Re: Answer to OJ

In article
<ad896bd1-1783-4782-8d6d-218bbb71f72f@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
WM <mueckenh@xxxxxxxxxxxxxxxxx> wrote:

On 31 Mrz., 06:23, Owen Jacobson <angrybald...@xxxxxxxxx> wrote:

One proof is as follows: The binary tree can be constructed by using
only paths for each of which we can find an index n_0 such that for n >
n_0 the nodes of this path all are zero.

Every infinitely-long binary string constructed this way has only
finitely many 1s. There infinitely-long binary strings which contain
more than finitely many 1s, but your construction will not create them
at any step, so you cannot inductively prove things from the properties
of the strings your construction creates to strings that your
construction omits.

What node=1 of the path 0.111... do you miss?

All those with infinitely many 1's in them.

More surprising, though, is the fact that your construction
demonstrably constructs at least one path through every node in the
tree!

And I can show that there is no node available that would allow to add
a further path.

But there are sets of nodes that allow more paths that WM has
constructed or can construct.

For any node, we can construct an initial path leading to it
using a finite number of 1s and 0s, and we can then proceed using
finitely many more 1s to any of its descendent nodes, but at some point
on each path, your construction "gives up" and generates infinitely
many zeros.

One key realization is that, from every node n, there are exactly as
many possible infinitely long paths through n's descendents as there
are from the root node's descendents.

None of them is missed by my construction.

Actually more are missing than are present.
All those with infinitely many 1's are missing.



Otherwise you should be
able to demonsaret what is missing.

All those with infinitely many 1's are missing.


Your construction never gets any
closer to finishing the tree - only further from starting. That does
not rule out simply starting from the complete tree with all of its
infinitely-many infinitely-long paths; however, if we simply assume all
of those paths exist then we must demonstrate that the results of that
assumption are consistent (a) with the axioms in use and (b) with each
other - which they are.

No, they are not. Compare my proof (A). All nodes of the infinite
binary tree are constrcuted by a countable set of infinite paths. If
you have started with uncounatbaly many paths, then you have a
contradiction because the whole tree is occupied.

There is no finite or countable limit to the number of paths that may
pass through (occupy) a given node, and in any complete infinite binary
tree, uncountably many pass through every node. The subtree of paths
rooted at any node is tree-isomorphic to the original tree.

Nevertheless after construction all paths, including 0.111... are in
the tree. After construction they creep in. This is strange.

Indeed. I suspect any recursive construction of paths in the tree will
never construct at least some paths in the tree at any step.
Nonetheless, those paths are present in the complete tree.

So you believe that there are more paths than lines?

What are lines, and how are they relevant to trees?





But if we accept it, then we must accept this also for Cantor's list. The
following list

0.0
0.1
0.11
0.111
...,

after completing construction, also contains the path 0.111....

The "completion" of that list contains only such paths whose count of
1's are members of N, but aleph_0 is not a member of N.



Paths bunches as all paths with finiely many nodes=1 and infinitely
many trailing zeros are also countable. Nevertheless it is assmed that
the path 0.111... creeps in. (It does not.)

It is in the complete tree but not in any of WM's constructed trees.
.

Relevant Pages

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