Virgil
Posts:
8,833
Registered:
1/6/11


Re: On the infinite binary Tree
Posted:
Dec 14, 2012 2:15 AM


In article <bb95833d4ace4cf78ce1943925a95d0f@r4g2000vbi.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 14 Dez., 01:13, George Greene <gree...@email.unc.edu> wrote: > > > There is simply NO "natural" passage to a "limit" here. EVERYthing WM > > is saying is being > > said ABOUT THE FINITE case. The infinite case is simply DIFFERENT. > > It is astonishing that this possibility is not considered in case of > Cantor's list, isn't it? > > In all applications of analysis the limit is defined solely by the > finite terms. > > > > For one thing, all the finite paths END and all the infinite ones > > DON'T. > > Then take all paths that cover as many nodes as possible.
Every path in an infinite tree "covers" infinitely many nodes. > > > All the finite paths TERMINATE AT ONE UNIQUE NODE and all the infinite > > ones don't. > > But they must be defined by more than the nodes, if there shall be > more than conutably many.
Nonsense! The countably infinite set of nodes necessarily has uncountably many subsets, and uncountably many of those subsets will be paths.
> And we all know that an infinite sequence is > never defined by its terms but only by a finite definition. Alas, > there are only countably many finite definitions.
Alas WM is provably wrong! Again!! As usual!!!
Known infinite sequences may require definition, but there is nothing that requires an infinite sequence to be known. Every one of the uncountably many infinite ordered subsets of the ordered set N is an infinite sequence and there are uncountably many of them, most of them "unknown". > > > > THE REAL ISSUE here is WM's equivocation on what it means for a > > collectionofpaths > > "To Cover" (that's my verb, not his) a collectionofnodes. > > You can also say to construct a collection of nodes, namely the > complete Binary Tree, by countably many paths.
While a set of countably many paths may cover all nodes, it necessarily omits more paths than it can include. > > > WM has been in here with this shit for over a decade. > > Two wrong assertions in one simple sentence.
Actually WM HAS been here with his shit for over a decade.
> Small wonder that you > don't understand that Cantor has already been refuted.
Cantor has not been refuted outside of Wolkenmuekenheim. > > I construct the complete infinite Binary Tree by means of countably > many paths.
You often have claimed so, but since there are many proofs, quite valid everywhere here outside of your Wolkenmuekenheim, that you are wrong, your "proofs" do not hold up here!
> that you ar won If there were more paths definable by nodes, then you > should be able to say which one.
> If not, then you should be able to > see that only finite definitions define paths like that of 1/3. Alas, > there are only countaby many finite definitions. Is it really that > hard to understand?
Where is it written that one must be able to name every member of a set in order to count its members? The set of reals is known to have more members than can be named, at least ouside of Wolkenmuekenheim > > > His premise is so blatantly bullshit that it really should NOT be > > taking you THIS long! > > But you cannot explain the error in my claim? Here it is again: > > I construct the complete infinite Binary Tree by means of countably > many paths. Since it is well known that, at last outside Wolkenmuekenheim, the set of all possible such paths trivially bijects with the uncountable set of all subsets of N, no countable set of paths can include every path.
At last outside Wolkenmuekenheim!
> If there were more paths definable by nodes, then you > should be able to say which one.
We can find lots of them that you have missed as soon as you tell us which ones you have included
> If not, then you should be able to > see that only finite definitions define paths like that of 1/3.
But the Cantor diagonal process, which proves any listing that you present must be incomplete,is totally independent of any need for or reliance on finite definitions.
If f(n) = m when the nth branching of a particular path is to the left and f(n) = w when the nth branching is to the right, then that f defines a sequence and defines that path.
Each such f defines a path different from every other such f, and there cannot be any surjection from N to the set of all such f's.
> Alas, > there are only countaby many finite definitions. Is it really that > hard to understand? There are perfectly legitimate ways to prove that there have to be more paths, and f functions, and subset of N, than your limited definitionism will cover. 

