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

## Lesen sie die antworten