mm-4079
===
Subject: Re: Cayley tables of finite loops of small order
> We are developing algorithms for determining properties of loops and
> quasigroups of small order. We are interested in obtaining Cayley
> tables of loops with the following properties (defined by identities):
> (1) C loops: ((xy)y)z = x((yy)z)
> (2) A_m loops: x((yx)(zx)) = ((xy)(xz))x
> (3) RIFF loops: (xy)(z(xy)) = ((x(yz))x)y
snip
I have added your property tests to my Mathematica package
GroupMLHoop.nb,
http:library.Wolfram.com/infocenter/MathSource/4894/
and included them in Ex. 28a of Section 4.4 of GroupLoopDemo.nb,
testing a number of tables with different properties. They all pass
the RIFF test; some pass and some fail the other two tests.
Because my main interest is in signed tables (multiplication
tables for algebras, in which some products are signed, e.g. Complex
with ixi=-1, Quaternion with ixj=-k), the tests are extended to cope
with signed entries in the Cayley table.
All my loops pass the RIFF test, possibly because my tables are all
quasigroups with the sorted elements (the unit) in the first row and
column (I call them ordered quasigroups).
My generalized Cayley-Dickson procedure cd[{gamma__}] creates many
ordered quasigroups, in addition to the conservative loops for which
it was designed. E.g.:-
Incantation size C loop A_m loop Associative
cd[-3., 1.,{2}] 8 yes yes yes C4C2
cd[-3.,-1.,{2}] 8 no yes no
cd[ 3.,-1, {2}] 8 no no no
cd[ 1, 1, {3}] 12 yes yes yes C3K
cd[ 1, -1, {3}] 12 no no no
cd[-1, -1, {3} 12 yes no no
cd[-1.,-1.,{3} 12 no yes no
Can you supply a reference for your identities, and say why they are
of interest? I am looking for an identity that is diagnostic for
Frobenius's conservative property, Det[A] Det[B]= +or- Det[AB];
Moufang Loop iff conservative appears to be true, but this does not
extend to signed tables.
Roger Beresford.
Studious of elegance and ease, Myself alone I seek to please. (John
Gay.)
===
Subject: Re: Cayley tables of finite loops of small order
> We are developing algorithms for determining properties of loops and
> quasigroups of small order. We are interested in obtaining Cayley
> tables of loops with the following properties (defined by identities):
> (1) C loops: ((xy)y)z = x((yy)z)
> (2) A_m loops: x((yx)(zx)) = ((xy)(xz))x
> (3) RIFF loops: (xy)(z(xy)) = ((x(yz))x)y
> snip
> I have added your property tests to my Mathematica package
> GroupMLHoop.nb,
> http:library.Wolfram.com/infocenter/MathSource/4894/
Apologies - my revisions have not yet appeared there. I did not expect
the moderator to be faster than Mathematica! Please have patience.
Roger Beresford.
More haste, less speed!
===
Subject: Re: Does this torus embed in 3-space?
Well, I found out my abstract torus with 3 holes does embed in 3-space!
It is called Dyck's regular map, described in 1888; a beautiful, very
symmetric embedding was found by Ulrich Brehm in 1987 (Mathematika, 34
(1987) 229-236. Also a very symmetric polyhedron (in 3-space) with 3
holes and only
10 vertices is described in the same issue, p. 237. And a polyhedron of
genus 4 with 11 vertices by Bokowski and Brehm, Geometriae Dedicata 29
(1989) 53-64, and a polyhedron with 7 triangles at a vertex,
24 vertices and 3 holes, by Schulte and Wills, J. London Math. Society
(2) 32 (1985) p. 539.
Apparently tori with small numbers of vertices *are* mostly
realizable in 3-space,
so much so that the first abstract polyhedron found that was not
embeddable in 3-space had as its edges the complete graph on 12 vertices,
with *6* holes (Bokowski and de Oliveira)
> Torus with 8 triangles at each vertex, 3 holes and 12 vertices:
> (1 5 6)
> (1 6 7)
> (1 7 8)
> (1 8 9)
> (1 9 10)
> (1 10 11)
> (1 11 12)
> (1 12 5)
> (2 11 10)
> (2 10 5)
> (2 5 12)
> (2 12 7)
> (2 7 6)
> (2 6 9)
> (2 9 8)
> (2 8 11)
> (3 10 9)
> (3 9 12)
> (3 12 11)
> (3 11 6)
> (3 6 5)
> (3 5 8)
> (3 8 7)
> (3 7 10)
> (4 9 6)
> (4 6 11)
> (4 11 8)
> (4 8 5)
> (4 5 10)
> (4 10 7)
> (4 7 12)
> (4 12 9)
===
Subject: Re: Domino tilings of a 3-dimensional lattice
danfux@my-deja.com (Dan) schreef:
> This link :
> http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cg
> i?Anum=A004003
> gives a formula for the number of domino tilings of 2n x 2n lattice .
> Is there also a familiar formula for the number of domino tilings of
> 2n x 2n x 2n lattice ?
Well, I'm not really an expert on this matter, but here's my
2 (euro)cents.
Tiling a lattice with dominoes is the dimer model. This is a
lattice model (with short-range interactions) from statistical
mechanics. You are asking for the partition function of this
model in a specified three-dimensional geometry. Finding exact
expressions for partition functions (and other properties) of
such models is a field of its own.
In one dimension solving such a model is fairly trivial. For
instance, counting the domino tilings of a 2n x k lattice for
some small value of k should pose no difficulties.
In two dimensions only special models appear to be solvable.
Solvability of a model is usually related the existence of a
solution to the Yang-Baxter Equation. The dimer problem on
a planar lattice was one of the first lattice models that was
solved exactly (this was before the time of the YBE).
In more than three dimensions there seem to be very few models
that can be solved exactly.
Moreover, you are interested in the partition function for a
finite geometry. For many solvable (two-dimensional) models,
the partition function has only been found in the thermodynamic
limit, that is, when the lattice becomes infinitely large. For
your (three-dimensional) case, denote the number of tilings of
a 2n x 2n x 2n lattice by Z(N) with N=(2n)^3. The quotient
(log Z(N))/N will have a finite limit as n and hence N tend to
infinity. I expect that even for this limit value no closed
form, or anything near, is known.
Alain.
===
Subject: Re: Factors of cyclotomic polynomials plus 1
> Let Phi(n) be the n-th cyclotomic polynomial.
> Algebraic factors of Phi(n)-1 follow well know patterns.
> However, is anything known about the algebraic factorisation of Phi(n)+1?
...
> There's no concrete pattern, as far as I can see.
But my sight was at the time limited.
Les Reid offlist volunteered the following rule :
Phi(4) | Phi(n)+1 iff n = 2^a . p^b . q^c
where
a in {0,1};
b,c >= 1;
p!=q primes ==3(4)
With an appreciation that at least 5 parameters might be required, I worked
out the following (using the noble art of trial and error).
Phi(8) | Phi(n)+1 iff n = 2^a . p^b . q^c . r^d
where
a in {0,1};
b,c,d >= 1;
p,q,r distinct primes;
p == 7(8);
q,r == 3(4)
or n = 2^a . p^b . q^c
where
a in {0,1};
b,c >= 1;
p prime ==5(8);
q prime ==7(8)
or n = 2^2 . p^b . q^c
where
b,c >= 1;
p!=q primes ==3(4)
I intend to try to work out equivalent rules for Phi(12) and beyond, and I
hope that a more general formula can be found. (Note, for example, that the
third rule for Phi(8) is the same as the rule for Phi(4) with the power of
2 bumped up to 2.) If anyone can provide any other input, I'd be most
grateful.
Both Les and I noticed that all known factors of Phi(n)+1 are themselves
cyclotomic, are there any known reasons for this?
Phil
===
Subject: Re: Factors of cyclotomic polynomials plus 1
> Both Les and I noticed that all known factors of Phi(n)+1 are themselves
> cyclotomic, are there any known reasons for this?
There appear to be many counter-examples to this. The smallest seems to be
n
= 21. I'm depending on the following Maple computation which should be
clear
even if you are not familiar with Maple.
> with(numtheory):
> cyclotomic(21,x)+1;
x^12-x^11+x^9-x^8+x^6-x^4+x^3-x+2
> factor(%);
(x^2+1)*(x^10-x^9-x^8+2*x^7-2*x^5+x^4+2*x^3-2*x^2-x+2)
# The only n such that phi(n) = 10 are 11 and 22 as shown by Maple's
# inverse of phi function:
> invphi(10);
[11, 22]
# But as the following shows neither the cyclotomic polynomial for 11
# nor 12 is equal to the second factor above:
> cyclotomic(11,x);
x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1
> cyclotomic(22,x);
x^10-x^9+x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1
--Edwin Clark
===
Subject: Re: Factors of cyclotomic polynomials plus 1
>> Both Les and I noticed that all known factors of Phi(n)+1 are themselves
>> cyclotomic, are there any known reasons for this?
>There appear to be many counter-examples to this. The smallest seems to be
n
>= 21. I'm depending on the following Maple computation which should be
clear
>even if you are not familiar with Maple.
You hardly need Maple for this. Since the constant term of Phi(n)+1 (for
n > 1) is 2, it obviously has at least some roots that are not on the unit
circle, and thus at least one factor that has such roots and so is not
cyclotomic.
Of course Les and Phil know this, so I'm assuming that it is not what
they really meant. I think the real question is whether, after dividing
out any cyclotomic factors, the remaining non-cyclotomic polynomial is
irreducible over the rationals.
===
Subject: Re: Factors of cyclotomic polynomials plus 1
> Both Les and I noticed that all known factors of Phi(n)+1 are
themselves
> cyclotomic, are there any known reasons for this?
>>There appear to be many counter-examples to this. The smallest seems to 
be
n
>>= 21. I'm depending on the following Maple computation which should be
clear
>>even if you are not familiar with Maple.
> You hardly need Maple for this. Since the constant term of Phi(n)+1 (for
> n > 1) is 2, it obviously has at least some roots that are not on the 
unit
> circle, and thus at least one factor that has such roots and so is not
> cyclotomic.
> Of course Les and Phil know this, so I'm assuming that it is not what
> they really meant. I think the real question is whether, after dividing
> out any cyclotomic factors, the remaining non-cyclotomic polynomial is
> irreducible over the rationals.
Yes, sorry, that is indeed exactly what we meant.
Caldwell, Unique (Period) Primes and factorization of cyclotomic
polynomials minus one (1997), has provided the best insight so far.
Therein it is proved that if Phi(x) | Phi(n)+r for r real,
then x==n or x|phi(n). It then derives a series of, IIRC, 6 rules
for deciding which of these Phi(x) divide Phi(n)-1.
My aim would be to codify the rules for the +1 side, of course.
Incidentally, brute force has provided several very large algebraic
factorisations on the +1 side. E.g. x such that Phi(x)|Phi(n)+1
n=5929, phi(n)=4620, x = 4 12 28 44 84 132 308 924
n=6125, phi(n)=4200, x = 8 40 280 1400
It appears that the largest proportion of Phi(n)+1 that can be so factored
is 1/6th. When put in the context of trying to prove the primality of
Phi(n,a) probable primes (the direction from which I've approached the
whole problem) this means that any algebraic factors thus found can
only contribute as helpers to a Pocklington-based proof, and never form
the basis for a Lehmer-Lucas-based proof.
Unfortunately I know of no PrPs which are currently just short of a KP/BLS
factorisation limit that could benefit from these helper factors, and thus
in practical terms this exercise appears to be somewhat useless. I might
contrive such a prime simply so that the data doesn't go to waste.
Phil
===
Subject: Re: gamma function
>can anybody tell me how to differentiate or integrate a gamma
>function?
> The derivative of Gamma(x) is Psi(x) Gamma(x); Psi is sometimes
> called the digamma function, and is defined (surprise!) by
> Psi(x) = Gamma'(x)/Gamma(x). The antiderivatives of Gamma(x)
> do not have a closed-form expression AFAIK (and as far as Maple
> and Mathematica know).
Gamma(y) = int_0^infty e^{-x} x^{y-1} dx
Gamma'(y) = int_0^infty e^{-x} x^{y-1} log x dx
holds for real y > 0. What about other y?
GC
===
Subject: Re: gamma function
>>can anybody tell me how to differentiate or integrate a gamma
>>function?
>> The derivative of Gamma(x) is Psi(x) Gamma(x); Psi is sometimes
>> called the digamma function, and is defined (surprise!) by
>> Psi(x) = Gamma'(x)/Gamma(x). The antiderivatives of Gamma(x)
>> do not have a closed-form expression AFAIK (and as far as Maple
>> and Mathematica know).
>Gamma(y) = int_0^infty e^{-x} x^{y-1} dx
>Gamma'(y) = int_0^infty e^{-x} x^{y-1} log x dx
>holds for real y > 0. What about other y?
One can use the other representations, or the
recursion formula.
For example,
Gamma(y) = prod (1+1/n)^y/(1+y/n)/y,
which is valid for all y other than non-positive integers,
with (1+1/n)^y = exp(y*ln(1+1/n)).
This gives
Psi(y) = -1/y + sum (ln(1+1/n) - 1/(n+y)).
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Deptartment of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: gamma function
>Gamma(y) = int_0^infty e^{-x} x^{y-1} dx
>Gamma'(y) = int_0^infty e^{-x} x^{y-1} log x dx
>holds for real y > 0. What about other y?
Note that for x > 0, |x^(y-1)| = x^Re(y-1). The integrals
converge for Re(y) > 0, uniformly on Re(y) >= epsilon for any
epsilon > 0. They diverge at 0 for Re(y) < 0 (Re(y)=0 is more
delicate, and I'm not quite sure what happens there). Being
limits of analytic functions, uniform on compacta, the integrals are
analytic functions on Re(y) > 0. The second is the derivative
of the first on the positive real axis, and therefore on all of
Re(y) > 0.
Alternatively, you could use the Lebesgue Dominated
Convergence Theorem applied to difference quotients
(Gamma(y+h)-Gamma(y))/h, together with a suitable estimate.
===
Subject: Linear equations in a Boolean ring
In a Boolean ring B, how can one solve the equation
a_1*x_1+a_2*x_2+...+a_n*x_n=b
where each a_i is a given element in B, b is a given element in B,
and one wants to find values for the x_1, x_2,...,x_n in B?
===
Subject: Markov chain question
This is a question on the relationship between, the zero and non-zero
elements of the transition probability matrix and the corresponding
transient and absorbing states.
Let T_a and T_b be the transition probability matrices for two finite state
discrete Markov chains A and B, respectively.
We are given that
a) i(T_a) = i(T_b), where i(X) is the indicator matrix of X
i.e., i(X) has 1.0 entries wherever X has non-zero entries and has 0.0
entries wherever X has zero entries.
b) T_a^n converges as n -> infinity. Let the limiting matrix be Ta^{inf}
Now, I think I can show that T_b^n also converges. However, I am trying to
find out if i(Ta^inf) = i(Tb^inf).
I have gotten this far:
Since i(T_a) = i(T_b), we have i(T_a^2) = i(T_b^2) and so on ...
I can show that for any finite k, i(T_a^k) = i(T_b^k),
but does this property hold in the limit?
In other words, if i(T_a) = i(T_b), then do A and B have the same transient
and absorbing states?
Kumar
===
Subject: Re: Markov chain question
>This is a question on the relationship between, the zero and non-zero
>elements of the transition probability matrix and the corresponding
>transient and absorbing states.
>Let T_a and T_b be the transition probability matrices for two finite
state
>discrete Markov chains A and B, respectively.
>We are given that
>a) i(T_a) = i(T_b), where i(X) is the indicator matrix of X
>i.e., i(X) has 1.0 entries wherever X has non-zero entries and has 0.0
>entries wherever X has zero entries.
>b) T_a^n converges as n -> infinity. Let the limiting matrix be Ta^{inf}
>Now, I think I can show that T_b^n also converges. However, I am trying to
>find out if i(Ta^inf) = i(Tb^inf).
>In other words, if i(T_a) = i(T_b), then do A and B have the same
transient
>and absorbing states?
For finite Markov chains, yes. The irreducible classes, transient,
recurrent and absorbing states, the periods of the classes (and in
particular whether the limiting matrix exists, as that is true if
and only if all recurrent classes are aperiodic) are all determined
by what you call the indicator matrix of the transition matrix. This
is, or should be, explained by most texts that deal with finite Markov
chains.
===
Subject: Re: Neighbor numbers forms a subset closed under bitwise OR
operation
> Suppose I have a line of binary number as below:
> {000 001 011 010 111 110 101}
Notice that each time you add a 1 at place k, you start with
(2^k - 1) and start going down:
0, 1, 3, 2, 7, 6, 5...
I'd try that idea. But do not be too sure... :) it is just a guess.
0, 1, 3, 2, 7, 6, 5, 15, 14, 13, 12, 11, 10, 9, 8, ...
(I HAVE NOT CHECKED even this example, but if this does not help then
it must be quite complicated).
Must have something to do with your M, depending on N, etc...
Best wishes,
Pedro.
===
Subject: Paper published by Algebraic and Geometric Topology
The following paper has been published:
Algebraic and Geometric Topology
URL:
http://www.maths.warwick.ac.uk/agt/AGTVol3/agt-3-24.abs.html
Title:
Fixed point data of finite groups acting on 3--manifolds
Author(s):
Peter E. Frenkel
Abstract:
We consider fully effective orientation-preserving smooth actions of a
given finite group G on smooth, closed, oriented 3-manifolds M. We
investigate the relations that necessarily hold between the numbers of
fixed points of various non-cyclic subgroups. In Section 2, we show
that all such relations are in fact equations mod 2, and we explain
how the number of independent equations yields information concerning
low-dimensional equivariant cobordism groups. Moreover, we restate a
theorem of A. Szucs asserting that under the conditions imposed on a
smooth action of G on M as above, the number of G-orbits of points x
in M with non-cyclic stabilizer G_x is even, and we prove the result
by using arguments of G. Moussong. In Sections 3 and 4, we determine
all the equations for non-cyclic subgroups G of SO(3).
Secondary: 57R85
Keywords:
3-manifold, group action, fixed points, equivariant cobordism
Author(s) address(es):
Department of Geometry, Mathematics Institute
Budapest University of Technology and Economics, Egry J. u. 1.
1111 Budapest, Hungary
Email: frenkelp@renyi.hu
===
Subject: Paper published by Geometry and Topology
The following paper has been published:
URL:
http://www.maths.warwick.ac.uk/gt/GTVol7/paper14.abs.html
Title:
Precompactness of solutions to the Ricci flow in the absence of injectivity
radius estimates
Author(s):
David Glickenstein
Abstract:
Consider a sequence of pointed n-dimensional complete Riemannian
manifolds {(M_i,g_i(t), O_i)} such that t in [0,T] are solutions to
the Ricci flow and g_i(t) have uniformly bounded curvatures and
derivatives of curvatures. Richard Hamilton showed that if the initial
injectivity radii are uniformly bounded below then there is a
subsequence which converges to an n-dimensional solution to the Ricci
flow. We prove a generalization of this theorem where the initial
metrics may collapse. Without injectivity radius bounds we must allow
for convergence in the Gromov-Hausdorff sense to a space which is not
a manifold but only a metric space. We then look at the local geometry
of the limit to understand how it relates to the Ricci flow.
Secondary: 53C21
Keywords:
Ricci flow, Gromov-Hausdorff convergence
Received: 9 December 2002
Proposed: Gang Tian
Seconded: John Morgan, Leonid Polterovich
Author(s) address(es):
Department of Mathematics, University of California, San Diego
9500 Gilman Drive, La Jolla, CA 92093-0112, USA
Email: glicken@math.ucsd.edu
===
Subject: Scanned Abramowitz and Stegun web site
Epigone-thread: trehtrahblee
A mirror site for Abramowitz and Stegun exists at:
http://jove.prohosting.com/~skripty/
- Tom Willis
===
Subject: which linearly ordered sets are sets of reals?
I have linearly ordered sets (of equivalence classes of probability
distributions, as it happens) and I want to know whether each item can
be labelled with a real number while preserving the ordering. I think
this will be true for all the cases I'm interested in, and so a
sufficient but not necessary condition may well be good enough.
topology might characterise subsets of the reals, but I don't know if
this is true, or if there are other useful characterisations.
-thomas
===
Subject: Re: which linearly ordered sets are sets of reals?
> I have linearly ordered sets (of equivalence classes of probability
> distributions, as it happens) and I want to know whether each item
> can be labelled with a real number while preserving the ordering. I
> think this will be true for all the cases I'm interested in, and so a
> sufficient but not necessary condition may well be good enough.
> topology might characterise subsets of the reals, but I don't know
> if this is true, or if there are other useful characterisations.
> -thomas
Regarding separable -- yes this is true.
The condition is: There is a countable set C such that there is an
element of C between any two elements of the basic set. Clearly this is
necessary. And if the condition is true then first map all elements of C
to rationals while preserving the order, and that will define an
order-preserving mapping of the whole basic set.
N
===
Subject: Re: which linearly ordered sets are sets of reals?
> I have linearly ordered sets (of equivalence classes of probability
> distributions, as it happens) and I want to know whether each item can
> be labelled with a real number while preserving the ordering. I think
> this will be true for all the cases I'm interested in, and so a
> sufficient but not necessary condition may well be good enough.
> topology might characterise subsets of the reals,
I theorem of Cantor says any two densely ordered sets
without endpoints that are countable are order-isomorphic.
Densely ordered means ordered in such a way that between
any two elements there is another. The proof is by the
back-and-forth method: since both sets are countable,
count them in one-to-one fashion, thus:
a_1, a_2, a_3, ...
b_1, b_2, b_3, ...
Now the problem is to construct an order-isomorphism, i.e.,
an order-preserving bijection. First let a_1 correspond
to b_1. Then let a_2 correspond to b_i, where i is the
smallest index among {2, 3, 4, ... } such that b_1 < or < b_i
according as a_1 < or > a_2. Then find the smallest index j
such that b_j has not been assigned a mate among the a's.
Let it corrspond to the a of smallest index among the
unmated a's such that everything is in the right order.
Keep going back and forth: To the first unmated b assign
the first unmated a that is compatible in the sense that
this correspondence is to be an strictly increasing, then
to the first unmated a assigne the first umated b that is
compatible, etc.
For example, in this way you can construct a strictly
increasing bijection between the rationals and the real
algebraic numbers.
A corollary is that having a countable dense subset
suffices for your purpose. -- Mike Hardy
===
Subject: Re: which linearly ordered sets are sets of reals?
>> I have linearly ordered sets (of equivalence classes of
>> probability distributions, as it happens) and I want to know
>> whether each item can be labelled with a real number while
>> preserving the ordering. I think this will be true for all the
>> cases I'm interested in, and so a sufficient but not necessary
>> condition may well be good enough.
>> order topology might characterise subsets of the reals, but I don't
>> know if this is true, or if there are other useful
>> characterisations.
>Regarding separable -- yes this is true.
>The condition is: There is a countable set C such that there is an
>element of C between any two elements of the basic set. Clearly this
>is necessary. And if the condition is true then first map all
>elements of C to rationals while preserving the order, and that will
>define an order-preserving mapping of the whole basic set.
Not necessary. Let the set be S = [0,1] union [2,3]. Tho there's no
c in C that's between 1 and 2, nevertheless S embeds into the reals.
Please do show, given C subset S with that condition, how its order
extends to S, that S has cardinality c and order embeds into the reals.
----
===
Subject: Re: which linearly ordered sets are sets of reals?
>> I have linearly ordered sets (of equivalence classes of
>> probability distributions, as it happens) and I want to know
>> whether each item can be labelled with a real number while
>> preserving the ordering. I think this will be true for all the
>> cases I'm interested in, and so a sufficient but not necessary
>> condition may well be good enough.
>> order topology might characterise subsets of the reals, but I don't
>> know if this is true, or if there are other useful
>> characterisations.
>Regarding separable -- yes this is true.
>The condition is: There is a countable set C such that there is an
>element of C between any two elements of the basic set. Clearly this
>is necessary. And if the condition is true then first map all
>elements of C to rationals while preserving the order, and that will
>define an order-preserving mapping of the whole basic set.
> Not necessary. Let the set be S = [0,1] union [2,3]. Tho there's no
> c in C that's between 1 and 2, nevertheless S embeds into the reals.
> Please do show, given C subset S with that condition, how its order
> extends to S, that S has cardinality c and order embeds into the reals.
> ----
By between I meant non-strict between (i.e. between-or-equal). So
here is the result in more formal terms.
The given linearly ordered set is S. The condition is:
There is a countable subset C of S such that for each x,y in S, x < y,
there exists z in C such that x<= z <= y.
The claim is: This condition holds for S if and only if S is order
isomorphic to a subset of reals.
N