Re: Cantor Confusion

In article <1167999736.624629.206820@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
Dik T. Winter schrieb:
> And my question is: Do these two trees, namely the complete tree and
> the union of all rational trees differ such that the one has edges or
> nodes which are missing in the other?

No.

But the distinction is apparently difficult (although about first year
at University for mathematics). The set of terminating binary expansions
is countable, the set of non-terminating binary expansions is no
countable. (You may replace terminating binary expansions with binary
expansions terminating with either a continuous stream of 0'z or of 1's.)

But these streams are *not* present in all paths of the union of all
rational trees! That has been overlooked, as it appears, even in the
last years of university mathematics - in the last 130 years.

Oh.

Can you really believe that a thinking brain will accept your assertion
that two absolutely identical systems of nodes and edges will supply
different systems of paths, i.e., strings of nodes and edges?

Yes, it entirely depends on how you find your paths. Consider a graph
consisting of three sets, (1) the edges, (2) the nodes and (3) the paths.
Consider the following three graphs:
x x x
/ \ / \
x x x---x x---x
where we identify the nodes from the different graphs and also the common
edges. It is clear that if we define a path in a graph as a sequence of
nodes connected by edges, we see that in each complete graph there are three
paths. So we may say that the left hand graph consistes of the three sets:
{n1, n2, n3}, {e2, e3}, {n2-e3-n1, n1-e2-n3, n2-e3-n1-e2-n3}
(e2 is opposite n2, etc.) When we take the union of the first two of
them we get the following graph:
{n1, n2, n3}, {e1, e2, e3},
{n2-r3-n1, n1-e2-n3, n3-e1-n2, n2-e3-n1-e2-n3, n3-e1-n2-e3-n1)
And if we take the union of all three we get the following graph:
{n1, n2, n3}, {e1, e2, e3},
{n2-e3-n1, n1-e2-n3, n3-e1-n2, n2-e3-n1-e2-n3, n1-e2-n3-e1-n2,
n3-e1-n2-e3-n1}
See, the same set of nodes and edges, but different sets of paths.
Moreover, they all miss one path from the complete graph with three nodes
and edges, the circular:
n2-e3-n1-e2-n3-e1-n2
We clearly see that the union of a set of (complete) graphs does not
necessarily give a complete graph.

It is quite similar with your tree. The union of all finite trees gives
a tree with all the nodes and edges, but not all the paths of the complete
tree. While the union of the sets of edges and nodes behave normal, the
union of the sets of paths is *not* the set of paths of the complete tree.
And you are arguing about the union of the sets of paths.

> > You are wrong. sqrt(2) has a pretty good representation: "sqrt(2)".
>
> That is a name. Name ist Schall und Rauch.

In the same way '2' is a name, '10' is a name. '10' is clearly a name
for a quantity that depends on the context. In the same way 'sqrt(2)'
is the name for a quantity that depends on the context. For instance,
the wedge that is used for the quantity 7 in Hyderabad Arabic is used
for the quantity 8 in Devanagari and for the quantity 6 in Javanese.

Therefore these wedges are not numbers but only names which can mean
anything we define.

Right. And so is every notation of numbers. Numbers are not concrete
entities, they are abstract entities. And we name them by symbols or
strings of symbols according to particular convention.

But in Devanagari and Hyderabad Arabic and Javanese it is clear what we
mean by

||||||
|||||||
||||||||

Well, even I do not know. I would say either 8 or 7. But try that in
a culture that has no idea about abstract entities. Yes, there is one
such culture, a tribe of Indians along the Amazone. They can not count,
and do not count, and do not understand counting at all. In their
culture only concrete entities are acceptable (and that also only if
the person telling about it has personal experience with it).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.

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