Das Kalenderblatt 120126

25/01/2012 - 11:08 von WM | Report spam
Das Kalenderblatt 120126

IW: The one thing I don't understand about these WM binary tree
threads is how WM, as an ultrafinitist who believes that numbers as
large as a googolplex don't exist {{Das ist ein Missverstàndnis. Jede
Zahl, die angegeben werden kann, exististiert. Und in den Diskussionen
zum Binàren Baum wird ohnehin die Existenz jeder individualisierbaren
Zahl vorausgesetzt.}}, can believe that an infinite tree even exists
in the first place, much less debate over whether it's countable or
not.
TL: That has been asked before. If I recall correctly, his answer
was along the lines of it merely being a premise for a reductio ad
absurdum argument.
IW: OK, I see. So it's sort of like "assume infinite sets exist.
Then the binary tree is both countable and uncountable -
contradiction. Hence only finite sets exist."
WM: Yes, you got it. But instead of saying "only finite sets exist"
which would most mathematicians immediately cause to stop reading, one
should say sets can be potentially infinite in the sense infinity was
applied over thousands of years: oo. (Unless MatheRealism is invoked.
Then sets of numbers are finite but values of numbers are not
limited.)
[I. Walker, Tim Little, "The complete infinite binary tree has only
countably many infinite paths", sci.logic, 26. 3. 2009]

CO: All paths will occur in the final tree (the union of what you
get by unioning all the infinite paths he adds).
WM: And no other path than those used for construction can appear
in the
tree. Because paths do not reproduce.
V: Your 'constructive' generation of such a tree always omits most
paths. You can constructively generate complete trees to any finite
depth, but not a complete infinite tree.
CO: No, it need not. There is a countable set of paths, such that
its union yields the entire tree, and thus all paths.
WM: Nice to see that at least few readers understand.
For that argument I deviced my proof (A). Paths may be imagined to
do this or that. They cannot but follow one line in the tree. And
there are all lines in the tree. There is no infinite sequence of bits
that is lacking. Neverteless all this is constructed by a countable
number of paths.
WM: Claimed but never proven, and all constructions so far
presented have been shown to fail.
CO: No, I don't think so. He has not rigorously provided the
construction, but it is quite trivial.
WM: So it is, but only after you have found just the thing.
CO: It adds only a countable number of paths, and there is no
infinite sequence of bits that is lacking. All of this is correct. The
point is that the paths interact. Again, suppose he adds no path
corresponding to .1111
He does add
.100000
.110000...
.111000...
So when they are all unioned, the path .111111... will also be there
along with an uncountable number of other paths that he never
*explicitly* added in his countable list.
WM: Do paths have sex together? Even only those ending by zeros?
[...] When you consider my proof (A), then you see that I add
always exatly one path.
V: The way you do it prohibits any path from having a path with
infinitely many branches in the opposite direction from the ones you
add.
CO: I think now that under the most natural way of understanding
what he is saying, this is incorrect.
All paths will occur in the final tree (the union of what you get
by unioning all the infinite paths he adds).
V: How does one get a path with infinitely many 1's (right
branchings) by unioning sets of paths none of which have infinitely
many 1's?
WM: How do you get an infinite set by unioning infinitely many
FISONs?
V: Your 'constructive' generation of such a tree always omits most
paths. You can constructively generate complete trees to any finite
depth, but not a complete infinite tree.
WM: No, it need not. There is a countable set of paths, such that
its union yields the entire tree, and thus all paths.
V: Such a union may 'cover' all nodes without 'covering' all
paths.
WM: That is a word! Let me put it as: Paths have souls. It is not
sufficient to cover all their bones (nodes).
I am collecting reasons why I am in error. Now I have already
three:
1) SUM 0 = infinite.
2) Paths reproduce themselves (even those with same endings).
3) Paths have souls.
I will ask my students and other bright people whether finished
infinity is worth to be paid by swallowing at least one of these
snags.
[Calvin Ostrum, Virgil, "The complete infinite binary tree has only
countably many infinite paths", sci. logic, 27. 3. 2009]

Gruß, WM
 

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#1 wernertrp
25/01/2012 - 12:09 | Warnen spam
On 25 Jan., 11:08, WM wrote:
Das Kalenderblatt 120126

   IW: The one thing I don't understand about these WM binary tree
threads is how WM, as an ultrafinitist who believes that numbers as
large as a googolplex don't exist {{Das ist ein Missverstàndnis. Jede
Zahl, die angegeben werden kann, exististiert. Und in den Diskussionen
zum Binàren Baum wird ohnehin die Existenz jeder individualisierbaren
Zahl vorausgesetzt.}}, can believe that an infinite tree even exists
in the first place, much less debate over whether it's countable or
not.
   TL: That has been asked before. If I recall correctly, his answer
was along the lines of it merely being a premise for a reductio ad
absurdum argument.
   IW: OK, I see. So it's sort of like "assume infinite sets exist.
Then the binary tree is both countable and uncountable -
contradiction. Hence only finite sets exist."
   WM: Yes, you got it. But instead of saying "only finite sets exist"
which would most mathematicians immediately cause to stop reading, one
should say sets can be potentially infinite in the sense infinity was
applied over thousands of years: oo. (Unless MatheRealism is invoked.
Then sets of numbers are finite but values of numbers are not
limited.)
[I. Walker, Tim Little, "The complete infinite binary tree has only
countably many infinite paths", sci.logic, 26. 3. 2009]

   CO: All paths will occur in the final tree (the union of what you
get by unioning all the infinite paths he adds).
   WM: And no other path than those used for construction can appear
in the
tree. Because paths do not reproduce.
   V: Your 'constructive' generation of such a tree always omits most
paths. You can constructively generate complete trees to any finite
depth, but not a complete infinite tree.
   CO: No, it need not. There is a countable set of paths, such that
its union yields the entire tree, and thus all paths.
   WM: Nice to see that at least few readers understand.
   For that argument I deviced my proof (A). Paths may be imagined to
do this or that. They cannot but follow one line in the tree. And
there are all lines in the tree. There is no infinite sequence of bits
that is lacking. Neverteless all this is constructed by a countable
number of paths.
   WM: Claimed but never proven, and all constructions so far
presented have been shown to fail.
   CO: No, I don't think so. He has not rigorously provided the
construction, but it is quite trivial.
   WM: So it is, but only after you have found just the thing.
   CO: It adds only a countable number of paths, and there is no
infinite sequence of bits that is lacking. All of this is correct. The
point is that the paths interact. Again, suppose he adds no path
corresponding to .1111
   He does add
   .100000
   .110000...
   .111000...
So when they are all unioned, the path .111111... will also be there
along with an uncountable number of other paths that he never
*explicitly* added in his countable list.
   WM: Do paths have sex together? Even only those ending by zeros?
   [...] When you consider my proof (A), then you see that I add
always exatly one path.
   V: The way you do it prohibits any path from having a path with
infinitely many branches in the opposite direction from the ones you
add.
   CO: I think now that under the most natural way of understanding
what he is saying, this is incorrect.
   All paths will occur in the final tree (the union of what you get
by unioning all the infinite paths he adds).
   V: How does one get a path with infinitely many 1's (right
branchings) by unioning sets of paths none of which have infinitely
many 1's?
   WM: How do you get an infinite set by unioning infinitely many
FISONs?
   V: Your 'constructive' generation of such a tree always omits most
paths. You can constructively generate complete trees to any finite
depth, but   not a complete infinite tree.
   WM: No, it need not. There is a countable set of paths, such that
its union yields the entire tree, and thus all paths.
   V: Such a union may 'cover' all nodes without 'covering' all
paths.
   WM: That is a word! Let me put it as: Paths have souls. It is not
sufficient to cover all their bones (nodes).
   I am collecting reasons why I am in error. Now I have already
three:
1) SUM 0 = infinite.
2) Paths reproduce themselves (even those with same endings).
3) Paths have souls.
   I will ask my students and other bright people whether finished
infinity is worth to be paid by swallowing at least one of these
snags.
[Calvin Ostrum, Virgil, "The complete infinite binary tree has only
countably many infinite paths", sci. logic, 27. 3. 2009]

Gruß, WM



Es gibt gedanklich Dinge die mit den Gedanken der Vernunft gedacht
werden können
aber nicht im Universum existieren.
Diese Dinge sind vernünftigerweise nach einer Logik aufgebaut.
Sind sie dazu noch wiederspruchfrei logisch aufgebaut desto besser.
Vielleicht gibt es auch Gedanken die nur in unendlich vielen
nacheinander durchführbaren
GedankenSchritten erfolgen müssen, um einen Beweis zu führen.
Aber auch solche gedanklichen Dinge werden nach gàngiger Mathematiker
Lehrmeinung
als existierend abgenickt.

Wie sieht der Sachverhalt aber aus, wenn diese Gedankengànge nur noch
von 1-2 Mathematikoryphàen
auf der Welt gehandhabt werden können ?
Akzeptieren andere Mathematiker diese als existierend ?

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