
Re: On the infinite binary Tree
Posted:
Dec 15, 2012 4:27 PM


On Dec 15, 9:08 am, forbisga...@gmail.com wrote: > On Friday, December 14, 2012 10:34:33 PM UTC8, Ross A. Finlayson wrote: > > On Dec 14, 10:17 am, forbisga...@gmail.com wrote: > > > > On Friday, December 14, 2012 6:30:11 AM UTC8, CharlieBoo 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 > > > > > > 00 > > > > > > 01 > > > > > > 10 > > > > > > 11 > > > > > > 000 > > > > > > 001 > > > > > > 010 > > > > > > 011 > > > > > Only 2 start with 1 and 6 start with 0. Shouldn't it be 5050? 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 > > > interval. > > > Now, consider diagramming the tree this way, in a breadthfirst > > > traversal, lay out each row in the yaxis 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 breadthfirst 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 > > > values. > > > There are infinitely many natural integers. > > > Regards, > > > Ross Finlayson > > And yet when you do this you never get to an > infinite binary tree because every leaf expanded so > far is finite. There is a dual representation of > (some?) rationals but not of any irrational. > > My concern is about the countability of the paths > through the infinite binary tree. Every path is > an infinite expansion. Some numbers have two paths > that represent them. I don't see how you get from > countable nodes to countable paths. Cantor showed > any algorith for counting paths can be shown to not > count many paths.
(....)
Let's look at what Cantor showed us: combined with that there isn't a function from naturals to reals in their usual order: none of the other functions are a bijection from naturals to a continuous line segment of the reals. A function from naturals to reals in their usual order doesn't see the same results.
For example, imagine lining up the real numbers from zero to one in their normal order. The expansions in binary would look like:
.0000... .0000... .0000... ... .1111... .1111... .1111...
Here, the ellipse in the middle has that the values go from zero, to one, here with EF or BT (EF_oo and BT_oo, where BT_oo is EF_2^oo).
Then, with the antidiagonal argument, starting from the beginning, and different at a place for each, "the" antidiagonal:
.111(1) = 1.0
is at the "end" of the list of those values.
With Cantor's first, also called nested intervals, with that there is a way to naturally or lexicographically order the expansions, these are the same elements as all that comprise the complete ordered field, because they are between zero and one, and there are none not in it. Here, via the total ordering of the elements in the path, and that they are only constructed in that order, they go from zero to one, and then as inputs to the algorithm of nested intervals: there aren't any.
Then of course there are the powerset results, for these there is a settheoretic recourse, about how the reals aren't just the complete ordered field, but the continuum, then also about how the infinite is.
Then, for each node in the tree, finitely distant from the root, and there are infinitely many: those are countable. Here, Cantor's results about the binary tree, are not via a direct anti diagonalization, instead, that the reals biject to to {0,1}^w as does P(N) that transitively they have the same cardinal (CantorSchroeder Bernstein theorem generally), and that the paths do to the Cauchy expansions of the unit line segment, of real numbers.
{0,1}^w = R_[0,1] = R = {0,1}^w = P(N) > N
It would be of interest to see an actual diagonalization of the tree, the constructive argument. Because, in the breadthfirst traversal of the paths, the order of the expansions is the same as the order of the natural integers.
Draw a line: that's what Cantor shows: the line is drawn without lifting the pencil. It's drawn: from the origin, its origin.
There are infinite integers, are there not: infinite integers?
In the news: http://www.newscientist.com/article/dn23003higgsbosonhavinganidentitycrisis.html
Regards,
Ross Finlayson

