Conference rusure::math

>> What positive integers are reachable in this fashion?
        
        i.e., as [ 2^n / 3^m ]		n and m *positive* integers
        
        First, why do you reject the m = 0 case?
        
        Anyway, to reach the integer k, you need to have
        
        	ln k  <=  n ln 2 - m ln 3  <  ln (k+1)
        
        Let epsilon be less than both ln k and ln (k+1) - ln k.
        Suppose there is a rational approximation p/q to ln 3 / ln 2
        which is slightly greater than ln 3 / ln 2 but is so
        close to it that 0 < p ln 2 - q ln 3 < epsilon.  Then there
        will be an integral multiple L of p ln 2 - q ln 3 such
        that
        	ln k  <=  Lp ln 2 - Lq ln 3  <  ln (k+1)
        
        Then n = Lp and m = Lq are a solution for k.
        
        It only remains to show that the condition:
        
        	Suppose there is a rational approximation
        	p/q to ln 3 / ln 2 which is slightly greater
        	than ln 3 / ln 2 but is so close that
        	0 < p ln 2 - q ln 3 < epsilon.
        
        can be satisfied.
        
        I think a combination of Louieville's [sp?] theorem about
        rational approximation to algebraic numbers, and
        continued fraction theory, combine (the first may not be
        needed, as ln 3 / ln 2 is not algebraic, but I threw it in
        in case it describes the counterexamples) to show that
        there are arbitrarily large p and q above or below ln 3 / ln 2
        such that | p/q - ln 3 / ln 2 | is less than a fixed
        constant multiple of 1/q^2.  If true this is sufficient
        to imply the condition above.  But I don't have my
        references here in the office and don't care to derive
        the necessary results in this reply.
        
        Dan

    All positive integers are reachable.
    
    Proof:  Let p = 2^n/3^m.  Then let q = log(p) = n*log2 - m*log3.
    
    We will show that the set generated by all the q's is dense in the
    real.  Consider the sequence: a(n) = n*log2 mod log3 for n >= 0.  Then
    since log2/log3 is irrational, the set {a(n): n >= 0} is dense in the
    interval [0, log3].  This means the set of all q's is dense in the
    real.  This implies that the set of all p's is dense in the positive
    half of the real line.  The desired result follows immediately from that.
    
    Eugene

Both of those proofs are hard for me to follow.  The second one seems to
depend on knowledge of exactly what "dense" means (no wisecracks please),
and in the first it looks like Dan "reduced" the problem to another
stipulation that may not be closer to home than the original.

Dan, I'm not quite clear how you chose epsilon to be both less than ln of
the first number and less than the difference of the lns of the two numbers.
Actually, for k greater than some very small integer, isn't any epsilon that
is less than the difference of the lns of k and k+1 going to be WAY less
than ln k anyway ?

By the way, the reason I demanded m>0 is so k's of the form 2^k don't get
a "free ride" by trivially dispensing with them by using 3^0 in the
denominator.

/Eric

        re .3,
        
>> Dan, I'm not quite clear how you chose epsilon to be both less than ln of
>> the first number and less than the difference of the lns of the two numbers.
        
	How is easy.  If 0 < a and 0 < b, then there are positive
        epsilon less than both a and b.  :-)  As for why ...
        
>> Actually, for k greater than some very small integer, isn't any epsilon that
>> is less than the difference of the lns of k and k+1 going to be WAY less
>> than ln k anyway ?
        
        ... just think of it as the mathematical equivalent to
        wearing both a belt and suspenders. :-)  I just wanted to
        make sure that an integral multiple of epsilon fell in
        [ ln k, ln k+1 ) so I specified being less the width of
        the range, as well as less than its start.
        
        Dan
        

    re .3,
    
    "A set X is dense in the set Y" means that if you chose any point y in
    Y and any interval around y (no matter how small), there is an x in X
    that is in the interval.  One example is that the set of rationals is
    dense in the set or the reals.  In .2, I proved that the set generated
    by 2^n/3^m is dense in the positive reals.  In other words, these
    numbers "jam" the entire positive real line.  A much stronger result
    than asked.
    
    Eugene

RE: .5 (Eugene)

    Its been a long time since Abstract Algebra in college but I thought
    that a set A is said to be dense in a set B iff there is an element of
    A between any two elements of B; and a set is said to be simply dense
    if it is dense in itself.  Is the definition you give equivalent,
    under some appropriate conditions, to this definition or have I just
    misremembered.  Your definition sounds like some kind of continuity
    property, and would seem to require at least a local distance metric.
    Density, as I (mis?) remember it is a property of a suitably *ordered*
    set (i.e, "in between" needs to be defined, but not "close" or even
    "half way between").

				    Topher

RE: .6 (me)

>    A between any two elements of B; and a set is said to be simply dense
							      ^^^^^^

    Rereading this I see the possibility of confusion.  "Simply" is a term
    in the meta language (English) not part of the name of the mathematical
    concept, I should of said...

*    ...; and a set is simply said to be dense

					    Topher

    
    I don't think ordering has anything to do with it. You can say, for
    instance, that the set of points with rational coordinates is dense in
    R2. What you need is some concept of "near", ie a topology.
    
    It's 12 years or so since I last looked at any of this, but I'd suggest
    the following definition:
    
    A set A is dense in a topological space B if A is a subset of B and any
    neighborhood of any point of B contains a point of A.
    
    (Oh, yes, I didn't define "neighborhood". Let's try this:
    
    A neighborhood of a point x of a topological space B is a subset of B
    that contains an open set containing x.
    
    )
    
    Andy.

    Now you've gone and done it!  You've got to define "topological space"
    if you're going to define "neighborhood".
    
    Though I wonder how much help Eric will get out of this.  He's not dense.
    
    :-)

RE .8 (Andy)

    As I said, I've never been particularly deep into this, and am rusty on
    what I did know, but if you define density in terms of topological
    neighborhood (which doesn't depend on a distance measure, as you point
    out), doesn't that require that you define open/closed, and doesn't
    that require a definition of boundary, and doesn't that require that an
    appropriate ordering exist?  Is there a definition for a boundary which
    does not require ordering?  What distinguishes elements in the boundary
    set from those in its complement?

    Your definition does sound like it might be provably equivalent to
    mine, though, and yet have the "flavor" (minus the analytic/distance-
    based dependency) of the definition of .5.  That resolves some of my
    confusion.

				    Topher

    re .10,
    
    The definition of "dense" has nothing to do with algebra.  It is a pure
    topological property.  However, for the purpose of this problem a much
    simpler definition will surfice.  Just think of it as something
    equivalent to "rationals are dense in the reals".
    
    Eugene

        To see the similarity in the definitions, for a linearly
        ordered set (X,<) define the order topology on X to be
        the topology generated by all of the sets {x in X | x < a}
        and { x in X | a < x} for any a in X.
        
        (The "topology generated by" a collection of subsets of X
        is all of the finite intersections and arbitrary unions
        of the empty set, X, and those subsets.)
        
        Now consider the ordered set (X,<) with the order
        topology on X, let Y be a subset of X, and see if the
        "order definition" of Y dense in X agrees with the
        topological definition.
        
        Dan

       It is very likely that all integer are reachble by Eric's 
       method.  In facts it is true if log(2)/log(3) is not rational.
       
       Let P = { n*log(2) + m*log(3) / n,m integers }
       Then P is a subgroup of R.  As any subgroup of R, it is either 
       discrete either dense.  In facts it is discrete if log(2) and 
       log(3) are integral multiples of a common divisor and dense 
       otherwise.  Thus if log(2)/log(3) is not rational, Eric's 
       method will produce every integer.
       
       Let me decipher slowly.

       
       1) About density
       
       The notion of order density and topological both exist.  There 
       is no reason to consider one prevailing the other.  In facts 
       they coincide on a set carrying both an order and a topology 
       provided tey are compatible.  'Compatible' means that all the 
       intervals ]a,b[ are open and generate the opens of the 
       topology.  It is the case for R.  Let's detail it.
       
       A subset P of R is said *order* dense if for any x<y in R, it 
       exists p in P such that x<p<y.
       
       A subset P of R is said *topologicaly* dense if its closure, 
       i.e. the smalest closed subset of R containing P, is R itself.  
       It is equivalent to say that for any point x of R and any 
       neighborhood V of x, there is at least one point which is both 
       in V and P.  Putting aside the referance to x, we see that P is 
       dense iff all open of R contains at least one point of P.
       
       The equivalence between order density and topological density 
       is almost obvious since V is open if, and only if, it contains 
       an interval ]a,b[.
       								      
       
       2) Dense subgroups.
       
       The notion of density can be localized.  P is said (locally) 
       dense around x if any neighborhood V of x contains at least one 
       point which is in both in V and in P.  In facts it is just an 
       other way of saying that x is adherent to P.  It is obvious 
       that P is dense iff it is dense everywhere.
       
       Now subgroups of a topological (or ordered) group have a 
       striking behavior: if subgroup is dense around one point only 
       it is dense everywhere.  This comes from the fact that, by 
       definition of topological (ordered) group, the translations  
       x-->x+a  are homeomorphisms (monotonous).
       
       Hence, to see that subgroup P is dense, it is only needed to 
       check that it is dense around 0.  In other words to see if for 
       all epsilon>0, there is x in P such that |x| < epsilon.
       
       
       3) Discrete subgroups.
       
       Discrete is the opposite of dense.  A subset P of a topological 
       space R is discrete if all its points are isolated.  A point x 
       is said isolated in P if its exists a neighborhood V of x which 
       has no other points that x in common with P.
       
       As before if a point of a subgroup is isolated, then by 
       translation all others are also isolated.  Therefore a subgroup 
       P is discrete in R as soon as one of its point (e.g., 0) is 
       isolated.  
       
       
       4) A subgroup of R it is either discrete either dense.
       
       Let P be a subgroup of R, and consider the smaler p>0 in P.  If 
       p doesn't exist it means that P is dense arround 0, which 
       implies that P is dense everywhere.  On the other hand if p 
       exist, then 0 is isolated, thus P is discrete.  In facts in 
       this case P = pZ = { p*n | n integer }.
       
       
       5) Back to the original problem.
       
       Consider the subgroup P = { 2^n + 3^m / n,m integers}
       
       If P is discrete then there is a real p such that P = pZ, hence 
       such that log(2) = pn and log(3) = pm.  In that case 
       log(2)/log(3) would be rational.  Taking for granted that it is 
       not, P is dense. Therefore for every x<y it exists two integers 
       n and m such that 
       
                        x < n*log(2) + m*log(3)) < y
       
       Take x and y as Dan in .1, and change the sign of m to get the 
       result.

	You start with an undefined concept called an open set.  You don't 
	need a metric for this.  Euclidean disks in R^n work fine as open sets,
	but so do lots of other things.  There are a bunch of axioms to make
	sure that they behave like we think they should (trust me).

	Now, if a point p has the property that every open set containing p
	also contains a point in the set A, then p is said to be a limit
	point of A.

	The closure of a set is the set of all of it's limit points.

	A set A is dense in B if the closure of A is B.

	Avery

	I posted .14 before I read all of .13.  It's definitions and approaches,
	e.g., for closure, are equivalent.

	-fab

    re .13,
    
    log2/log3 is irrational.
    
    Proof:  log2/log3 = log 2.  Suppose it is rational.  Then it is equal
                           3                                         (p/q)
    to p/q where p and q are integers and q > 0.  This implies that 3     = 2.
         p   q
    ==> 3 = 2 .  This can't happen since 2 and 3 are relative primes.
    
    Therefore we conclude that log2/log3 is irrational.
    
    Eugene
    

        re .14,
        
>>	The closure of a set is the set of all of it's limit points.
        
        The closure of a set is the union of the set and all of
        its limit points.
        
        Dan

    re .16,
    
    Thanks, you've closed the last open part.

    Alain

Re .17

>        The closure of a set is the union of the set and all of
>        its limit points.
        
	Yes, but every point in a set is also a limit point of the set,
	so the original definition is correct and more concise (and, by
	Ocum's razor, the prefered definition :-).

	A related concept:  One definition of the boundary of a set A is 
	the set of points such that every open set containing the point 
	contains both a point in A and a point not in A.  Clearly, every 
	point in the boundary is a limit point of A.  As I recall, except
	in certain pathological topologies and sets, the boundary of an 
	open set B is disjoint from B.

	-fab

        No, by definition p is a limit point of A if every open
        set containing p contains a point of A distinct from p.
        
        Not every member of a set is by definition a limit point
        of that set.  Think of the definition of a perfect set: a
        closed set every point of which is a limit point of the
        set.  So Cantor's set or [0,1] is perfect, while the set
        {0} union {1/n : n = 1,2,3,...} is closed but not perfect
        (these three as subsets of the Euclidean real line).
        
        Dan

	Yes, it's coming back to me.  Isolated points of a set are not 
	limit points, just as you would expect.	

	Sorry, time has given me a lobotomy.

	Avery

        What does your personal name mean?
        
        	"Irohanihohetochirinuruowakayotarezotsunenaramu..."
        
        Dan

	It isn't even math related, so it's WAY off the topic, but 
	since you asked, see the Japanese notes conference, Nihongo
	(JIT081::NIHONGO), note number 128.1 (This reply contains a 
	pointer to the conference).  It is pretty long, and the part
	describing my personal name is at the end.

	-fab