Conference rusure::math

	I don't think that this works.   I don't think you've covered the
real line.

	What we need to do is to hijack part of Liouville's constructive
proof of the existence of transcendental numbers.   He had a lemma:

	If x is algebraic of degree n, then given any K > 0, and any r > n,
then the set of rationals, p/q, such that:
		| p/q - x | < K/q^r
is finite.   (p/q are not necessarily assumed to be in lowest terms here.)

	Now, the ordering of Q described in the base note essentially sets
up a function f: Q -> Z+, such that the covering interval for rational p/q
has length t/2^f(p/q), where t = epsilon.   (Here p/q *are* in lowest
terms.)

	Let me propose a particular ordering on Q.   Nothing outrageous.
Say that the rationals are 0/1 & p/q where q is a positive integer, and p
is a non-zero integer.   Let's say that that f(p/q) < f(r/s) iff
		(1) |p| + q < |r| + s
	or	(2) |p| + q = |r| + s, & q < s
	or	(3) |p| + q = |r| + s, & q = s, p > 0

	This is a total order on the integers, as can be easily seen.   The
f which is generated is quite complciated, owing to the lowest term
restriction on p/q, but what we can say is that:

	f(p/q) >= f(|p|+q-1/1) > (|p|+q-1)�

	Now, let's apply the lemma, with x = sqrt(2).   Thus n = 2, and we
will take K = t, and r = 3.   For only a finite set of rationals p/q:

	| p/q - x | < K/q^3

	Identify all such rationals (this task terminates, because it turns 
out there is an explicit upper bound on q.)   Say that there are N of them.
We compute d as the nearest that any of them comes to x, and then define
t' = min(t,d).   Now using the same ordering of the rationals as before,
but with the new scaling factor t', none of the N intervals can cover x.   The
longest interval that exists is t'/2.

	For every other rational p/q:

	| p/q - x | >= K/q^3

	= t/q^3

	> t/2^(q^2)		since 2^(q^2) > q^3

	> t/2^((|p|+q-1)^2)

	> t'/2^f(p/q)		by previous inequalities

	In summary:

	| p/q - x | > t'/2^f(p/q)

	So the gap is between p/q and x is greater than the interval, and x
is not covered.

	So x is not covered by the interval corresponding to any rational in
the sequence.   So we haven't covered R.

Regards,
Andrew.

    re .0,
    
    All you have done is showing that the set of rational numbers have zero
    measure.
    
    Eugene 

    Re: .2 Yes, but .0 also seemed to show the rationals had measure
    infinity.  Given that each rational point was surrounded by an actual
    open interval containing many irrationals, one might easily conclude
    that almost every irrational was by chance convered by an interval
    assigned to a "nearby" rational.
    
    In fact, almost all irrationals are not covered.
    
    Is there a counter-version of the problem?  Imagine trying to cover the
    irrationals with "very small intervals" (though their total measure
    will of course be infinite).  Can we cover all the irrationals and
    still not cover, oh, say a dense subset of the rationals?  I guess no.
    
      John

re .3
>    Is there a counter-version of the problem?  Imagine trying to cover the
>    irrationals with "very small intervals" (though their total measure
>    will of course be infinite).  Can we cover all the irrationals and
>    still not cover, oh, say a dense subset of the rationals?  I guess no.
    
      No, that won't work for the irrational numbers.  The fundation of 
    the proof that the rationals have zero measure is the countability of
    the rationals (as a matter of facts, all countable subsets of real
    numbers have zero measure).
    Eugene 

Reply .1 makes it too complicated.  Answer .2 is correct.  Just because you
have covered the rationals, which are dense in R1, it DOES NOT follow that you
have covered the reals.  This is somewhat counter intuitive to the uninitiated,
because it seems that every irrational has to fall into one of those intervals.
It simply means that there are uncountably many irrationals r such that any 
sequence of rationals converging to r is covered by  intervals that do not
contain r.  There is no contradiction.

I'll make the next one harder.

Avery

    re: .4 Eugene, I'm not requiring the "covering" have finite measure.

>Reply .1 makes it too complicated.  Answer .2 is correct.

	Reply .1 is at least a *proof* that you haven't covered the reals,
albeit written in assembler rather than higher-level measure theory.
Eugene's clearly hit the nail on the head in his admirably brief .2, but
he *doesn't* address the question of proof.

	I would be interested to see that expressed in measure theoretic
terms.

Regards,
Andrew-the-algebraist-playing-away-from-home-here.

re .3 .6,
>    Is there a counter-version of the problem?  Imagine trying to cover the
>    irrationals with "very small intervals" (though their total measure
>    will of course be infinite).  Can we cover all the irrationals and
>    still not cover, oh, say a dense subset of the rationals?  I guess no.
    
     Hmmm...  I didn't read the above carefully before I wrote .4 (blush blush).
On the other hand, after re-read the above, I found it uh... not that hard
to show there is no such dense set because the complement of any interval, no 
matter how small the interval is, is never dense in the original set.

re .7,

Theorem:  If S is a countable subset of R, then m(S) = 0.

Proof:  Since S is countable, S = {s }.  Let e > 0.  Let 
                                    n 
                                            -(n+1)
C = {A : A = (a  , b  )  and b  - a   < e (2      ) and s  is in (a , b ) }
      n   n    n    n         n    n                     n         n   n

                                                 oo   -(n+1)
        Obvisouly B= UA  covers S, but m(B) = e (SUM 2      ) < e
                       n                         n=1 

Since e is arbitrary small, and m(S) < m(B) < e, we conclude that 
m(S) = 0.

Eugene

>	Let me propose a particular ordering on Q.   Nothing outrageous.
>Say that the rationals are 0/1 & p/q where q is a positive integer, and p
>is a non-zero integer.   Let's say that that f(p/q) < f(r/s) iff
>		(1) |p| + q < |r| + s
>	or	(2) |p| + q = |r| + s, & q < s
>	or	(3) |p| + q = |r| + s, & q = s, p > 0

You have proved that there are reals that are not covered using this one
particular ordering.  It has to hold for an arbitrary ordering.  The fact that
there is a bijection from one ordering to any other doesn't help-- a different
order means the intervals will be arranged differently.

The point is that .0 is no proof at all.  It makes the statement that "The
rationals are dense in R1, so you have just covered the entire real line with
..."  Intuitively this makes sense, but it is pure nonsense -- there is no
proof that you have covered all reals, just because they are limit points of
the rationals.

As I said, it is "proof" by intimidation.  The counter proof is simply that it 
violates at least one of the axioms of measure theory, e.g., that the measure
of the union of two disjoint sets is the sum of their measures, or as Eugene
said, that it implies R1 has zero measure (I don't have any of my math books
here in Japan, so I can't come up with any better ones.)

However, your approach has merit as a pedagogical device because in
introductory measure theory courses when you use .0 as a proof that a countable
set has Lebesque measure zero,  there are bound to be students who think this
shows the reals also have measure zero, and don't have enough faith in the
system to accept the counter proof (so what are they doing in a measure theory
course?)   The most convincing existence proof is to come up with an example,
as you were doing in .1.  As I said, I don't think it works as is, but I am 
sure you could fix it up to work for an arbitrary ordering.

Avery

>You have proved that there are reals that are not covered using this one
>particular ordering.  It has to hold for an arbitrary ordering.

	True.   It wasn't a point which I wanted to make, because I couldn't
see how to prove the general statement.   :-).

>The counter proof is simply that it 
>violates at least one of the axioms of measure theory,

	Logically, you would also need to prove that the reals constitute
a model of the axioms of measure theory.   Granted, this is probably a
straightforward exercise, but it needs to be mentioned.
	
> As I said, I don't think it works as is, but I am 
> sure you could fix it up to work for an arbitrary ordering.

	Not as it stands.   The set of algebraics is countable.   So we
order them a_i, and then take each r_i to be sufficiently close to the
relevant a_i to cover it.   So you need to enter the transcendentals to
get the proof for an arbitrary ordering.

Regards,
Andrew.

    re: .8
    
>>    Is there a counter-version of the problem?  Imagine trying to cover the
>>    irrationals with "very small intervals" (though their total measure
>>    will of course be infinite).  Can we cover all the irrationals and
>>    still not cover, oh, say a dense subset of the rationals?  I guess no.
>    
>     Hmmm...  I didn't read the above carefully before I wrote .4 (blush blush).
>On the other hand, after re-read the above, I found it uh... not that hard
>to show there is no such dense set because the complement of any interval, no 
>matter how small the interval is, is never dense in the original set.
    
    Eugene, you can call me a thick-headed lout if you like, but I still don't
    get it.  Suppose we accept your reasoning and then apply the same logic
    back to the original problem.  Complement any covering of the rationals,
    and you have NOT covered a dense subset of the irrationals.  Isn't that
    what follows?  Doesn't this say there exists an uncountable set of reals
    (of infinite measure) that is nowhere dense?  Isn't that wrong?
    

John, Didn't mean to call you a thick head.  If I sounded I did, I apologize.
However, I think I mis-read your statement again, so it may just be me that 
qualifies to be a thick head.  Anyway, the problem remains:

>    Is there a counter-version of the problem?  Imagine trying to cover the
>    irrationals with "very small intervals" (though their total measure
>    will of course be infinite).  Can we cover all the irrationals and
>    still not cover, oh, say a dense subset of the rationals?  I guess no.

Your guess is right.  Let S be an open cover of the irrationals.  
Let T = R\S.  Let RT be the set of rationals in S.  Now let x be in R.
Let e > 0.  Since S is dense, there exists s in S such that ||s - x|| < e/100
Since S is open, let N be an open interval in S center at s of length less
than e/100.  Now N must contain a rational r.  With some calculations, you 
will get ||r - x|| < e.  Since r is in RT, RT is dense in R.

Eugene

>>    Is there a counter-version of the problem?  Imagine trying to cover the
>>    irrationals with "very small intervals" (though their total measure
>>    will of course be infinite).  Can we cover all the irrationals and
>>    still not cover, oh, say a dense subset of the rationals?  I guess no.
				 ^^^^^
				 ^^^^^

	John, what do you mean by "dense"?  If you take all of the
	rationals in [0,1], then that is a dense subset of [0,1],
	but it is not a dense subset of [0,2], and it is not a
	dense subset of the real line.  Look again at .8:

>     Hmmm...  I didn't read the above carefully before I wrote .4 (blush blush).
>On the other hand, after re-read the above, I found it uh... not that hard
>to show there is no such dense set because the complement of any interval, no 
>matter how small the interval is, is never dense in the original set.

	All he is saying is that if you are covering all of the
	irrationals, then your cover is certainly non-empty, and
	so contains at least one interval (a,b).  Then the complement
	excludes (a,b) and therefore excludes all of the rationals
	in (a,b) and therefore the complement cannot cover a dense
	subset of the rationals [i.e. of *all* of the rationals].

	If you had asked about not covering (i.e., the complement
	covering) an infinite subset of the rationals, that is
	possible:  {(n,n+1) | n in Z} covers the irrationals but
	excludes infinitely many rationals.

	I think the question you were trying to ask is, is it possible
	to have an open cover of the irrationals, for which there is
	at least one tiny little interval (a,b) such that the cover
	excludes a dense subset of [the rationals in] (a,b)?  And the
	answer is still "no", as there is an irrational x in (a,b),
	which is covered by some (c,d), and so the set excluded by the
	cover is missing all of the rationals in the little interval
	(c,d) and so isn;t dense in (a,b)

		a        c  x  d            b
			  ^^ ^^
			  ^^ ^^
		all rationals in here are covered

	So perhaps what you were really trying to ask was one of:

		is there an open cover of the irrationals, for
		which the complement (a subset of the rationals)
		has

			a limit point?

			infinitely many limit points?

			uncountably many limit points?

			a closure which contains an interval (a,b)?

	(Does anyone disagree with "yes", "yes", "???", and "no"?)

	Dan

    Dan, you are a mind-reader!  Actually it was the question of a limit
    point that got me to ask initially, but that seemed so obvious that
    I changed the inquiry on the fly to "dense", by which I meant
    "somewhere dense", i.e., a set whose closure was a finite interval (a,b).
    Bit of a humpty-dumpty act there (OK, thin-skinned, not thick-headed :-).
    
    I agree with Eugene's analysis, and Dan's excellent summary:
    
>			a limit point?
>			infinitely many limit points?
>			uncountably many limit points?
>			a closure which contains an interval (a,b)?
>	(Does anyone disagree with "yes", "yes", "???", and "no"?)
    
    I wonder if "???" is equivalent to the Continuum Hypothesis...
    
      John

	Silly me.  If we have an open cover U of the irrationals,
	then its complement A = R - U is both closed and a subset
	if the rationals.  Since A is closed, it contains all of
	its limit points, and its closure is itself.  Since it is
	also a subset of the rationals, its closure (itself) cannot
	contain a non-empty interval (a,b), and the set of its
	limit points (a subset of itself) cannot be uncountable.
	Seen in this light, most of the questions become easy to
	answer.  In particular, the "???" of .-2 is "no".

	Dan