who can prove that the set of paths in an infinite binary tree and the set of infinite binary strings are both uncountable sets, have no need of the sort of religious faith that bolsters WM's unprovable claims.
From: imaginatorium on 23 Jan 2007 14:20 David Marcus wrote: > As for Cantor, his great idea was that it made sense to compare sets by > whether you could biject or inject them. Before him, people thought that > this idea didn't work. I don't understand what you mean by this - I thought that "before Cantor", people just assumed that the only size "beyond any finite size" would be "infinite". You could always line up two infinite sets side by side (i.e. in a bijection), and start counting, and neither would ever end, so you couldn't say that either was bigger than the other. Do you mean that they thought something else entirely? Brian Chandler http://imaginatorium.org
From: Virgil on 23 Jan 2007 14:22 In article <1169545975.865624.242740(a)d71g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169396052.939963.194070(a)q2g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > Therefore, by induction, all finite n-level trees for all n in N exist. > > > > But that is all that standard induction allows one to conclude. > > > > > > It is enough. Or should there one level be missing? Please specify > > > which remains to be included. > > > > Anything that concludes anything about infinite trees. > > Conclusions are not members or subsets of trees. Standard induction does not justify your /conclusion/ that your conjunction of finite trees is an infinite tree. If you wish to justify such a conclusion, you must look elsewhere.
From: Andy Smith on 23 Jan 2007 14:32 G. Frege <nomail(a)invalid.?.invalid> writes >> >> If you see this as straightforward it is because your mindset has been >> conditioned by your education to see this as normal. >> >Right. That's what a _mathematical education_ is for. (Won't you think >so?) > >> >> I can safely say that if your concepts of infinite sets was placed in >> front of the population at large 99% would think that [...] >> >Who cares?! 99% (or more) are no mathematicians (or at least seriously >concerned with set theory). > I am just observing that all of this is not intuitively obvious (i.e definitely not in the same way as e.g. Euclid, or even basic calculus is). >> >> Cantor provided a perspective to view infinity, and his insight >> underpins, as I understand it, modern set theory. >> >I guess, that's a reasonable point of view. > >> >> It may be consistent, ... >> >It most certainly is. > >> >> but I don't see that the philosophical rational is trivial ... >> >It isn't. > >You might try to get a copy of > > Rudy Rucker, Infinity and the mind. The science and > philosophy of the infinite. > >A very nice book. > I bought a copy 25 years ago, and am just re-reading it now. But my impression is that a lot of people on this NG would take issue with Rudy Rucker particularly in respect of achieving actual infinity, not to mention his Zeno based speed-up scheme to count to infinity and beyond (David Marcus please comment?) >> >> ...and, as I understand it, in Cantor's day there were many eminent and >> far from stupid mathematicians who couldn't get a handle on it. >> >There were SOME of them, that's right. > >A quote from Herb Enderton's Elements of Set Theory: > >"Cantor's work was well received by some of the prominent >mathematicians of his day, such as Richard Dedekind. But his >willingness to regard infinite sets as objects to be treated in much >the same way as finite sets was bitterly attacked by others, >particularly Kronecker. There was no objection to a 'potential >infinity' in the form of an unending process, but an 'actual infinity' >in the form of a completed infinite set was harder to accept." > Yep, that is Aristotle's heritage. >> >> From posts on this site I can see that their descendants are still here >> and active ... >> >...NONE of which is a professional mathematician. (Does that ring a >bell? ;-) > Well, for what it is worth and if you haven't guessed, I am certainly not. Signal processing is what I do for a living, Taylor expansions all valid, approximations etc, no problem ... >> >> If I was asked to sum it up, at present I would say that my >> understanding is that you can't have an actually infinite integer, but >> reals can be defined as having an [...] infinite binary >> representation ... So no surprise that the reals are "uncountable". >> >Well, actually, it WAS a surprise. (Some still aren't able to get it. >:-) > Well me to, struggling with the validity or otherwise of Cantor's diagonalisation logic. And, arguing about the infinite gets contaminated with perceptions/lines of argument easily derived from the finite, and it is still not clear to me (short of taking a course in axiomatic set theory) quite what inferences can be legitimately made concerning infinite as opposed to finite situations. But, to cut to the chase, if you don't have enough integers to define even one real, what chance of counting them all? -- Andy Smith
From: Virgil on 23 Jan 2007 14:42
In article <1169547646.733664.94830(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Therefore there is no uncuntable set of path in the union tree. But > there are all paths in the union trees. One can show that for every list (sequence) of infinite paths in the infinite tree, there is a path in that tree not in that list. That was, in essence, Cantor's original diagonal proof. So Cantor proved uncountability of the set of paths. And that set has been uncountable ever since. > > It is clear how to produce the union of the finite trees to all who > discuss here. Virgil, William, and you agree that this union contains > all nodes and all edges and all levels. What we do not agree is that it is a "union" of trees. The word "union" has a meaning which is incompatible with WM's conjoining process. We also do not agree that the set of paths of any infinite binary tree in which all paths are endless is countable, as we can prove otherwise. > > > > > T1 as the countable union of all finite sets of finite paths contains > > > only a countable set of finite paths. > > > > No. It *is* the countable set of finite paths. > > Right. It is the union of all nodes If T1 is, as stated above, the union of sets of paths, then it is a set of paths containing only those paths which appear in at least one of the sets being unioned. And as none of those sets contain any infinite paths, neither can the union have any infinite paths as members. In mathematics, one goes by the definition. If the above definition is correct, then WM is wrong. If WM wants to have a different definition which gives the result he wants, let him provide it, but until then WM remains wrong. > > > > No. T1 is a set of paths, it is not a tree. T1 does not contains nodes > > or edges as elements. Only paths. > > The nodes can be enumerated in various ways. Here is a very simple > method: For the defintion given, sets of nodes are irrelevant to what the union will be. > > This is the finite tree (of the kind defined by m) with n levels . > The union of trees is the ordered union of their nodes. Not according to the definition given above. > The union of above elements exists for any set of n in N. This union is > a tree, i.e., an ordered set of nodes as given above. Wrong! The union for the definition given is a set of paths. > > > > > > > Here it is (for trees of the kind weeping willow): Taking the union of > > > two trees corresponds to taking the union of their sets of paths. > > > > Why? The union of the sets of paths is a set of paths, not a tree. > > The union of trees is a union of their nodes. Irrelevant when the definition in use refers only to unions of sets of paths. > The paths are merely > subsets which are defined by the nodes and the special kind of tree. Dead wrong when the "union" is a union of sets of paths. > > > > > Then > > > we have:The set of paths in the union is the set of paths in the larger > > > tree. > > > > If you define a tree by the set of paths in it. Yes. Once again, WM demonstrates that he is mathematically inept. |