Re: Two results of set geometry

William Hughes wrote:
On Sep 28, 10:24 pm, Tony Orlow <t...@xxxxxxxxxxxxx> wrote:
*** T. Winter wrote:
In article <1190631570.757076.95...@xxxxxxxxxxxxxxxxxxxxxxxxxxx> WM <mueck...@xxxxxxxxxxxxxxxxx> writes:
> On 20 Sep., 04:33, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> > > There is a bijection between the set of all paths representing real
> > > numbers of the interval (0, 1] and the set of all nodes representing
> > > natural numbers:
> > > Every node with value 0 is mapped on that path which after passing
> > > this node passes only nodes with value 1. Every node with value 1 is
> > > mapped on that path which after passing
> > > this node passes only nodes with value 0.
> > Again you give an injection from the nodes to the paths. So you have
> > to *prove* that it is a bijection. Let us continue.
> > > Why is this a bijection? Because every node of every path, and
> > > therefore every path, is in the bijection.
> > Prove, *please*. Where in that "bijection" is the path "0.10101010..."?
> If it is in the tree, then it is in the bijection. Do you think that
> 0.101010... has nodes or initial segments which are not in my mapping?
What are you thinking? The mapping you gave is from nodes to paths. Where
in that mapping do initial segments figure? And *where* in that mapping
figures a specific node mapping to the path "0.1010101010..."?
Oh ***, the question is whether 2/3 actually exists in WM's tree!

If the levels are all finite, then the value so far achieved through
traversal of the tree still differs by some finite amount from 2/3. You,
he, and I, we all know that. Right? So, there is no finite level, and
that's the only kind there are in this tree, at which we can fully
express 2/3, or any such number.

So, the question then becomes,

Is there a last level? The answer is no.

Recall. A path is the limit of its
finite initial segments.

Gee, and I thought a path was a sequence of nodes. When did it become the limit of what potentially exists within it?


So, the statment 2/3 is in the tree means

Given any epsilon > 0 we can find a level
at which the approximation to 2/3 is closer than epsilon.

That would be a finite level, with a finite epsilon, would it not? Is 2/3-epsilon equal to 2/3?


One way to show that 2/3 in the tree is to show there is a level
at which we can fully express 2/3. However, this is
not the only way.

That is the only way to show it is fully included in the tree.


So the fact that there is no level at which we can fully express 2/3
is true but beside the point. The two statements

Given any epsilon > 0 we can find a level
at which the approximation to 2/3 is closer than epsilon.

There is no level at which we can fully express 2/3

do not contradict one another. However, the only way they
can both be true is if there is no last level.

Sure. And there isn't. The limit of the sequence is 2/3, but that value does not exist within the nodes of the tree. Every node is finitely distant from the root, so every node is differs from the value of an infinite path by some finite value. The infinite paths are never finished, so those values requiring an infinite string don't fully exist in the tree, but only as a limit of initial segments.


The statement

There is no finite level at which we can fully express 2/3

does not imply that 2/3 is only in the tree if the tree has
infinite levels.


- William Hughes



I disagree.

Peace,
Tony
.