A lesson for set theorists

The complete infinite binary tree contains, by definition, all real

numbers between 0 and 1 as infinite paths, i. e., as infinite

sequences { 0, 1 }^N of bits.

0,

/ \

0 1

/ \ / \

0 1 0 1

/

0 ...

The set { a_k | k in N } of nodes a_k of the tree is countable.

a0,

/ \

a1 a2

/ \ / \

a3 a4 a5 a6

/

a7 ...

If every node a_k is covered by an arbitrary infinite path p_k

containing that node, then no further node is remaining to distinguish

any further path from the paths p_k of the countable set P = { p_k | k

in N }. This proves that it is impossible to distinguish more than

countably many paths by infinite sequences of bits.

FIRST COUNTER ARGUMENT

All nodes of a path are covered by other paths. Therefore covering all

nodes of the binary tree by a countable set of paths does not show

that all paths of the binary tree form a countable set.

Here is an example for the counter argument: We can cover every node

of path p = 0.111... by means of covering every path p# of the set

{

0.000...

0.1000...

0.11000...

0.111000...

...}

No. Every path p# of the form 0.111...111000... has at least one node

0 at a position where p = 0.111... has a node 1. So when covering, in

the binary tree, all nodes of path p none of the paths p# has been

covered. In other words: every p# has an uncovered node after all

nodes of p have been covered.

Vice versa, when all p# have been covered, then p has not yet been

covered. Every path p# deviates from p leaving at least one of its

nodes 1 uncovered.

Of course this is only an existence proof. After covering all paths

p#, an uncovered node of p cannot be named. But if no such node

existed, p would have been written down (= covered in the tree) by

writing down (= covering in the tree) all paths p#.

Then, however, Cantor's diagonal proof would fail because then the

possible antidiagonal p = 0.111... of the above list was in the list.

SECOND COUNTER ARGUMENT

In order to save Cantor's diagonal proof, it must be claimed that

although all nodes of p are covered by the set of paths p#, there is

no path p# = p. That means:

I) All nodes 1 of p = 0.111... are in the list of all paths p# above.

II) No line p# contains all these nodes 1, i.e., the limit p is not in

the list.

These two assertions cannot be satisfied simultaneously.

Proof by complete induction. If the existence of all nodes 1 in

different lines of the list is claimed, then at least two lines must

be shown which, with respect to their contents of nodes 1, cannot be

replaced by one line. This is obviously impossible.

On the contrary, it can be proved by complete induction, hence for

every line of the list, that all nodes which are in that line and in

all previous lines, also are in the next line. (Complete induction

holds for all natural numbers, though not for the set of them. But

that ist not required here.)

Proof by construction of an infinite set. Construct the above

list, but remove always line number n after having constructed the

next line number n + 1. Then the list shrinks to a single line but

this single line contains the same as the list because never anything

has been removed that had not been added before to an existing line.

Therefore this single line contains all nodes 1. It is the limit p 0.111... If, however, line number n is not removed after line number n

+ 1 has been constructed, there cannot be an effect of this marginalia

on line n + 1 and later lines.

If a single line p = 0.111... can be constructed at all (otherwise it

was impossible to construct a complete Cantor list) then this line

must also be in the complete list.

When imaging all nodes 1 in one single line, then nobody will complain

about all nodes existing in one single line. When imaging all nodes 1

written as in the above list, then the existence of all nodes 1 in a

single line is denied.

This result appears paradoxical. But the only paradoxical is the

assumption that infinity can be finished and that thousands of rather

intelligent people have been believing that for more than 100 years.

THIRD COUNTER ARGUMENT

All this binary tree twaddle does not apply because real numbers are

only limits and Dedekind-cuts.

Fine. Limits and Dedkind-cuts have to be defined. The number of all

definitions is countable, in fact it is finite and will remain so for

ever.

[This lesson was inspired by a discussion with William Hughes in

sci.math on June 9, 2010.]

http://groups.google.com/group/sci....ing=d&
Gruß, WM

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