Re: infinity
- From: Virgil <ITSnetNOTcom#virgil@xxxxxxxxxxx>
- Date: Fri, 02 Sep 2005 12:14:08 -0600
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.
> > 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.
> 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".
> , 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.
If TO would learn to read he cold avoid makeing such as as of himself.
> 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 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.
Since the tree is the same for both, however labeled, TO's objection
fails.
>
> 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.
And false for binary trees with infinitely many branches.
.
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