Re: Cantor and the binary tree

In sci.math, mueckenh@xxxxxxxxxxxxxxxxx wrote:
> Yes, for any pair. Now consider all pairs. If E' is countable, then the
> set of all pairs is countable, hence the set of all bunches is
> countable, namely P.

Right. Then E' is equal to the set of nodes (or branches), because at
any node, a 'first difference' between some two paths might occur.
Right?

>> > Let P be the cardinal number of the set bunches
>>
>> So P is the cardinal number of the set of nodes, right? Since each
>> bunch is uniquely defined by a single node.
>
> Yes, but better say branches.

So E' = P.

>> > and let R be the cardinal number of the set of single paths which is
>> > a subset of the set of bunches.
>>
>> No, it isn't. Each bunch contains infinitely many paths, so there
>> cannot be a bunch with just a single path in it. Each path is /in/
>> some bunch, but that is something completely different. The 'bunches'
>> group the paths together in great big lumps, and it is not very
>> surprising that there are many more lumps than paths.
>>
>> > Then we have E >= E' >= P >= R
>>
>> Only the first of these equalities is correct. The second is
>> obviously untrue, since E' = 1 and P is very big.
>
> E' >= P is the number of bunches.
> R is a subset of P.

You are confusing the cardinalities with the sets in your notation,
but I guess your meaning is clear.

R is /not/ a subset of P, because P is a set of bunches, and R is a
set of paths. And, as I said, no bunch contains a single path, so
there is no trivial embedding. /Every/ bunch contains infinitely many
paths.

> If a binary representation is uniquely defined before a bit number oo
> appears, then all the paths separate in the tree.

Well, any /two/ paths seperate at some point. But at /any/ point in
the tree, there are still infinitely many paths together, because it
is alway possible to branch at a later stage.

> You cannot imagine
> that, but the tree is infinitely long. There is a unique prescription
> for 1/3 = 0.010101..., for instance, as a path in the tree.

Sure, but that is something completely different.


Jan

.

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