Date: Dec 15, 2012 1:34 AM
Subject: Re: On the infinite binary Tree
On Dec 14, 10:17 am, forbisga...@gmail.com wrote:
> On Friday, December 14, 2012 6:30:11 AM UTC-8, Charlie-Boo wrote:
> > On Dec 12, 5:26 am, Zuhair <zaljo...@gmail.com> wrote:
> > > Lets take the third degree binary tree
> > > 0
> > > / \
> > > 0 1
> > > / \ / \
> > > 0 1 0 1
> > > Now this has 7 nodes BUT 8 paths, those are
> > > 0-0
> > > 0-1
> > > 1-0
> > > 1-1
> > > 0-0-0
> > > 0-0-1
> > > 0-1-0
> > > 0-1-1
> > Only 2 start with 1 and 6 start with 0. Shouldn't it be 50-50? If
> > you start with 0 then all start with 0. You seem to be inconsistent
> > in your manipulations of the tree.
> four of his paths start at second level nodes.
> > You need to diagonalize the binary tree rather than the list of real
> > numbers - and account for multiple trees representing the same
> > number. Or just ignore the real number interpretation and deal with
> > binary strings without regard to numbers.
> > My many examples of diagonalization in different contexts is a model
> > to do just that. It's always good to generalize.
> My concern is how one shows there are countably many paths through
> the infinite nodes. I can't move to the second path though infinity
> until I've completed the first, that is unless an algorith is proposed
> that lets me idenify which infinite paths I've taken without completely
> taking them. I might identify a path by its representation as a rational
> number. Unfortunately that leaves out the irrational numbers.
In the binary tree, of nodes of finite distance from the root, it
takes all the nodes to represent the rationals, of the unit
Now, consider diagramming the tree this way, in a breadth-first
traversal, lay out each row in the y-axis at the corresponding x-
coordinate of the row, with its y value being the rational value of
the expansion. Then, the maximum value of this point set at each x is
max(x) = 1- 1/2^x. In the limit that's one. Now, the paths are
ordered lexicographically, in the breadth first traversal, for each
they are. And, at each row, the difference and distance between paths
is constant. Then, the infinite binary tree, would have that, in the
limit or asymptotically, there are are infinitely many paths,
lexicographically ordered, in the unit interval, in their natural
ordering as representations of reals, or rationals and irrationals, in
the unit interval. These paths are dense in the unit interval. And,
for any node, there is the simple branch to the left or right that
represents a rational.
Now this is all without some completed infinity or "at least
infinitely many" number of rows or here columns. Yet, as a function,
modeled by each row, the function so standardly modeled has range
[0,1], here without simply declaring that at the point at infinity the
range is the continuous line segment. Via Cauchy, these expansion are
all the real numbers of the unit interval, and then some, with dual
representation of real in their binary expansions.
Then, one way to look at the rows of the binary tree is as a family of
functions BT_p = b^p for b as the domain from zero to 2^p. Then,
connecting the dots of each BT_p(n) to BT_p+1(2(n+1)) and BT_p(2(n
+1)-1), or along those lines, the resulting diagram, has that the
infinite trees are rooted in those.
So, there doesn't exist a breadth-first traversal of the infinite
binary tree. Yet, that is a countable enumeration, of the nodes in
breadth first order, and the paths in their natural order, of their
There are infinitely many natural integers.