Re: Two results of set geometry

On 18 Sep., 01:15, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Sep 13, 5:22 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On 12 Sep., 23:36, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:

There is no order preserving bijection between omega
(the set of finite sequences of 1's in the rows) and omega (the set of
1's of the first column) of the following matrix:

1
11
111
...

WHAT order is not preserved by an obvious bijection between N and the
set of strings of 1 of finite length?

The normal order of natural numbers.

There is such an order preserving bijection"

For n in w, let F(n) = the n-length string of 1's.

If we take the null string to be string of 0 length, then:

length(F(n)) < length(F(m)) <-> n < m.

Please reread the thread "A
problem of set geometry". There you have put a lot of questions which
I answered there.

I gave up there as I gave up, for some time, on you when at the same
time in another thread I couldn't get you to understand that for f:NxN->{0 1}, the notation {f(<i j)> | i in N and j in N} just equals {0} or

Oh I thought you had understood my explanation. You cannot take part
in the discussion as long as you can't understand that the sets I am
dealing with are different *positions* of 1 (as I repeatedly
explained).

{1} or {0 1} (depending on f), while {<<i j> f(<i j)> | i in N and j
in N} does NOT loose the matrix. And, so excruciatingly ironically,
you came back to say that I am the one who would lose the matrix
thereby.

Then, even upon any inclination I might have had to return to your
"geometry" thread, it was closed by the moderator anyway.

That's a pity, to blame on Virgil, who started an argument that
reality does not exclude infinity. Such questions are not desired by
the moderators.

There is an infinite column of omega 1's. There is a diagonal of omega
1's which can be projected on the first column but there is no finite
sequence of 1's in any line on which the diagonal could be projected.

If you're talking about

1
11
111
1111
....

then, yes, what you just said is correct in this sense:

The first column is a denumerable sequence of 1's. The diagonal is a
denumerable seqeunce of 1's. There is a bijection between the
aforementioned sequences. And each row is a finite sequence of 1's and
there is is no finite sequence of 1's that can be bijected with a
denumerable sequence of 1's, so there is no row that can be bijected
with the first column or with the diagonal.

And no proof of a contradiction have you shown.

The contradiction is that the diagonal consists of (those elements of
the matrix which are identified by) last 1's of the rows. So the rows
are mapped on the diagonal. The diagonal with all its elements and
initial segments comes into being only by this mapping. But the
inversion of this mapping is impossible.

Regards, WM

.

Relevant Pages

  • Re: Two results of set geometry
    ... There is such an order preserving bijection" ... If we take the null string to be string of 0 length, ... 1's which can be projected on the first column but there is no finite ... The first column is a denumerable sequence of 1's. ...
    (sci.math)
  • Re: Two results of set geometry
    ... As I understand it, he doesn't believe in the reals that we would call irrationals, because they cannot be represented in any finite string. ... You claim there are more paths> than nodes, and insist that he demonstrate an injection from nodes to> paths. ... > tiny bit of reflection will confirm that this is the set of all strings> which, after some finite position, have an infinitely repeating finite> string of digits following. ... But, as I mentioned, there are plenty of> other unused nodes, such as 110, available for the bijection. ...
    (sci.math)
  • Re: Two results of set geometry
    ... Finite segments of the first column are representing the finite ... Sets can only be in bijection with sets. ... Therefore the set of rows is not sufficient to fill an infinite set. ... with *all* three possible non-empty segments of the first column. ...
    (sci.math)
  • Re: Two results of set geometry
    ... I am expanding on WM's stated tree. ... string of digits following. ... Those that are an infinitely repeating finite ... other unused nodes, such as 110, available for the bijection. ...
    (sci.math)
  • Re: Two results of set geometry
    ... So the elements of the first column are sequences of '1's? ... how can you talk about a bijection between the elements of the first ... > and the initial segments are those of the first column. ... segments and an infinite initial segment while the rows are all finite ...
    (sci.math)