Re: Two results of set geometry

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
{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.

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.

MoeBlee




.

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)
  • Merry Christmas!
    ... dictionaries and every variant possibility has a separate "word" entry. ... The byte string of the "word", whose length is specified by a four ... match is found for a source byte sequence in the dictionary. ...
    (rec.arts.sf.written)
  • Merry Christmas! Linux RULES! New applications to develop!
    ... dictionaries and every variant possibility has a separate "word" ... Each entry in the dictionary contains: ... The byte string of the "word", whose length is specified by a four ... addresses whose entry is selected by the first byte of the sequence. ...
    (comp.os.linux.misc)
  • Re: user defined function that converts string to float
    ... > I need user defined function that converts string to float in c. ... initial, possibly empty, sequence of white-space characters (as ... point character, then an optional exponent part as defined in ... then a nonempty sequence of hexadecimal digits ...
    (comp.lang.c)
  • Re: Coin tossing guessing strategy...
    ... You need a precise string in exactly one order, ... with respect to sequences of n flips, is to note that if some sequence ... In fact, assuming a fair coin, all 10 coin sequences are equally ... heads and tails. ...
    (sci.math)