Re: infinity

Virgil said:
> In article <MPG.1d823fca46f8063298a1ca@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
>
> > Virgil said:
> > > In article <MPG.1d7f9000c6e111a298a19f@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> > >
> > > > > > > I mapped paths from the same tree to sets of naturals so as
> > > > > > > to establish a bijection between them and the power set of
> > > > > > > N.
> > > > > >
> > > > > > Now you are trying to change things, but fine, let's examine
> > > > > > what you are now proposing. You claim to have branches mapped
> > > > > > to all the naturals in a 1-1 correspondence, and therefore
> > > > > > countable. Then you claim that the set of paths is the
> > > > > > powerset of that set, and therefore uncountable? You really
> > > > > > should draw pictures before writing words. Then you won't
> > > > > > waste 1,000 of them sounding stupid. Here's your tree, with
> > > > > > each number being a branch/node:
> > > > > >
> > > > > > .
> > > > > > 1 2
> > > > > > 3 4 5 6
> > > > > > 7 8 9 10 11 12 13 14
>
> This form of labelling has nothing to do with me.
>
> The only labels I use are on branches: "Left branch" and "Right
> branch", and the construction of my bijections are based entirely on
> those labels and nothing else.
Bull. You are constructing a bijection with the elements of the tree and the
elements of the naturals and of the powerset of the naturals.
>
>
> > > The disrepancy is between what I described and what TO
> > > misunderstood me to have described. If he could actually read,
> > > perhaps such discrepancies would not occur, at least so often.
> > >
> > > Each path in an infinite binary tree consists of a sequence of
> > > branches which can be uniquely numbered with the infinitely many
> > > finite naturals, so that all paths have a branch 1 and a branch 2
> > > and a branch 3 and so on without end. Each path determines a set by
> > > including the numbers for each right branch and excluding the
> > > numbers for each left branch (the reverse would work equally well).
> > >
> > > This clearly bijects paths with subsets of N (the infinite set of
> > > finite naturals).
> > >
> > > Each node in the same tree can be matched with a unique natural in
> > > binary notation as follow: The root node is 1. For each left branch
> > > on the finite path from the root node to a given node, append 0,
> > > and for each right branch append 1. So the children of the root
> > > will be numbered 10 and 11, the grandchildren 100, 101, 110, and
> > > 111, and so on. the number of binary digits will be one more than
> > > the number of branches between the root and the node which matches
> > > the binary number.
> > >
> > > This clearly bijects the set of nodes of that same maximal binary
> > > tree with N (the infinite set of finite natural numbers).
> > >
> > > So that unless TO can construct a bijection between N (the infinite
> > > set of finite natural numbers) and P(N), there are more (in the
> > > sense of Cantor) paths than nodes.
> > >
> > Grrrrr.....you moron! That's exactly the way I described your proof
> > to begin with, which you denied.
>
> If TO thinks that what he said and what I said are the same, he has
> even worse dyslexia than previusly thought.
No, it's exactly as I remembered and described, and you are spewing nonsense to
defend your nonsense.
>
> > You have one tree where you label each node with a natural number
>
> I never label any nodes with anything in my construction. The only
> "labeling" is done to branches, "left" or "right".
You never mentioned "left" and "right". You mentioned bijections between the
paths and the elements of the power set, and between the nodes or branches and
the natural numbers. In the first, each branch denotes inclusion of a natural,
but the entire row of branches corresponds to that natural, not a signle
branch.
>
> > , and then another tree where you label each node with a bit
> > representing the membership of a natural in a set which is a member
> > of the powerset.
>
> Same tree, same labels on branches, no labels on nodes.
>
No, you have two different trees.
>
> > In your first tree, where each node is a natural,
>
> That is TO's tree, not mine, as I did not label any nodes at all.
>
> In the second, each path represents a unique
> > set of naturals, and the tree represents the power set of the
> > naturals, but each node does not represent a natural number.
>
> But each branch determines whether a particular natural is in, or not
> in, the set represented by that path.
Each set of branches at a given level correspond to a natural, not each branch.
>
>
> Each ROW
> > represents a natural number, and for row n, all of the 2^n nodes in
> > the row correspond to that natural. So, you did do exactly what I
> > said you did, which is change your interpretation of the tree
> > mid-proof.
>
> Even if I had changed the labeling, which I did not, unless I also
> changed the underlying tree, both constructions remain valid FOR THAT
> TREE. There is no rule that one cannot have more than one label for a
> thing if it is convenient to do so.
Convenience being the operative motivation.
>
> Since the tree is the same for both, however labeled, TO's objection
> fails.
Binary trees are binary trees, but they can be interpreted in a number of ways.
Perhaps the most common is as analog to the binary number system, with each
branch being one of two choices for a next digit, and each path representing a
nuique whole number. If we allow infinite binary strings in our whole numbers,
then we can construct a bijection between the paths of the tree and the natural
numbers. We can also construct a bijection between the paths of the binary tree
and the powerset of the natural numbers, as you suggested. Since, as you said,
bijection is transitive, this amounts to a bijection between the whole numbers
and the power set of the whole numbers, which are clearly different sizes.
> >
> > Let us try inductive proofs regarding the relationship between
> > branches and paths.
> >
> >
> > Proof: In a maximal binary tree, number of paths is half the number
> > of branches, plus 1.
>
> False for the tree consisting of only a root node, and false where the
> number of branches is odd.
The root node has one path, one end. That's half of the nonexistent branches,
plus 1.

Oh, and maximal means with all leaf nodes at the same level, so there could
never be an odd number of branches. Sorry.
>
> And false for binary trees with infinitely many branches.
No, proven true, inductively. Give it up. You're wrong.
>

--
Smiles,

Tony
.

Relevant Pages

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