Re: Cantor and the binary tree

Virgil said:
> In article <MPG.1d012f6c868bb0b9989d6d@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
>
> > Virgil said:
> > > In article <MPG.1cffea4895c6e40a989d4e@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> > >
> > > > > In a maximal binary tree, the set of nodes easily bijects to N and the
> > > > > number of unleafed paths easily bijects to P(N), and Card(N) <
> > > > > Card(P(N))
> > > > >
> > > > Show me the proof so I can show you the mistake. It probably has to do
> > > > with
> > > > some unjustified assumption about your imaginary "unleafed paths". How do
> > > > you
> > > > get a path without a node at the end of it anyway?
> > >
> > > One gets a path with no last node the same way one gets a binary or
> > > decimal fraction with no last digit.
> > >
> > > (1) Given any finite path, let the root node be labeled 1 and
> > > thereafter, let each left node be labeled 0 and each right node labeled
> > > 1.
> > > Then the last node in that finite path is represented by the binary
> > > integer of the digits for all those nodes taken in order from root to
> > > the node in question.
> > >
> > > And if the binary tree is maximal, having no leaf nodes, there will also
> > > be a node for each binary integer.
> > >
> > > Thus, I have created a bijection from the set of nodes to the set of
> > > naturals, N = {1,2,3,...}. QED
>
> > Sure, each node represents a natural number. Given your belief that the
> > naturals are all finite, you conclude that the nodes are too, as far as
> > distance from the root.
>
> Precisely.
> > >
> > > (2) Given any infinite path in a maximal binary tree,
> > > create a subset S of N as follows:
> > > if the nth child node of the tree is on a right branch from
> > > its parent node, n is to b a member of s, but if on a left branch,
> > > n is to be excluded.
> > >
> > > It is easily seen that
> > > every infinite path defines a subset of N,
> > > different infinite paths define different subset of N
> > > every subset of N is created by some infinite path
> > >
> > > Thus I have constructed a bijection from the set of infinite paths to
> > > the set of all subsets of N, i.e., to P(N). QED.
> > >
> > > Both bijections constructed as requested.
> > >
>
>
> > Actually you have a problem here. Every infinite path represents a single
> > member of N, not any larger subset.
>
> But I am not talking about finite paths, but paths which are distinctly
> NOT finite.
That doesn't matter. 0.111111111..... is represented by an infinite path, and
represents a single number value. A path is a sequence of branches that define
the digits in the number, whether finite or infinite. You can't just suddenly
change the definition of things when you get to infinity to please your fancy.
Try to be consistent.
>
> > Cantorians should work with binary trees and lists during
> > their apprenticeships. It's like a sailor knowing how to swim.
>
> Then it is TO who is drowning. An unbounded path in a maximal binary
> tree is not a bounded path, so that what holds only for bounded paths is
> irrelevant for unbounded ones.
That is so wrong so as to be criminally stupid. Binary trees are binary trees
and paths are paths. Their properties don't suddenly morph into something
totally different when infinite. That's moronic.
>
> TO is going down again. Is it for the third time?
>

--
Smiles,

Tony
.

Relevant Pages

  • Re: Cantor and the binary tree
    ... >>> naturals are all finite, you conclude that the nodes are too, as far as ... Every infinite path represents a single ... >>> Cantorians should work with binary trees and lists during ... criminally stupid is a measure of how close to criminal stupidiy he is ...
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  • Re: Cantor and the binary tree
    ... >> the root node, but no node is the end node of an infinite path from the ... And if the binary tree is maximal, having no leaf nodes, there will also ... I have created a bijection from the set of nodes to the set of ... Given any infinite path in a maximal binary tree, ...
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  • Re: Cantor and the binary tree
    ... But I later supplied that easy bijection to "found" my assertion. ... > In a maximal binary tree, the set of nodes easily bijects to N and the ... And if the binary tree is maximal, having no leaf nodes, there will also ... Given any infinite path in a maximal binary tree, ...
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  • Re: Cantor and the binary tree
    ... >>> In a maximal binary tree, the set of nodes easily bijects to N and the ... > Given any infinite path in a maximal binary tree, ... > Thus I have constructed a bijection from the set of infinite paths to ...
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  • Re: Cantor and the binary tree
    ... > There are no finite paths. ... If there are not nodes along the entire infinite path, ... > the node N of that branching. ... Nevertheless this is obviously a bijection. ...
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