Re: Paths
- From: WM <mueckenh@xxxxxxxxxxxxxxxxx>
- Date: 25 May 2007 00:10:20 -0700
On 24 Mai, 11:58, Franziska Neugebauer <Franziska-
Neugeba...@xxxxxxxxxxxxxxxxxxx> wrote:
WM wrote:
On 24 Mai, 00:14, Franziska Neugebauer <Franziska-[...]
Neugeba...@xxxxxxxxxxxxxxxxxxx> wrote:
In case (C) is wrong, then removing p from T would no change T,
proving that p does not exist.
non sequitur. The right conclusion is
U(P) = U(P \ { p }). ("U" is the letter "u").
And the letter U denotes, as we agreed upon:
U(P) := { k | E p (p e P -> k in p) }
Yes.
Your conclusion is correct and it shows that {p} is the empty set
or p is not in the tree.
I don't see any conclusion which states that { p } is the empty set
or p is not in the tree. To state that justifiably you need to show
that for a given p from
U(P) = U(P \ { p }}
follows
~E p [(i)]
or
~(p in CIBT). [(ii)]
If A \ B = A then either B is empty or B is not in A. Any other
possibility?
Relevance? Again: I can't see any valid conclusing leading to (i) or
(ii).
Consider the union of all finite paths by means of the EIT. Write the
finite initial segments of 0.111... in form of entries of the EIT:
0.1
0.11
0.111
....
U(P) = U(P \ {0.111...}) = U({0.1}, {0.11}, {0.111}, ... \ {0.111...})
This means: There is not the slightest difference concerning the
hardware of the EIT, namely the 1's (nodes) whether or not the
diagonal 0.111... of the EIT is one of the entries of the EIT too.
Both is correct to assume because it is impossile to decide. But if
the path 0.111... is in the EIT as an entry, then Cantor's diagonal
argument fails.
No. If (C) is proven wrong, we need no further claim. In the tree
there is no possibility to identify p.
This is a different issue which shall be discussed under a different
subject.
Why different subject? That is the central issue: What lets p exist?
This is a different issue. If you want to ask "What lets the path p
exist?" or generally "What lets the infinite sequence q exist?" or even
more generally "What lets a set r exist?" you should do so in a new
thread with the respective subject.
This topic belongs to our discussion: A bunch of paths exists on its
own if, and as far as, it can be distinguished from another bunch of
paths by at least one node. If this is in principle impossible for a
bunch with only one single path, then such a bunch does not exist. A
set with only this path cannot be formed, because this path does not
exist. (Every element which exists can be taken to be a set.)
A "retroactive effect", invalidating (p), requires a _valid_ (C),
i.e. a proof of (C).
You are mislead.
I don't think so.
That is as little a retroactive effect as the exclusion of the
existence of a fraction being equal to sqrt(2). p simply did never
exist. It was only assumed to exist.
1. Even if it was only assumed to exist you have not shown that from
this assumption we run into a contradiction.
We see that all the paths or bunches of paths existing in the tree as
static sets of nodes form a countable set.
2. The fraction which is equal to sqrt(2) is non-existent as can be
easily proven.
The path p is inexistent as meanwhile has been proven by U(P) = U(P \
{ p }). The tree consists of nodes and edges and of nothing else.
These nodes and edges say: We don't know of a p. If it exists, it does
not exist in our tree.
3. If you want to judge the question _whether_ it exists you have take a
look into formal set theories (e.g. ZFC) in order to recognize what is
meant by "existence" in those systems.
No religion please! And least such religion which considers itself to
be "eaxct"!
p e P is equally false.
Now we see that nothing in the world is changed when p is removed.
Of course we see a difference: P* = P \ {p} means that
p e P*
is false.
You cannot prove the existence of a natural number p between 2 and 3
by stating that p is not an element of the common natural numbers P*
but is in some P as a natural number.
..
A valid (C) would prove the existence of p.
Even if so an invalid (C) does not "disprove" its existence.
What kind of structure (verifiable by nodes or digits, not imagined
only) do you propose to prove that p exists?
Within the framework of modern set theories it is rather tedious but
nonetheless surely feasible to prove formally the existence of a given
p,
We are discussing the binary tree.
There exist some facts proving my existence. But there exists no fact
proving the existence of p.
Then you should ask for a formal proof of the existence of a given p
under a new subject in a new thread.
I ask for a formal proof of the existence of p in the binary tree
here. A formal proof in the binary tree consisting of nodes and edges
has to take into account these elements of the structure of the tree
and no religious beliefs.
You *can* define: p is the union of all finite intersections p n q
with paths q =/= p (as N is the union of all its finite initial
segments).
I have already defined what p is: p = (p_n) is an infinite sequence of
nodes of the binary tree having the properties:
1. p_0 = 0 (the root node)
2. Every ordered pair (p_i, p_(i+1)) is edge in CIBT.
And we have seen that this is nothing but the description of the union
of all finite paths pointing in this direction. Well, this union may
exist. But it does not lead to any uncountable set.
Then *p is a union* of countable elements. But p (and every
other infinite path) *is not* a power set. Therefore the set P of all
infinite paths is a countable union of countable unions of countable
sets and as such P is countable.
I see no necessity to define a path differently from that given above.
Then you should see the countability of all paths too.
I propose "existence of sets on their own" as a new subject for a new
thread where we can formalize your statement and check whether it holds
in modern set theoretic frameworks.1
You may create such a thread which would fit very well in the group
"aspects of the infinite". But please note that I will be absent for a
while (about one week) and therefore will not be able to react. This
does not mean that I reject you contribution. Also in this thread I
will not answer for a while. This does neither mean that I agree with
your position nor that your position were true.
erratum
MID <4654bcaf$0$97229$892e7...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
From U(P) = U(P*) we can validly infer
1. U(P) \ U(P*) = {}
2. U(P*) \ U(P) = {}
3. U(P) =c U(P*)
4. U(P*) =c U(P)
Ok, this conclusion is correct. A counterexample of your previous one
would be
A = {1, 2} =/= B = {1, 2, 3}, A \ B = {}
So we have now agreement that U(P*) is U(P), i.e., the CIBT is nothing
but the union of all finite paths. This is a countable set of paths.
Regards, WM
.
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