Re: abundance of irrationals!)

In article <1116089133.695757.49600@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

> Virgil wrote:
>
> > > The set of nodes is countable. The set of nodes is larger than the
> set
> > > of paths.
>
> > For an "unbounded" binary tree in which no path has a last node, it
> is
> > no longer the case that the set of nodes is "larger' than the set of
> > (maximal unbounded) paths.
>
> And how and where, do you think, the "overtaking" in cardinality takes
> place?
> >
> > Each node in such a tree is the terminal node to some path, so the
> nodes
> > correspond to the set of terminating paths, which do form a countable
>
> > set, but the set of non-terminating paths correspond to the set of
> > non-terminating strings of binary digits, which a diagonal
> construction
> > can easily show not to be countable.
> >
> > Given any listing of such strings of characters,
> > with each b_ij either "0" or "1":
> >
> > 1 : b_1 = b_1_1 b_1_2 b_1_3 ...
> > 2 : b_2 = b_2_1 b_2_2 b_2_3 ...
> > 3 : b_3 - b_3_1 b_3_2 b_3_3 ...
> >
> > so that b_m_n denotes the n'th character in the m'th string, for each
> m
> > and n in N.
> >
> > Chose string c with c
> _n from {""0","1"} so that c_n =/= b_n_n, and we
> > have a path (or string) not in that list.
> >
> > Thus every list is incomplete.
>
> Wrong again. In List 2, there *are* all possible bit sequences.

The only "list 2" I have seen presented has only one bit sequence of
alternating "0"s and "1"s. If I am misinterptreting it, give me a
reference to it or present it again to show me where I am worng.

> In order to form your antidiagonal number, you need too many digits.

I only need 1 digit for each n in N. How does that make too many? Isn't
that how many infinite strings of digits have?

> You see it easily by the following finite list:
>
> .00
> .01
> .10
> .11
>
> There is no square from. Therefore there is no diagonal which could
> cross all lines.

Any terminating sequence of binary digits can be arbitrarily extended
with zeros without changing the values, so the above is equivalent to

..0000
..0100
..1000
..1100

which is in square form.
>
> By the way, do you know why Cantor's theorem fails?

As it doesn't, I don't.
>
> Look at the new thread "CANTOR's theorem"
>
> Regards, WM
.