A problem of set geometry
WM
pairwise surjective and injective is called a trijection. So the
trijectiion is a set of triples (a_i, b_i, c_i) where a_i is in A, b_i
is in B, and c_i is in C.
Consider the following infinite matrix
1111111...
1100000...
1110000...
1111000...
...
For all n in omega including n = omega there is a trijection between
the initial segments of the first column, the diagonal and the first
line
(a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n)
such that all elements belonging to an initial segment of column,
diagonal, and line are 1's.
But there is no such trijection for n in omega including omega between
the first column, the diagonal and the n-th line
(a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_n1, .., a_nn)
Because there is no line (except the first) with omega 1's. The
diagonal gets its omega 1's from omega lines none of which contains
omega 1's. The question is how can this be understood geometrically?
Regards, WM
MoeBlee
> Definition: A ternary relation on three sets A, B, and C
A relation on three sets? By your following comments, I take it you
mean a relation that is a subset of AxBxC.
> which is
> pairwise surjective and injective is called a trijection.
If the relation is on AxBxC and is an injection, then it is a function
from AxB into C. So, what does "pairwise surjective and injective"
mean here?
Please give a clear and rigorous mathematical definition of 'R is a
trijection with respect to A, B and C'. That is a definition of the
form:
R is a trijection with respect to A, B and C <-> [fill in here]
where [fill in here] is a formula using only previously defined terms
and whose only free variables are among 'R', 'A', 'B' and 'C'.
Let's take that first. Then, once your definition is satisfactory,
I'll look at the next step in your argument.
> So the
> trijectiion is a set of triples (a_i, b_i, c_i) where a_i is in A, b_i
> is in B, and c_i is in C.
MoeBlee
WM
> On 21 Jul., 08:55, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Definition: A ternary relation on three sets A, B, and C
>
> A relation on three sets? By your following comments, I take it you
> mean a relation that is a subset of AxBxC.
A ternary relation T on three sets A, B, and C is a set T of ordered
triples t_n such that for any t_n in T there exist a_i in A, b_j in B,
and c_k in C such that
t_n = (a_i, b_j, c_k).
> > which is
> > pairwise surjective and injective is called a trijection.
>
> If the relation is on AxBxC and is an injection, then it is a function
> from AxB into C. So, what does "pairwise surjective and injective"
> mean here?
A ternary relation T on three sets A, B, and C is called a
trijection, if it is pairwise injectice and surjective, i.e., if for
a_i, a_i' in A, b_j, b_j' in B, and c_k, c_k' in C
1) every binary subrelation of the ternary relation is an injective
function
a_i T b_j & a_i' T b_j ==> a_i = a_i',
a_i T b_j & a_i T b_j' ==> b_j = b_j',
b_j T c_k & b_j' T c_k ==> b_j = b_j',
b_j T c_k & b_j T c_k' ==> c_k = c_k,'
c_k T a_i & c_k' T a_i ==> c_k = c_k'.
c_k T a_i & c_k T a_i' ==> a_i = a_i,'
and
2) every element a_i of A, every element b_j of B, and every element
c_k of C belongs to at least one element t_n of T.
So we can summarize: A set T is a trijection on three sets A, B, and C
if all elements t_n of T are ordered triples (a_n, b_n, c_n), and if
every a_i in A, every b_j in B, and every c_k in C belongs to one and
only one such triple t_n.
In the second case to be considered, the three sets A, B, and C
consist of all initial segments of the first column, of all initial
segments of the diagonal, and of all sequences of 1's in the lines,
respectively, of the matrix given in my first posting. When talking
about sequences, it is convenient to express the trijection by
A <--> B <--> C or by
a_n <--> b_n <--> c_n such that we get
(a_11, ..., a_n1) <--> (a_11, ..., a_nn) <--> (a_11, ..., a_1n).
Remark: The statement "for n in omega including omega" does not imply
a last element in N but is a simple notation for the limiting
"infinite sequence including all natural numbers as indexes", i.e., a
sequence with aleph_0 elements.
Regards, WM
MoeBlee
trees seem to come down to your principle that if all proper initial
segments of a sequence are finite, then the sequence is finite.
I AGREE that set theory is in contradiction with that principle. So IF
your latest argument here is going to end up coming down to that
principle or your principle that there do not exist "actually"
infinite sets, then PLEASE, save me the trouble and just tell me now
whether you're going to be invoking either of those principles, since
if either of those are principles that are going to be used in your
argument here, then there is no need for me to review this argument as
I already AGREE that either of those principles contradict set theory.
Again, for special emphasis, PLEASE save me the trouble right now if
you are going to use either of those principles (or even some other
principle other than first order logic with identity applied to the
axioms of ZFC, as you might as well state now what such principle(s)
are).
On Jul 25, 1:09 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 24 Jul., 20:27, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On 21 Jul., 08:55, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > Definition: A ternary relation on three sets A, B, and C
>
> > A relation on three sets? By your following comments, I take it you
> > mean a relation that is a subset of AxBxC.
>
> A ternary relation T on three sets A, B, and C is a set T of ordered
> triples t_n such that for any t_n in T there exist a_i in A, b_j in B,
> and c_k in C such that
> t_n = (a_i, b_j, c_k).
Just as I said:
T is a ternary relation on A, B, and C <-> T subset of (AxB)xC.
We don't need subscripts to say that.
> > > which is
> > > pairwise surjective and injective is called a trijection.
>
> > If the relation is on AxBxC and is an injection, then it is a function
> > from AxB into C. So, what does "pairwise surjective and injective"
> > mean here?
>
> A ternary relation T on three sets A, B, and C is called a
> trijection, if it is pairwise injectice and surjective, i.e., if for
> a_i, a_i' in A, b_j, b_j' in B, and c_k, c_k' in C
> 1) every binary subrelation of the ternary relation is an injective
> function
> a_i T b_j & a_i' T b_j ==> a_i = a_i',
> a_i T b_j & a_i T b_j' ==> b_j = b_j',
> b_j T c_k & b_j' T c_k ==> b_j = b_j',
> b_j T c_k & b_j T c_k' ==> c_k = c_k,'
> c_k T a_i & c_k' T a_i ==> c_k = c_k'.
> c_k T a_i & c_k T a_i' ==> a_i = a_i,'
There you write as if T is a binary reltion on AxB then a binary
relation on BxC, then a binary relation on CxA. That makes no sense
since T is a relation on (AxB)xC.
> and
> 2) every element a_i of A, every element b_j of B, and every element
> c_k of C belongs to at least one element t_n of T.
>
> So we can summarize: A set T is a trijection on three sets A, B, and C
> if all elements t_n of T are ordered triples (a_n, b_n, c_n), and if
> every a_i in A, every b_j in B, and every c_k in C belongs to one and
> only one such triple t_n.
If I'm not mistaken, this is what is specified by your summary.
T is a trijection on A, B, and C
<->
T is a ternary relation on A, B, and C &
AxeA EyeB EzeC(<x y z>eT & AveB AweC(<x v w>eT -> (v=y & w=z))) &
AyeB ExeA EzeC(<x y z>eT & AjeA AweC(<j y w>eT -> (j=x & w=z))) &
AzeC ExeA EyeB(<x y z>eT & AjeA AveB(<j v z>eT -> (j=x & v=y)))
> In the second case to be considered, the three sets A, B, and C
> consist of all initial segments of the first column, of all initial
> segments of the diagonal, and of all sequences of 1's in the lines,
> respectively, of the matrix given in my first posting.
That matrix was illustrated by:
1111111...
1100000...
1110000...
1111000...
...
Normally, that illustration would be sufficient for us to take it as
either a function on NxN onto {0 1} or as a denumerable sequence of
denumerable binary sequences, but in other threads you've given very
odd definitions of what a matrix is. So I want to be clear just what
YOU intend the above to be an illustration of. Since you seem inclined
toward NxN, I'll suggest:
M = {<<n k> v> | neN &
keN &
ve{0 1} &
(n=0 -> v=1) &
(~n=0 -> (v=0 <-> k>n))}.
If you mean something else, then please define it.
> When talking
> about sequences, it is convenient to express the trijection by
> A <--> B <--> C or by
> a_n <--> b_n <--> c_n such that we get
> (a_11, ..., a_n1) <--> (a_11, ..., a_nn) <--> (a_11, ..., a_1n).
I don't know what you mean '<-->' to stand for.
> Remark: The statement "for n in omega including omega" does not imply
> a last element in N but is a simple notation for the limiting
> "infinite sequence including all natural numbers as indexes", i.e., a
> sequence with aleph_0 elements.
Using an expression "n in omega including omega" is HORRIBLY
MISLEADING (at best horribly confusing) if all you want to say is that
a sequence has all natrual numbers as indexes. I won't go along with
such horribly misleading terminology as "omega in omega". There is no
need to directly clash with the ordinary meaning of "in", especially
in context of what is in omega so that we say "omega in omega", just
to say that a sequence is on N. If you just want to say that a
sequence has all natural numbers as indexes, then it suffices just to
say "the sequence is denumerable" or, more specifically, "the sequence
is on N" or "the sequence is an omega-sequence" (where 'is an omega-
sequence' or 'is an w-omega sequence' are defined as 'is a function on
N'). And if you want to use the word 'limit' in the sense of 'union'
then that's okay as long as we are clear that that is the precise
sense being used.
MoeBlee
WM
> In another thread, I suggested that your arguments about matrices and
> trees seem to come down to your principle that if all proper initial
> segments of a sequence are finite, then the sequence is finite.
>
> I AGREE that set theory is in contradiction with that principle.
Do you allude to the fact that the observable (and utilizable)
universe is finite? I can assure you that I shall not infer any
consequences from that fact for this discussion. You could have
convinced yourself by a quick glance at the matrix presented in my
first contribution (which you repeated below). Its columns and rows
and diagonal are actually infinite. Even the first line contains
infinitely many 1's. All I use is the principle that actually infinite
sets exist and the principle of bijection.
>
> > > If the relation is on AxBxC and is an injection, then it is a function
> > > from AxB into C. So, what does "pairwise surjective and injective"
> > > mean here?
>
> > A ternary relation T on three sets A, B, and C is called a
> > trijection, if it is pairwise injectice and surjective, i.e., if for
> > a_i, a_i' in A, b_j, b_j' in B, and c_k, c_k' in C
> > 1) every binary subrelation of the ternary relation is an injective
> > function
> > a_i T b_j & a_i' T b_j ==> a_i = a_i',
> > a_i T b_j & a_i T b_j' ==> b_j = b_j',
> > b_j T c_k & b_j' T c_k ==> b_j = b_j',
> > b_j T c_k & b_j T c_k' ==> c_k = c_k,'
> > c_k T a_i & c_k' T a_i ==> c_k = c_k'.
> > c_k T a_i & c_k T a_i' ==> a_i = a_i,'
>
> There you write as if T is a binary reltion on AxB then a binary
> relation on BxC, then a binary relation on CxA.
I hoped to express in an understandable manner that for a trijection
the use of (AxB)xC is equivalent to the use of Ax(BxC) and is also
equivalent to the use of AxBxC.
> That makes no sense since T is a relation on (AxB)xC.
This sentence is certainly wrong or shows one-way thinking. You _can_
understand a ternary relation as a function with domain in AxB and
range in C, but that is by no means obligatory. In particular for a
trijection as I described above, there is no reason for any asymmetry
between A, B and C.
>
> > and
> > 2) every element a_i of A, every element b_j of B, and every element
> > c_k of C belongs to at least one element t_n of T.
>
> > So we can summarize: A set T is a trijection on three sets A, B, and C
> > if all elements t_n of T are ordered triples (a_n, b_n, c_n), and if
> > every a_i in A, every b_j in B, and every c_k in C belongs to one and
> > only one such triple t_n.
>
> If I'm not mistaken, this is what is specified by your summary.
>
> T is a trijection on A, B, and C
> <->
> T is a ternary relation on A, B, and C &
> AxeA EyeB EzeC(<x y z>eT & AveB AweC(<x v w>eT -> (v=y & w=z))) &
> AyeB ExeA EzeC(<x y z>eT & AjeA AweC(<j y w>eT -> (j=x & w=z))) &
> AzeC ExeA EyeB(<x y z>eT & AjeA AveB(<j v z>eT -> (j=x & v=y)))
I would prefer to separate the variables by commas and to use
parentheses instead of the larger-than symbols, because multitudes of
the latter appear already as indicators of authorship on my screen.
>
> > In the second case to be considered, the three sets A, B, and C
> > consist of all initial segments of the first column, of all initial
> > segments of the diagonal, and of all sequences of 1's in the lines,
> > respectively, of the matrix given in my first posting.
>
> That matrix was illustrated by:
>
> 1111111...
> 1100000...
> 1110000...
> 1111000...
> ...
>
> Normally, that illustration would be sufficient for us to take it as
> either a function on NxN onto {0 1} or as a denumerable sequence of
> denumerable binary sequences, but in other threads you've given very
> odd definitions of what a matrix is.
Perhaps you misunderstood the following point (which is important for
the further argument):
Independent of the values of the elements of a matrix or /before/
assigning any values to the elements of the matrix, we can conclude
that {a_ik | i = k & i, k e N} c {a_ik | i >= k & i, k e N} c {a_ik |
i, k e N} where "c" denotes "is a subset of". That means, the set of
positions of diagonal elements form a set which is a subset of the
positions of the lower left triangle (including the diagonal) which
again is a subset of the set of all positions of elements.
> So I want to be clear just what
> YOU intend the above to be an illustration of. Since you seem inclined
> toward NxN, I'll suggest:
>
> M = {<<n k> v> | neN &
> keN &
> ve{0 1} &
> (n=0 -> v=1) &
> (~n=0 -> (v=0 <-> k>n))}.
>
> If you mean something else, then please define it.
I think the definition of what a matrix is should be clear to every
participant of sci.math.research. Your definition describes just the
matrix under discussion, except that in my trijection I started to
enumerate the elements of the matrix by 1 and not by 0. For matrices I
prefer to have a first line and first column but not a zeroth one.
>
> > When talking
> > about sequences, it is convenient to express the trijection by
> > A <--> B <--> C or by
> > a_n <--> b_n <--> c_n such that we get
> > (a_11, ..., a_n1) <--> (a_11, ..., a_nn) <--> (a_11, ..., a_1n).
>
> I don't know what you mean '<-->' to stand for.
It is simply a more transparent notation for the triple
((a_11, ..., a_n1), (a_11, ..., a_nn), (a_11, ..., a_1n)).
>
> > Remark: The statement "for n in omega including omega" does not imply
> > a last element in N but is a simple notation for the limiting
> > "infinite sequence including all natural numbers as indexes", i.e., a
> > sequence with aleph_0 elements.
>
> Using an expression "n in omega including omega" is HORRIBLY
> MISLEADING (at best horribly confusing)
It is a definition like others more, a definition which, for instance,
Cantor repeatedly used in his paper Ueber eine Eigenschaft des
Inbegriffes aller reellen algebraischen Zahlen, Crelles Journal f.
Mathematik Bd. 77, S. 258 - 262 (1874), a_oo, b_oo etc. for lim{n -->
oo} a_n. But if you find this abbreviation too confusing, then we can
reformulate, although somewhat more clumsy:
There is a trijection between the initial segments of the first
column, the diagonal and the first
line, for A n in N
(a_11, ..., a_n1) <--> (a_11, ..., a_nn) <--> (a_11, ..., a_1n)
including the complete first column, the complete diagonal and the
complete first line
(a_11, a_21, a_31, ...) <--> (a_11, a_22, a_33, ...) <--> (a_11, a_12,
a_13, ...)
such that all elements a_ik are 1's.
But there is no such trijection for n in omega between the first
column, the diagonal and the n-th line
(a_11, ..., a_n1) <--> (a_11, .., a_nn) <--> (a_n1, ..., a_nn)
including the complete first column, the complete diagonal and another
than the first line
(a_11, a_21, a_31, ...) <--> (a_11, a_22, a_33, ...) <--> ?
such that all elements a_ik are 1's.
Because there is no line (except the first one) with omega 1's. The
WM
> trees seem to come down to your principle that if all proper initial
> segments of a sequence are finite, then the sequence is finite.
>
> I AGREE that set theory is in contradiction with that principle.
Do you allude to the fact that the observable (and utilizable)
universe is finite? I can assure you that I shall not infer any
consequences from that fact for this discussion. You could have
convinced yourself by a quick glance at the matrix presented in my
first contribution (which you repeated below). Its columns and rows
and diagonal are actually infinite. Even the first line contains
infinitely many 1's. All I use is the principle that actually infinite
sets exist and the principle of bijection.
> > > > Definition: A ternary relation on three sets A, B, and C
I hoped to express in an understandable manner that for a trijection
the use of (AxB)xC is equivalent to the use of Ax(BxC) and is also
equivalent to the use of AxBxC.
> That makes no sense since T is a relation on (AxB)xC.
This sentence is certainly wrong or shows one-way thinking. You _can_
understand a ternary relation as a function with domain in AxB and
range in C, but that is by no means obligatory. In particular for a
trijection as I described above, there is no reason for any asymmetry
between A, B and C.
>
> > and
> > 2) every element a_i of A, every element b_j of B, and every element
> > c_k of C belongs to at least one element t_n of T.
>
> > So we can summarize: A set T is a trijection on three sets A, B, and C
> > if all elements t_n of T are ordered triples (a_n, b_n, c_n), and if
> > every a_i in A, every b_j in B, and every c_k in C belongs to one and
> > only one such triple t_n.
>
> If I'm not mistaken, this is what is specified by your summary.
>
> T is a trijection on A, B, and C
> <->
> T is a ternary relation on A, B, and C &
> AxeA EyeB EzeC(<x y z>eT & AveB AweC(<x v w>eT -> (v=y & w=z))) &
> AyeB ExeA EzeC(<x y z>eT & AjeA AweC(<j y w>eT -> (j=x & w=z))) &
> AzeC ExeA EyeB(<x y z>eT & AjeA AveB(<j v z>eT -> (j=x & v=y)))
I would prefer to separate the variables by commas and to use
parentheses instead of the larger-than symbols, because multitudes of
the latter appear already as indicators of authorship on my screen.
>
> > In the second case to be considered, the three sets A, B, and C
> > consist of all initial segments of the first column, of all initial
> > segments of the diagonal, and of all sequences of 1's in the lines,
> > respectively, of the matrix given in my first posting.
>
> That matrix was illustrated by:
>
> 1111111...
> 1100000...
> 1110000...
> 1111000...
> ...
>
> Normally, that illustration would be sufficient for us to take it as
> either a function on NxN onto {0 1} or as a denumerable sequence of
> denumerable binary sequences, but in other threads you've given very
> odd definitions of what a matrix is.
Perhaps you misunderstood the following point (which is important for
the further argument):
Independent of the values of the elements of a matrix or /before/
assigning any values to the elements of the matrix, we can conclude
that {a_ik | i = k & i, k e N} c {a_ik | i >= k & i, k e N} c {a_ik |
i, k e N} where "c" denotes "is a subset of". That means, the set of
positions of diagonal elements form a set which is a subset of the
positions of the lower left triangle (including the diagonal) which
again is a subset of the set of all positions of elements.
> So I want to be clear just what
> YOU intend the above to be an illustration of. Since you seem inclined
> toward NxN, I'll suggest:
>
> M = {<<n k> v> | neN &
> keN &
> ve{0 1} &
> (n=0 -> v=1) &
> (~n=0 -> (v=0 <-> k>n))}.
>
> If you mean something else, then please define it.
I think the definition of what a matrix is should be clear to every
participant of sci.math.research. Your definition describes just the
matrix under discussion, except that in my trijection I started to
enumerate the elements of the matrix by 1 and not by 0. For matrices I
prefer to have a first line and first column but not a zeroth one.
>
> > When talking
> > about sequences, it is convenient to express the trijection by
> > A <--> B <--> C or by
> > a_n <--> b_n <--> c_n such that we get
> > (a_11, ..., a_n1) <--> (a_11, ..., a_nn) <--> (a_11, ..., a_1n).
>
> I don't know what you mean '<-->' to stand for.
It is simply a more transparent notation for the triple
((a_11, ..., a_n1), (a_11, ..., a_nn), (a_11, ..., a_1n)).
>
> > Remark: The statement "for n in omega including omega" does not imply
> > a last element in N but is a simple notation for the limiting
> > "infinite sequence including all natural numbers as indexes", i.e., a
> > sequence with aleph_0 elements.
>
> Using an expression "n in omega including omega" is HORRIBLY
> MISLEADING (at best horribly confusing)
It is a definition like others more, a definition which, for instance,
Cantor repeatedly used in his paper Ueber eine Eigenschaft des
Inbegriffes aller reellen algebraischen Zahlen, Crelles Journal f.
Mathematik Bd. 77, S. 258 - 262 (1874), a_oo, b_oo etc. for lim{n -->
oo} a_n. But if you find this abbreviation too confusing, then we can
reformulate, although somewhat more clumsy:
There is a trijection between the initial segments of the first
column, the diagonal and the first
line, for A n in N
(a_11, ..., a_n1) <--> (a_11, ..., a_nn) <--> (a_11, ..., a_1n)
including the complete first column, the complete diagonal and the
complete first line
(a_11, a_21, a_31, ...) <--> (a_11, a_22, a_33, ...) <--> (a_11, a_12,
a_13, ...)
such that all elements a_ik are 1's.
But there is no such trijection for n in omega between the first
column, the diagonal and the n-th line
(a_11, ..., a_n1) <--> (a_11, .., a_nn) <--> (a_n1, ..., a_nn)
including the complete first column, the complete diagonal and another
than the first line
(a_11, a_21, a_31, ...) <--> (a_11, a_22, a_33, ...) <--> ?
such that all elements a_ik are 1's.
Because there is no line (except the first one) with omega 1's. The
Virgil
WM <muec...@rz.fh-augsburg.de> wrote:
> Definition: A ternary relation on three sets A, B, and C which is
> pairwise surjective and injective is called a trijection. So the
> trijectiion is a set of triples (a_i, b_i, c_i) where a_i is in A, b_i
> is in B, and c_i is in C.
>
> Consider the following infinite matrix
>
> 1111111...
> 1100000...
> 1110000...
> 1111000...
> ...
>
> For all n in omega including n = omega there is a trijection between
> the initial segments of the first column, the diagonal and the first
> line
Except that if n is IN omega it cannot equal omega, which destroys the
whole edifice.
>
> (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n)
>
> such that all elements belonging to an initial segment of column,
> diagonal, and line are 1's.
>
> But there is no such trijection for n in omega including omega between
> the first column, the diagonal and the n-th line
But there is no need for one either, since for all m, n IN omega, and
all m <= n, a_mn = a_nn = a_nm.
WM's problem is that he wants endless sequences to have two ends, when
they don't.
Virgil
WM <muec...@rz.fh-augsburg.de> wrote:
> Do you allude to the fact that the observable (and utilizable)
> universe is finite?
That is something which one must assume, as it can never be proven
beyond the possibility of falsification. There is nothing about the
physical world that can be proved beyond the possibility of
falsification, at least if one accepts scientific methods rather than
mere faith as one's standard of proof.
WM
> In article <1185033318.897116.179...@w3g2000hsg.googlegroups.com>,
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > Definition: A ternary relation on three sets A, B, and C which is
> > pairwise surjective and injective is called a trijection. So the
> > trijectiion is a set of triples (a_i, b_i, c_i) where a_i is in A, b_i
> > is in B, and c_i is in C.
>
> > Consider the following infinite matrix
>
> > 1111111...
> > 1100000...
> > 1110000...
> > 1111000...
> > ...
>
> > For all n in omega including n = omega there is a trijection between
> > the initial segments of the first column, the diagonal and the first
> > line
>
> Except that if n is IN omega it cannot equal omega, which destroys the
> whole edifice.
Please excuse the ambiguity of language. I did not want to express
that omega is included in omega but that omega is included in the
trijection.
>
> > (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n)
>
> > such that all elements belonging to an initial segment of
column,
> > diagonal, and line are 1's.
>
> > But there is no such trijection for n in omega including omega between
> > the first column, the diagonal and the n-th line
>
> But there is no need for one either, since for all m, n IN omega, and
> all m <= n, a_mn = a_nn = a_nm.
>
> WM's problem is that he wants endless sequences to have two ends, when
> they don't.-
The complete set of natural numbers has omega elements. It is a valid
statement in set theory that a set with omega elements (like the set
of all positions of the first column) can be extended to a set of
order omega + 1 by adding one element. This feature can be used to
investigate the trijection between the positions of 1's of the first
column, the diagonal, and the lines in the matrix M
1000...
11000...
111000...
...
If there is a trijection, then it should remain a trijection after
extending the sets involved by one element each. Extending the first
column of M by one element we obtain a sequence of order type omega +
1 (because it has the order type omega). After adding one 1 to each
sequence of 1's in the lines of M no sequence of 1's has order type
omega + 1 (because each one has a finite order type). This can be
repeated infinitely often, such that the order type of the first
column becomes omega*2 (notation from Cantor, 1895) while the order
type of the set of lines cannot surpass omega. For the diagonal the
case remains undecided. But whatever the diagonal may be, there is no
trijection between the sequences of 1's in the first column, the
diagonal, and the lines - not finally and, therefore, also not
initially.
Regards, WM
WM
> In article <1185616827.525976.322...@k79g2000hse.googlegroups.com>,
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > Do you allude to the fact that the observable (and utilizable)
> > universe is finite?
>
> That is something which one must assume, as it can never be proven
> beyond the possibility of falsification. There is nothing about the
> physical world that can be proved beyond the possibility of
> falsification, at least if one accepts scientific methods rather than
> mere faith as one's standard of proof.
Many mathematicians support the view that questions involving the
hardware of mathematics do not belong to the realm of mathematics.
Therefore this topic does not fit extremely well into
sci.math.research. Nevertheless I will try to briefly answer your
concern.
Let us define a mathematician as a device which, by means of CPU and
memory, is connecting sequences of symbols using fixed rules like
(FOP) logic to obtain sequences of symbols. Then there is undisputable
physical evidence that the spatial and temporal volume at the
mathematician's disposal is finite - however extended the
mathematician may be (without loosing the connection of its parts).
Therefore only finite sequences of symbols can be treated and only
finite sequences of symbols can be derived.
This statement does not depend on open questions concerning the
structure and maximum density of matter or whether or not the universe
is closed. It merely relies on some laws of relativity theory, quantum
mechanics, and thermodynamics (information processing) which are so
fundamental (like the lever principle, the shape of our earth or the
origin of species) that they cannot be circumvented, even by most
advanced technologies.
Regards, WM
Virgil
WM <muec...@rz.fh-augsburg.de> wrote:
> On 10 Aug., 19:54, Virgil <vir...@comcast.net> wrote:
> > In article <1185033318.897116.179...@w3g2000hsg.googlegroups.com>,
> >
> >
> >
> >
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > Definition: A ternary relation on three sets A, B, and C which is
> > > pairwise surjective and injective is called a trijection. So the
> > > trijectiion is a set of triples (a_i, b_i, c_i) where a_i is in A, b_i
> > > is in B, and c_i is in C.
> >
> > > Consider the following infinite matrix
> >
> > > 1111111...
> > > 1100000...
> > > 1110000...
> > > 1111000...
> > > ...
> >
> > > For all n in omega including n = omega there is a trijection between
> > > the initial segments of the first column, the diagonal and the first
> > > line
> >
> > Except that if n is IN omega it cannot equal omega, which destroys the
> > whole edifice.
>
> Please excuse the ambiguity of language. I did not want to express
> that omega is included in omega but that omega is included in the
> trijection.
If you mean that the unbounded initial segments are inclused in the
"trijection" along with the bounded ones, that is quite obvious, but
equally irrelevant.
> >
> > > (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n)
> >
> > > such that all elements belonging to an initial segment of
> column,
> > > diagonal, and line are 1's.
> >
> > > But there is no such trijection for n in omega including omega between
> > > the first column, the diagonal and the n-th line
> >
> > But there is no need for one either, since for all m, n IN omega, and
> > all m <= n, a_mn = a_nn = a_nm.
> >
> > WM's problem is that he wants endless sequences to have two ends, when
> > they don't.-
>
>
> The complete set of natural numbers has omega elements. It is a valid
> statement in set theory that a set with omega elements (like the set
> of all positions of the first column) can be extended to a set of
> order omega + 1 by adding one element.
Not at all. It is only true that a well ordered set of omega elements
after having a new element appended to its ordering is a new set of of
order type (omega+ 1). Anything stated less precisely need not be hold.
> This feature can be used to
> investigate the trijection between the positions of 1's of the first
> column, the diagonal, and the lines in the matrix M
>
> 1000...
> 11000...
> 111000...
> ...
>
> If there is a trijection, then it should remain a trijection after
> extending the sets involved by one element each. Extending the first
> column of M by one element we obtain a sequence of order type omega +
> 1 (because it has the order type omega). After adding one 1 to each
> sequence of 1's in the lines of M no sequence of 1's has order type
> omega + 1 (because each one has a finite order type). This can be
> repeated infinitely often, such that the order type of the first
> column becomes omega*2 (notation from Cantor, 1895) while the order
> type of the set of lines cannot surpass omega.
It is as easy to append a 1 to any endless line (as filled in to
endlessness by spaces or zeros) as to any endless column or diagonal.
If every column stars in line 1, then there is a space in every line for
every column, whatever may be filling that space, and appending to such
a line must be done following all such spaces.
And all WM's horses and all WM's men can't put his broken argument back
together again.
Virgil
WM <muec...@rz.fh-augsburg.de> wrote:
> On 10 Aug., 20:00, Virgil <vir...@comcast.net> wrote:
> > In article <1185616827.525976.322...@k79g2000hse.googlegroups.com>,
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > Do you allude to the fact that the observable (and utilizable)
> > > universe is finite?
> >
> > That is something which one must assume, as it can never be proven
> > beyond the possibility of falsification. There is nothing about the
> > physical world that can be proved beyond the possibility of
> > falsification, at least if one accepts scientific methods rather than
> > mere faith as one's standard of proof.
>
>
> Many mathematicians support the view that questions involving the
> hardware of mathematics do not belong to the realm of mathematics.
The "hardware" of mathematics is formal logic.
> Therefore this topic does not fit extremely well into
> sci.math.research. Nevertheless I will try to briefly answer your
> concern.
>
> Let us define a mathematician as a device which, by means of CPU and
> memory, is connecting sequences of symbols using fixed rules like
> (FOP) logic to obtain sequences of symbols.
Improper definition. Mathematicians only derive their sequences of
symbols from given sequences of symbols. Never from nothing.
Other than the validity of some form of logic, no mathematician declares
absolute truths independent of anything else, only relative ones,
deduced from various sets of assumptions.
It is only non-mathematicians like WM who claim knowledge of such
absolutes.
> Then there is undisputable
> physical evidence that the spatial and temporal volume at the
> mathematician's disposal is finite - however extended the
> mathematician may be (without loosing the connection of its parts).
> Therefore only finite sequences of symbols can be treated and only
> finite sequences of symbols can be derived.
"Set S is infinite" is a finite set of symbols.
>
> This statement does not depend on open questions concerning the
> structure and maximum density of matter or whether or not the universe
> is closed. It merely relies on some laws of relativity theory, quantum
> mechanics, and thermodynamics (information processing) which are so
> fundamental (like the lever principle, the shape of our earth or the
> origin of species) that they cannot be circumvented, even by most
> advanced technologies.
Since no mathematician requires an infinite set of symbols in order to
define or discuss infinite sets, WM's analysis is irrelevant.
It is mathematicians' finite sets of symbols that seems to be upsetting
WM.
[ Moderator's note:
I'd like to bring this thread to a close. Whether the
"hardware" is logic or the mathematician's brain is irrelevant.
I hope we can all agree that mathematicians are finite, and
they make statements that consist of finite strings of symbols,
but those statements may assert that something is infinite
or discuss the properties of such an object. And anyone who
can't agree on that is welcome to take the discussion elsewhere.
-RI ]