Re: The complete infinite binary tree has only countably many infinite paths.

On Mar 26, 7:57 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

Every edge leads to one and only one node. If we call the edge by E(x,
y) where (x, y) is the node, then the edge is uniquely defined. There
is not much work to do. Further if we work with edges, then we need
not work with nodes at all. Therefore even the E can be dropped. Why
do you see problems where they are not?

It is not what you *call* the edges, it is what they *are*. An
edge is essentially something between two nodes. So, to
specify *what it is*, you must give the two nodes. You don't
do that, do you? You just say, E(x,y) is what we call the edge
between x and y. Fine, but what are x and y? They are just
formal names. You must give the set of {(x,y)}, a subset
of NxN, in order to even define the tree. The problem I see
is that you haven't defined the tree at all here.

(This is assuming, again, what we should be able to do,
define the tree T in one shot, as a set of edges between
nodes, rather than building it up as a limit, which you have
been trying to do all along, but which is leading
into confusion)


Before goning on, please anwer the question: Do you agree that I
construct all the nodes respectively all the edges of the tree by my
prescription?

Above, you have said *nothing* about the tree. Let us
look yet again at the limit method. You start with
an empty tree, or maybe a tree containing the
edges for the path you call 0.0000..., and at each step
you add another infinite path to tree T(n) to get
tree T(n+1) (you do not say clearly
which path). Then, I believe, you say that T =
the union of the T(i), i in N. (Or more simply,
if the infinite set of edges added at step i is p(i),
the tree then is union p(i), in in N.

This is not a complete specification until you specify
precisely what each p(i) is.

   The only way to add  all paths is to
add edges one at a time in the right way.  You will never get
all paths by trying to add a countable number of infinite
paths.   You cannot construct the tree by adding paths.

Why not?

Because there are an uncountable number of paths, and
the process has only a countable number of steps, one
for each i in N.

However, if you add them correctly, the key point is
that you are also adding edges. And it is these edges,
not the paths, that end up making up all the other paths
that you have not added. So the edges do the job in
the end. And all these other paths sneak in although
you did not add them (at any given step).

S(i), for some strictly increasing function S of i.  So notice:
these infinite paths do not occur in any of the finite trees made
during the construction, because no such tree has all the edges
required!  Only the final tree has all the edges required.

Therefore I do not add edges but infinite paths in my proof (A)

That is your problem. You must add edges. You do, by
adding paths, and the edges, not the paths, do the job.

All paths simply do not fit into the tree. Nevertheless they are in
it, after infinity has been finished (and given they all exist).

What? They don't fit in, yet they are there? Well, that
makes no sense. But you appear to be coming close
to saying the right thing. You do not PUT all the paths
into the tree yourself, one at a time. You only put in
a countable number of paths in total. But in the end
the uncountable number are there.

How is that possible. It is because of the edges. When
you take an infinite set of paths that you put in, and take
one edge only from each of them, what you get is a new
path that you NEVER put in, among the COUNTABLE paths
you put in. That path is in the final tree, contributing
to the UNCOUNTABLE number in the tree.

Consider the path you denote .111111111....
Well, you never put this path in, of course. But it
is there ONLY in the infinite tree. It is there because
it takes one edge from
.10000000.....
one edge from
.11000000.....
one edge from
.11100000....

etc.

It is the single edges that do the work, not the puny number
of infinite paths that you added one by one.

If the tree has all nodes, after the end of construction, then there
are all paths.

Right. You added explicitly only a countable number of
paths, but there are an uncountable number that sneak
their way in by the end, as a result of them picking one
edge from each of an infinite number of the puny
countable infinity of paths that you explicitly put in.

So, there is no contradiction is there? You add
a countable number of paths, but they break into
edges and the edges recombine (only in the final
tree) to become many many more paths.

.

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