Conference rusure::math

    Aleph-null (that's Hebrew by the way, not Arabic) is the cardinal
    number of the set of integers N.
    
    Aleph (no subscript) is the cardinal number of the reals, R.
    
    Probably what you're thinking of is a sequence of increasingly large
    cardinal numbers, starting with Aleph-null, and thereafter each
    cardinal number is the number of the "power set" of the preceding
    cardinal, as in
    
    aleph-null, 2 (raised to the aleph-null), 2 (raised to 2 to the
    aleph-null), etc ...
    
    John 

    Aleph times aleph equals (the same) aleph.  So you didn't one-up him.
    Now if you had said "2 to the aleph ``IS NOT''", you'd have scored.
    
    Alephs are cardinal numbers, the sideways-8 infinity is not.  But your
    pal was speaking in cardinal (counting) terms, so feel free to
    interpret his "infinity" as Aleph-null.
    
    There are _gobs_ of infinities, each one demonstrably larger than the
    preceding.  Not just by repeated power-set applications, either.
    A good book on the subject, written in layman's terms, is titled
    something like _Mathematics and the Infinite_, by Rudy Rucker.
    
      John

        I have never seen "Aleph (no subscript)" used as the
        cardinality of the reals R.  Usually "c" is used as the
        cardinality of "the continuum" (the reals).
        
        Also, it is independent of the other axioms of set theory
        whether the cardinaly of 2^(Aleph x) is Aleph x+1.  The Beth
        sequence is defined so that Beth-null is Aleph-null, and
        Beth-n+1 is 2^Beth-n, etc..  So Beth-1 is c, but Aleph 1
        is not necessarily c.
        
        Dan
        

Here's how I like to think of it.  We start with a question such as:

	Are there more counting numbers or more even numbers ?

One person says

	Well, the even numbers skip every other counting number, so "obviously"
	although there's an infinite of each set, there are "really" twice
	as many counting numbers.

But the other person says

	Well, we can pairwise match up the counting numbers with the even
	numbers like this:

		1	2
		2	4
		3	6
		4	8
		5	10
		...

	Obviously, there's a match for each "all the way up", so there's the
	*same* number of counting numbers as even numbers.

I believe this second argument is what math folks agree with, and hence
counting numbers and even numbers represent two infinities of the "same size".

But, now consider comparing all the decimal numbers between 0 and 1 with the
counting numbers.  Which infinity is larger now ?  Well, we could start
matching them like this:

		1	.1
		2	.2
		3	.3
		...
		10	.10

Whoops!  This wasn't a good matching, since 10 and 1 both matched .1.  Let's
try to do it another way:

		1	.111111...
		2	.222222...
		3	.333333...
		...
		10	.10101010....
		11	.1111111.....

Oh well, that didn't work either, since 1 and 11 both matched the same value.

How about

		1	.01
		2	.001
		3	.0001
		...

Hey, that looks like it actually works !  Clearly, we've found a way to match
every counting number with a unique number between 0 and 1.  But we've obviously
left out "quite a few" of the numbers between 0 and 1, since for example, no
counting number will match .2 or .33.  So, there's many *more* numbers between
0 and 1 than counting numbers, and hence the "infinity" of numbers between
0 and 1 is larger than the "infinity" of counting numbers.

But something doesn't quite sit right with me about this proof yet.  Just because
this *particular* matching obviously doesn't hit alot of the numbers from 0
to 1, maybe other matchings do.  What right have I do judge the values of
infinities due to one simple matching I've chosen ?


Is this helpful ?

/Eric

> I believe this second argument is what math folks agree with, and hence
> counting numbers and even numbers represent two infinities of the "same size".
    
    It's a matter of definition, not agreement.  Two sets have the same
    cardinality if their elements can be placed into 1-to-1 correspondence.
    Period.  There may be many ways -not- to place elements into a 1-to-1
    correspondence, but all that is needed is ONE successful way, and they 
    are the "same size".
    
> But something doesn't quite sit right with me about this proof yet.  Just because
> this *particular* matching obviously doesn't hit alot of the numbers from 0
> to 1, maybe other matchings do.  What right have I do judge the values of
> infinities due to one simple matching I've chosen ?
    
    Yep, you can't appeal to failure to make the case.  What you have to do
    is prove by contradiction, i.e., assume a 1-to-1 correspondence exists
    and then show it does not really contain all of the elements of the
    "bigger" set.  Not hard to do once you know the trick...
    
      John

    re. .3
    
    See A.G. Hamilton's "Numbers, Sets and Axioms", pp 73-78.
    
    It may not be the common notation.
    
    John

    For what it is worth, there are different possible mathematical
    concepts around which can be loosely translated into ordinary English
    as describing the "size" of an infinite set.

    What we have been discussing is "cardinality".

    Measure theory is a different branch of mathematics, and as I
    understand it, it gives an answer closer to our ordinary understanding
    when we ask "which is 'larger' the set of integers or the set of
    even integers".  It says that the integers is the larger set (has
    larger measure) of the two -- in fact, that it is twice as large.

    You can think of this as based on answering the question, "if I select
    an integer at random what is the ratio of the probabilities that it
    is even vs that it is an integer."  The answer is, of course, 1/2.

    This is actually backwards.  Modern treatments of probability theory
    when applied to such questions as what does it mean to "select any
    integer at random" (as well as similar but even more tricky questions
    dealing with reals) generally depend on measure theory to describe what
    is going on.

    Don't ask for details, that's pretty much the extent of my knowledge.
    (Or ask if you want, I guess, but you are unlikely to get an answer
    from me).

				    Topher

    re: .7
    
    Intuition aside, how do you get a probability of one-half?
    The measure of the set of even integers is the same as the measure
    of the set of all integers: zero.
    
      John

Re .4  to prove that there are more reals than integers, use "Cantor
diagonalization" (alluded to in .5).

Assume you have a 1-1 mapping R(I) of the positive integers to the reals
between 0 and 1, and derive a contradiction -- namely, a real number that
isn't included in the mapping.  Simply pick a number that differs from R(1)
in its first decimal digit, and differs from R(2) in its second decimal digit,
etc.  Clearly it is different from any number already in the mapping.

RE: .8

    As I said, I don't know anything much about measure theory.  But if you
    read my "think of it as if" you'll see that I'm talking about the
    measure of the integers and even integers in the *integers* where their
    measure is definitely not zero.

    Even if we were talking about their measure in the reals, however, my
    scientific-american-level knowledge of measure theory says (if it is
    not way off base) that you can speak of ratios of measures which are
    zero overall.  This is done by means of limits, and described in terms
    of relative "density".  As I said, don't ask me for the details (one
    of these days, I have promised myself, I'm going to get a solid
    introduction to measure theory).  A familiar example is that the
    probability of drawing a sample from the standard normal of exactly 0
    is 0, as is the probability of drawing a sample of exactly 1.0, but we
    can still say that there is a sense (based on limits and exemplified by
    the probability density function) in which their ratio is exp(-.5).

					Topher

o.k. I picked an integer at random.  But I don't yet know if it's even.  It
seems to be quite a large integer and I'm reading it, and I haven't reached the
end yet to see if it ends in an even digit...


(just to make the point that truly "picking an integer at random" can produce
some quite large ones.  We tend to forget that in our limited world of
pseudorandomness)


/Eric

        Measures can be defined on countable sets, and can even be
        defined in such a way that both the even natural numbers and
        the odd natural numbers have measure 1/2.
        
        Dan

>        Measures can be defined on countable sets, and can even be
>        defined in such a way that both the even natural numbers and
>        the odd natural numbers have measure 1/2.
    
    No quarrel here.  Do you think you could also define a measure so that
    the evens have measure 1/3 and the odds have measure 2/3?
    
      John

        I think that is relatively straightforward.  Let m be any
        measure on the natural numbers, let E and O be the evens and
        odds.  Just define m' to be
        
                        1 m(A intersect E)   2 m(A intersect O)
        	m'(A) = - ---------------- + - ----------------
                        3       m(E)         3       m(O)
        
        Dan
        

    Thanks, Dan.  I knew an odd integer was more likely to come up than
    an even one.  :-)
    
      John (wondering if Eric needs another RA92 to store his random integer)

>                        1 m(A intersect E)   2 m(A intersect O)
>        	 m'(A) = - ---------------- + - ----------------
>                        3       m(E)         3       m(O)
        
        Well, not for *any* measure m, but for any measure for which
        neither m(E) nor m(O) is zero. :-)
        
        Dan

re: .7,.11

  Consider the following sequence:
   1, 3, 2, 5, 7, 4, 9, 11, 6, 13, 15, 8, 17, 19, 10, 21, 23, 12, ...
         ^        ^         ^          ^           ^           ^

  Picking a random element of this sequence, what's the probability of
  the picked element to be odd? Isn't it 2/3? On the other hand, isn't
  picking a random element of this sequence equivalent to picking a
  random (positive) integer? My point is, "picking a random integer"
  or, more generally, "picking a random element" of an infinite set,
  is not really defined.

  BTW, Alef does not realy resemble 4 (what reply was that?), it looks
  like:
                 \  |
                 /\ |
                 | \'
                 |  \

  Is there any information about "infinities" which are "larger" (by
  definition) than Alef-null, but "smaller" than Alef (or C)? Do they
  exist, proved not to exist, or unknown?

-- Nitsan

>  Is there any information about "infinities" which are "larger" (by
>  definition) than Alef-null, but "smaller" than Alef (or C)? Do they
>  exist, proved not to exist, or unknown?
    
    Also known as The continuum hypothesis.  I think it's an open problem.
    Or possibly independent of the rest of mathematics.
    
      John

RE: .17 (Nitsan)

>  Consider the following sequence:
>   1, 3, 2, 5, 7, 4, 9, 11, 6, 13, 15, 8, 17, 19, 10, 21, 23, 12, ...
>         ^        ^         ^          ^           ^           ^
>
>  Picking a random element of this sequence, what's the probability of
>  the picked element to be odd? Isn't it 2/3? On the other hand, isn't
>  picking a random element of this sequence equivalent to picking a
>  random (positive) integer?

    As I understand it, the answer is "No".  Measure theory is based
    on the concept of (open or closed) intervals within sets.  By changing
    the order you have changed the definition of what constitutes an
    interval, and therefore, what the probability is.  The probability
    is defined on the *sequence* of natural numbers, not the set.

    I'm going to attempt a hand-waving definition to give what I understand
    to be the idea.  In fact, there is some severe circularity here, which
    is why you need a fairly sophisticated kind of mathematics (measure
    theory) to put the whole thing on a firm footing.

    Imagine that you select an arbitrary interval of length N.  Here is
    where I handwave since I am speaking of "arbitrary" mostly to avoid
    saying "random".  (On the integers, the concept of length is pretty
    clear -- but, measure theory was first invented, I believe, to put the
    concept of geometric "measures" such as length, area, and volume on a
    firm, formal basis -- its, rather tricky otherwise).  On any such
    interval, I have a clear, unambiguous definition (frequentist) of
    choosing an even integer.  What that value is will depend on N and
    on how I chose which interval.  But if I let N grow indefinitely, I
    will find that the probability of even integers among the integers
    in my intervals will converge to 1/2.  This is pretty much independent
    of how I choose my intervals, as long as I choose them "reasonably"
    ("reasonable" being well-defined in the theory; in fact, for this
    specific case, I don't think I need "reasonable", since I can't think
    of a method of choosing which will prevent convergence).

>My point is, "picking a random integer"
>  or, more generally, "picking a random element" of an infinite set,
>  is not really defined.

    If this were the case, then most of statistics -- which routinely
    depends on taking a finite sample from infinite sets -- would be on
    shaky grounds.  This is precisely the problem that Kolomogorov solved
    by applying measure theory to probability theory's foundations.  It is
    true that one random sampling of an arbitrary infinite set, especially
    a non-countable set, is not well defined.  But sampling an infinite set
    with the right structure (divisible into intervals) *is* well defined
    when a measure (specifically a measure which meets the additional
    requirements of a "probability measure") can be imposed.  Different
    structures -- corresponding to different definitions of a "sample"
    on the unstructured set -- will result in different probabilities.  I
    believe, however, that the results are unambiguous if you use intervals
    on the integers or reals which are based on the "standard" ordering,
    if you restrict yourself to probability measures, and if you require
    that they "degenerate" to the finite population frequentist definition
    of probability values when the populations become finite.

                                              Topher

>.17
>  Is there any information about "infinities" which are "larger" (by
>  definition) than Alef-null, but "smaller" than Alef (or C)? Do they
>  exist, proved not to exist, or unknown?
        
>.18
>    Or possibly independent of the rest of mathematics.
        
        Cantor asked if any cardinals existed between that of the
        integers and that of the reals.  The statement that no such
        cardinal exists is known as CH, the Continuum Hypothesis.
        
        Godel showed that if the usual axiomitization of set theory
        is consistent, then it remains consistent if you also add
        AC (the Axiom of Choice) and CH (and even GCH, the Generalized
        Continuum Hypothesis, which states that there is no cardinal
        properly between any infinite cardinal X and the power set of
        X).
        
        Cohen showed that if the usual axiomitization of set theory
        is consistent, then it remains consistent if you also add
        AC and not-CH (or if you add not-AC and "the reals cannot be
        well-ordered").

        Dan
        

>    As I understand it, the answer is "No".  Measure theory is based
>    on the concept of (open or closed) intervals within sets.  By changing
>    the order you have changed the definition of what constitutes an
>    interval, and therefore, what the probability is.  The probability
>    is defined on the *sequence* of natural numbers, not the set.
    
    I disagree with this and the reasoning that follows.
    
    One of the main properties of a measure is countable additivity,
    i.e., �[U Ai] = sum(�[Ai]).  I.e., for pairwise disjoint Ai the
    measure of their union is the sum of the individual measures.  This is
    for any countable collection Ai of measurable sets.
    
    What then is the probability of choosing, say, 47 from the integers?
    All along we have been implicitly assuming a uniform distribution.
    
    �({47}) cannot be 0, since by countable additivity we would rapidly
    conclude the probability of selecting any integer is 0, but we know the
    probability is �(Z) = 1 by definition of a probability measure.
    
    �({47}) cannot be less than 0 (again by definition) nor greater than 0
    because then �(Z) would be infinite (by countable additivity over the
    individual points), but we know �(Z) = 1.
    
    Conclusion:  single points are not measurable sets when the probability
    measure is uniform over the integers.
    
    But the entire limiting process in .19 depends on the concept of being
    able to measure single points, forming a sequence of probability measures.  
    Which is fine for any interval of length N, but not fine "in the limit" 
    since the sequence of measures does not converge to a measure.
    
      John

RE: .21

    As I have said before, I know very little about measure theory -- I
    only know a bit about what I have been told about how it has been
    applied and what problems it solves.

    First off, let me say that I agree completely that there is no such
    sampling distribution as the discrete uniform distribution over all
    integers.  And I completely agree that the probability of drawing any
    integer -- or any finite set of integers -- from such a distribution if
    it existed would be zero (we can see this by taking the limit, for
    example).  I think that your reasoning is wrong here, however.

    Keep in mind that probability is defined in terms of measure theory. I
    have reversed the dependency for rhetorical convenience.  Measure
    theory is, though, more general.  Though there is no (uniform)
    probability distribution over the integers more general measure theory
    may be applied.

    Let's start with sampling from the normal distribution.  Any time I
    do so, I am taking a single point -- with probability 0 -- from an
    infinite set.  Which would *appear* to be a counter-example to your
    argument but is not.  The reason is that the set is non-countable so
    "countable additivity" you invoked in your argument does not apply.
    But wait...

    Let's derive another distribution from the normal one.  I'll call it
    the RatNormal distribution.  It is simply the normal distribution
    restricted to rational values.  This is a countable -- though not
    discrete -- distribution.  One can clearly integrate this (over the
    rationals) and get a value of 1.  One can clearly sample from this
    distribution and get out a rational whose individual probability
    is zero.  And yet we are summing a countable number of zero
    probabilities and getting a finite sum.

    Here is a complementary example which illustrates a problem with your
    argument.  Is there a *continuous* uniform distribution of infinite
    width?  No more than for the discrete uniform distribution family.
    But your argument would, since we are not dealing with countable
    sums, not exclude this pseudo-distribution.  Is it excluded for
    completely different reasons than the discrete case?  I think not --
    I think that the same argument applies to both cases.

    A "point" (single number) can be viewed as a limit to a sequence of
    intervals as the width of that interval goes to zero.  This is the
    justification for the probability density function of a distribution
    for example.  In fact, in formal analysis, this is how real numbers
    are defined (though rationals do not depend on that -- in fact, the
    intervals have rational end-points/width).  The problem with the
    infinite uniform distributions is not that points have measure zero,
    but that all *finite intervals* have measure zero.  This is where
    both the Normal and RatNormal distributions are different.

    Countable additivity has a discontinuity when adding up zeros.
    Aleph-null*0 is, in general, ill-defined.  Depending on how you go
    about adding them up you can get different values.

    As I understand it, measure theory allows you to assign meaning, under
    some circumstances to ratios of zero-measures.  This makes sense when
    you remember that while  a/0 is undefined for a ~= 0 (since there is
    no x which solves the equation a = x*0); the expression 0/0 is
    "meaningless" rather than undefined (since *any* x will solve the
    equation 0 = x*0).  If one carefully defines what one means then
    you can attach a meaning to a ratio of things with measure zero.  I
    think that this may require non-standard analysis, however.

				Topher

>    discrete -- distribution.  One can clearly integrate this (over the
>    rationals) and get a value of 1.  
    
    Not true.  The integral of any function over a set of measure 0 is 0.
    
>    Countable additivity has a discontinuity when adding up zeros.
>    Aleph-null*0 is, in general, ill-defined.  Depending on how you go
>    about adding them up you can get different values.
    
    I don't follow this.  Surely sum(i=0..oo)[0] = 0, which was the
    principle I was using.  You can play games with the summand and force
    some odd-looking results (e.g., the Grandi series +1-1+1-1...) but
    that's due to fiddling with the summand.  If you add 0 indefinitely,
    you'll get a 0 sum.
    
    It is certainly true that there are countless :-) distributions where
    one samples a value that a priori had probability 0 of being chosen.
    In fact, with probability 1 one selects an element that had probability
    0 of being chosen.  I don't think measure theory enters into this
    apparent paradox, though.
    
    One can define a series of rational numbers that approach, say, sqrt(2)
    in the limit.  Yet the limiting point is not a rational number.  
    I think that's the sort of thing you're doing by developing a series
    of measures on finite intervals and then �viola� taking the limit and
    saying "see, the answer must be �".
    
      John

RE: .23 (John)

>>    discrete -- distribution.  One can clearly integrate this (over the
>>    rationals) and get a value of 1.  
>    
>    Not true.  The integral of any function over a set of measure 0 is 0.

    Terminology mismatch here.  You are talking about, what I would
    describe as taking the integral over the reals but with irrationals
    having zero probability.  When I speak of doing integration over the
    rationals, I'm talking about integration with the rationals as their
    domain (the range is real, however).  It isn't zero at irrationals
    its undefined.  The measure of the rationals in the
    rationals is not zero.  That the reals have measure 0 in the complex
    numbers does not mean that integration over the complex plane means
    that all integration over the reals is zero.

>    I don't follow this.  Surely sum(i=0..oo)[0] = 0, which was the
>    principle I was using.  You can play games with the summand and force
>    some odd-looking results (e.g., the Grandi series +1-1+1-1...) but
>    that's due to fiddling with the summand.  If you add 0 indefinitely,
>    you'll get a 0 sum.

    The Grandi series is a good example.  You can define *how* you want to
    sum the Grandi series and get a reasonable -- even consistent (as long
    as you don't violate your assumptions about how the addition is to be
    done) -- result.  But without that additional information about how
    the sum is to be totalled the Grandi sequence has no limit.

    If you add 0 indefinitely (an unspecified finite number of times,
    however large) you *will* get 0, but if you add 0 an aleph0 number of
    times you need to specify how the addition is to be done to make any
    sense of it.  Adding zero an aleph0 number of times is equivalent to
    multiplying 0 by aleph0 which is meaningless in the same sense as 0/0
    is meaningless.

    Now, as I understand it, if you restrict yourself to standard analysis
    and its associated calculus, then you can get away with infinitely
    summing zero to zero.  And in fact, that is sufficient for handling
    probability theory over finite sets and over the integers.  But you
    need measure theory (based on non-standard analysis) to deal with the
    reals.

>    It is certainly true that there are countless :-) distributions where
>    one samples a value that a priori had probability 0 of being chosen.
>    In fact, with probability 1 one selects an element that had probability
>    0 of being chosen.  I don't think measure theory enters into this
>    apparent paradox, though.

    It doesn't, perhaps, "enter into it", but, according to Kolomogorov it
    is required to resolve it on a firm, formal basis.  Your actual sample
    events (reals) have zero probability.  Overall you need the total
    probability to sum to one.  To do so you use measure theory to define
    your proability measure in terms of intervals and your final outcomes
    in terms of limits of those intervals.  That justifies the probability
    density function which allows us to integrate to the cumulative
    probability function (at least for continuous CPFs).  The PDF makes
    meaningful such statements as "the probability of 0 relative to 1 from
    the normal distribution is exp(-�)".

>    One can define a series of rational numbers that approach, say, sqrt(2)
>    in the limit.  Yet the limiting point is not a rational number.  
>    I think that's the sort of thing you're doing by developing a series
>    of measures on finite intervals and then �viola� taking the limit and
>    saying "see, the answer must be �".

    Guilty as charged.  I didn't prove anything.  I illustrated the process
    by which the concept of the density of infinite (countable) sets in
    other infinite (ordered countable) sets is defined.  I have been told
    that this definition can be formalized and proven consistent using
    measure theory.  Any finite interval of length at least two will have
    associated with it a real number representing the ratio of even
    integers to total integers (i.e., the intervals size) in that interval. 
    If you allow those intervals to expand in size you will get a perfectly
    classical series to sum.  If you reqire "intervals" to be based on
    locality based on the standard ordering of the integers, those series
    will consistently converge and will consistently converge to the same
    value (although the specific series will depend on how you select the
    position of the interval of each size).

    We are dealing with perfectly classical series convergence -- not
    improperly extrapolating a property (membership in the rationals) to
    their infinite sum.  The rationals are not closed under addition.  The
    limit of a series of quite proper probability distributions is
    not necessarily a proper probability distribution; e.g., the uniform
    distribution of infinite width, or the delta distribution (which among
    other definitions is the limit of uniform distributions as the width
    goes to 0).  Sometimes such improper distributions are handy though,
    if you are careful with them, and clean up the mess afterwards.  Non-
    standard analysis was in large part motivated by the delta function/
    distribution.

					Topher

>Re .4  to prove that there are more reals than integers, use "Cantor
>diagonalization" (alluded to in .5).
>
>Assume you have a 1-1 mapping R(I) of the positive integers to the reals
>between 0 and 1, and derive a contradiction -- namely, a real number that
>isn't included in the mapping.  Simply pick a number that differs from R(1)
>in its first decimal digit, and differs from R(2) in its second decimal digit,
>etc.  Clearly it is different from any number already in the mapping.


My first reaction, of course, was that this was kind of clever.

But now I'm a bit troubled again.  Going back to the mapping from counting
numbers to even numbers again:

	e(1)	0
	e(2)	2
	e(3)	4
	e(4)	6
	e(5)	8
	...

We ask the same question now, namely "have we hit all the even numbers" ?

Well, why can't we do the diagonal thing here too.  Namely, construct an
even number that differs from e(1) in first digit, and differs from e(2)
in second digit, and differs from e(3) in third digit etc.  Clearly, this
new even number is not in our list, and hence we haven't provided a complete
mapping.

/Eric

RE: .25 (Eric)

    You have constructed an "integer" with an infinite number of digits.
    There is no such beast -- all integers have a finite number of
    digits.  As a consequence, the number you have constructed which is
    not on the list, doesn't belong on the list.

    Although essentially correct the process of diagonalizing the reals is
    actually a little trickier than the description given.  You have to be
    careful because what you are constructing in the diagonalization is a
    real *representation* that isn't on the list of representations. Since
    many reals (specifically those rationals whose reduced numerator will
    divide into a power of the representation radix) have more than one
    representation you may discover an value which is on the list, just
    represented in another way.  You have to set the whole thing up
    carefully to avoid this problem.

				    Topher

        I recall reading that Cantor's argument that the reals are not
        countable did not use decimal or binary representation.
        
        Here is one proof which uses only order properties of the reals.
        If a(0), a(1), a(2), ..., a(n), ... is a countable sequence of
        reals, then define b(0), c(0), b(1), c(1), b(2), c(2), ... such
        that
        
        	b(0) < b(1) < b(2) < ... < c(2) < c(1) < c(0)
        
        Simply take b(0) to be a(0), and "cross out" all a(n) <= b(0).
        If the entire sequence of a's was just crossed out, then
        obviously the sequence of a's didn't contain any reals greater
        than b(0), and so could not be a list of all of the reals.
        Otherwise, some a's remain, and the first one remaining in the
        sequence is taken as c(0), and all a's still remaining which
        are >= c(0) are now crossed out.
        
        Repeat this process for k = 1, 2, 3, ...  If no a's remain,
        then the original sequence of a's did not include any reals
        between b(k-1) and c(k-1).  If a's remain, select the first
        in the sequence as b(k), and cross out all a's <= b(k).  If
        all the a's are gone, then the reals between b(k) and c(k-1)
        were missing from the original enumeration.  Otherwise, take
        c(k) as the first a in the sequence still remaining.
        
        The above process may stop because the sequence of a's did
        not include any reals between the largest (most recently
        chosen) b and the smallest (also most recently chosen) c.
        Or it may continue "forever" and allow us to define the
        b and c sequences such that
        
        	b(0) < b(1) < b(2) < ... < c(2) < c(1) < c(0)
        
	Let B be the least upper bound of the b's, and let C be the
        greatest lower bound of the c's; we have
        
        	b(0) < b(1) < b(2) < ... < B <= C < ... < c(2) < c(1) < c(0)
        
        Any x such that B <= x <= C cannot have been among the a(n) -- if
        it did, it would have eventually been chosen as one of the b's or c's.
        
        So either way, the list of a's does not enumerate all the reals.
        
        I think Cantor's original diagonal argument applied to a set A
        and its power set P(A) = { B | B is a subset of A }
        
        	Given any f : A -> P(A), then the "diagonal set"
        	D = { x in A | x is not in f(x) } is a subset of A
        	which is not in the range of f.
        
        This shows that no set A can be mapped onto all of P(A),
        and gives an explicit subset D of A which is not "touched"
        by f.  If A is the positive integers, you can picture this
        as
        
        	x	0 in f(x)?	1 in f(x)?	2 in f(x)?	...
        	-----------------------------------------------------------
        	0         yes             no              yes
        	1         no              no              yes
        	2         no              yes             yes
        	.
        	.
        	.
        
        Then run down the diagonal inverting the yes/no's to see what
        is in the diagonal set D.
        
        Dan
        

RE: .27 (Dan)

>        I recall reading that Cantor's argument that the reals are not
>        countable did not use decimal or binary representation.

    Could be, but the diagonalization proof using decimal representation is
    usually presented as Cantor's proof.

>        I think Cantor's original diagonal argument applied to a set A
>        and its power set P(A) = { B | B is a subset of A }

    In case anyone has lost track -- this is not intended as an alternate
    way of showing that the reals are not countable (i.e., equal in "size"
    to the integers).  One can consistently assume either that the size
    of the power set of the integers (aleph-sub-one) is equal to the
    size of the set of reals (a.k.a., the size of the continuum, or C) --
    which is refered to as the continuum hypothesis, or that they are
    not equal.

					Topher

.24>    The measure of the rationals in the rationals is not zero.  
    
    Tell me more.  What IS "the measure of the rationals in the rationals",
    and how is it calculated?  What sets of rationals are measurable?  Bearing 
    in mind that measure must be translation invariant and countably additive.
    
.24>    That the reals have measure 0 in the complex
>	numbers does not mean that integration over the complex plane means
>	that all integration over the reals is zero.
    
    The integral over a set of measure zero is always 0.  Now that's a fact.
    (Well, at least if the function is integrable over the set).
    
    Speaking colloquially:  the rationals have "length" 0, so any integral
    (i.e., any area) with one side 0 will necessarily be 0.  Likewise the
    reals have "area" 0 so any volume based on area 0 will also be volume 0.
    
      John

RE: .29 (John)

    Well, I keep saying that I'm out of my depth on this, and perhaps I
    completely misunderstood what was explained to me, but you haven't
    convinced me of that yet.

>    Tell me more.  What IS "the measure of the rationals in the rationals",
>    and how is it calculated?  What sets of rationals are measurable?  Bearing 
>    in mind that measure must be translation invariant and countably additive.

    If it is a probability measure then the measure of the rationals in the
    rationals is one.  In other words if you are selecting a single
    rational from the rationals -- however you do it -- the probability
    that you will end up with some rational will be 1.

    You cannot, of course, do a Reimann (sp?) integral over the rationals
    since it is only defined over R^d, but you can do a Lebesgue integral
    to get the "area" under the rationals in such a way as to completely
    ignore the reals.

    Because of additivity you cannot assign a zero measure to individual
    rationals and end up with a total measure of the whole set with
    anything other than zero.  But you can refuse to assign a meaning to
    the measure of an individual point and instead talk about measures of
    finite, positive-lengthed rational intervals and (countable) unions of
    those intervals.  The limit of the measure of a sequence of such
    intervals goes to zero as the length of the interval goes to zero, but
    that doesn't mean that the measure on the point itself is zero.  We
    can talk about the measure of infinite (obviously countable) sets of
    rationals.

    Imagine we sample the rationals with the following restriction: the
    probability that the value is less than a rational value R is the same
    as the probability that a value sampled from the normal distribution.
    That's my RatNormal distribution.  Do you really claim that no such
    distribution can exist?  I would say that if the foundations of
    probability theory are insufficient to handle this than they need work.

				    Topher

RE: .28 (Me)

>    One can consistently assume either that the size of the power set of
>    the integers (aleph-sub-one) is equal to the size of the set of reals
>    (a.k.a., the size of the continuum, or C) -- which is refered to as the
>    continuum hypothesis, or that they are not equal.

    Dan has sent me mail informing me that this is not true -- that C is
    equal to the size of the power set of the integers.  After massaging my
    aging memory for a while, I realize that he is quite correct, and that
    I confused two theorems of Cantor (I believe that he failed to prove
    either one of them).  The first, which remained unsolved for a long
    time, was the continuum hypothesis -- that there exists no cardinality
    (size) between the cardinality of N (natural numbers) and the
    cardinality of P(N) (power set of N).  Gauss showed that you could
    consistently assume either the continuum hypothesis or its converse. 
    (This was later extended to the generalized continuum hypothesis which
    is about infinite cardinalities between any infinite set and its
    power-set).  The other hypothesis -- related to the continuum
    hypothesis by Cantor -- was the hypothesis that C (the number of reals)
    was strictly less than the size of P(N).  This was, however, shown to
    be false -- that the cardinality of P(N) is equal to C.

    Sorry for confusing things with my faulty memory.

				    Topher

        Cantor certainly knew and showed that the cardinality of the
        reals ("c") was the same as the cardinality of the power set
        of the naturals ("two to the aleph-null").  As he looked at
        the cardinalities of various subsets of the reals, he found
        they came in three types, finite, infinite and countable, and
        infinite of cardinality c.
        
        For example, every nonempty perfect subset of the reals has
        cardinality c (examples are the well-known Cantor set, or any
        interval [a,b] with a < b).  Every open subset includes an
        interval [a,b] and so has cardinality c.  Every closed subset
        can be written as a disjoint union of a possibly empty perfect
        set and a set which is finite or countably infinite (this actually
        uses a little bit of the axiom of choice).
        
        So Cantor formulated the question CH, which stands for the
        "no" answer to the question of whether there are any infinite
        sets with cardinality strictly between aleph-null and c.  I've
        been told that Cantor was good enough that he never actually
        mistakenly thought he had proved or disproved CH.  Godel first
        showed that if the other usually accepted axioms of set theory
        were consistent, then CH (and GCH) plus those axioms are
        consistent as well.  Cohen showed that with the same assumption,
        not-CH plus those axioms are also consistent. So CH turns out to
        be independent of the other axioms of set theory.
        
        Dan

.30>>    Tell me more.  What IS "the measure of the rationals in the rationals",
.30>>    and how is it calculated?  What sets of rationals are measurable?  Bearing 
.30>>    in mind that measure must be translation invariant and countably additive.
>
>    If it is a probability measure then the measure of the rationals in the
>    rationals is one.  In other words if you are selecting a single
>    rational from the rationals -- however you do it -- the probability
    
    The objection in .29 was that you (Topher) defined a function and then
    used various results from probability/measure theory to justify certain
    results.  The problem I had with that is that you failed to demonstrate
    that your RatNorm actually qualifies as a probability measure.  You
    need to demonstrate that it satisfies a certain number of axioms, or
    properties if you dislike that use of the word "axiom", in order to
    qualify as a probability measure.
    
    Perhaps it is time to take this discussion offline.
    
      John