Re: Multiple infinities  one more look
 From: "Ross A. Finlayson" <raf@xxxxxxxxxxxxxxx>
 Date: Fri, 7 Dec 2007 22:58:18 0800 (PST)
On Dec 7, 9:59 pm, Venkat Reddy <vred...@xxxxxxxxx> wrote:
On Dec 8, 8:33 am, "Mike Terry"
<news.dead.person.sto...@xxxxxxxxxxxxxxxxxxx> wrote:
"Venkat Reddy" <vred...@xxxxxxxxx> wrote in message
news:052a8b2692db46a98caf1ba9c7300927@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I see. I think order is related to how you choose to represent the
number. If the real number is represented by a digit sequence of
infinite length, we immediately have an order for all reals, since
digit position have an order. For example, for the length of 6 digits,
one can generate these sequences by writing all possible permutations
and combinations of digit sequences in an orderly fashion. This can be
continued for lager length of digit sequences without limit. All
possible digit sequences are guaranteed to appear somewhere in the
list.
No. This process only generates finite digit sequences. Real numbers have
infinite digit sequences...
Instead, if you choose to represent a real number by an algebraic
equation, here also we have an order in representing the equation
itself, so the resulting reals have an order. Likewise for
transcendental as well  once you have a way of identifying the
number, we have a way of ordering them.
It seems you are thinking of something like "computable numbers", and these
are indeed countable. Real numbers need not be computable in this sense
(i.e. having a finite program to output their digits in sequence).
You may say you want to represent the real number as a point on the
line such as the interval [0,1]. Here also we have an order. Cut the
line into half and mark the 0.5 as your first real number. Cut the two
pieces in to two equal parts, mark 0.25 and 0.75 as new real numbers.
Continue the process for ever and you have an ordering of all points.
No. Real numbers need not be one of these points you construct, although
your process will construct real numbers arbitrarily close to any given real
number.
Let's identify a real number as a unique label for a cut in a line
segment (spatial continuum) at a random location. The fact is, such
cuts or splits or points exist only after you imagine them. The
continuum doesn't already have them ready for us to count. This is
similar to fact that we also do not have "ones" ready to count as
natural numbers, but only after you imagine some discrete items.
The natural ordering of such cuts, is exactly similar and opposite to
ordering of natural numbers. Natural number sequence keeps introducing
a new "one" in an attempt to fill the "empty continuum". A reciprocal
process for this would be to introduce a new "zero" or split in an
attempt to empty the "full continuum". However, introducing the splits
wouldn't be by adding one at a time, but 2^n of those at a time. This
follows from the observation that in natural number sequence we are
adding a "one" to every "empty continuum", and hence we need to add a
split to every "full continuum", that is, the pieces resulting from
the previous cuts.
This ensures that we can always imagine 2^n number of splits for every
imaginable natural number n. When the natural number hits infinity,
then we have 2^oo as the number of points, all are perfectly ordered
or sequenced.
 venkat
One might be quick to point that this process can only hit the real
numbers of the form of multiples of 1/(2^n), and misses the numbers
like 1/3, 1/pi, 1/sqrt(2). My answer for this is, have a
representation of these numbers as points on the real line and show me
why my process can't hit these points, if you really continue the
process for ever.
Better yet, have a list of all possible representations  digit
sequences, algebraic equations, geometric models etc. For each of
these representations, have a process to generate all possible numbers
in that representation as I have shown above. That should cover all
real numbers in an orderly fashion, though in a 2dimensional order.
Well, you've given 3 suggestions above, and all 3 suggestions are wrong and
putting them together doesn't make things any better! :)
Ask for any number, one of these processes is guaranteed to have it in
its list (I would even say every process have that number in its list,
but at different positions, but I don't need to depend on this
anyway).
Also, since each of these processes have a countable list, and the
number of processes is also countable, the reals must be countable.
Did I miss anything?
Each of your processes covers only a proper subset of the real numbers.
Regards,
Mike.
 venkat Hide quoted text 
 Show quoted text 
In reference to something along the lines of 1, 3, 5, ..., then 2, 4,
6, .... A notion about mapping that to a sequence is to map 1 to 1, 3
to 2, 5 to 3, etcetera, in the order of the natural integer sequence,
then there is a question about all of those values as constants being
used already. Then, with some notion of a transfinite sequence, then
there is a consideration of "omega+1, omega+2", etcetera. So,
consider omega a constant of sorts, yet all the previous constants are
used already. So, omega is some infinitary constant (label), where
each of the first half of the sequence have leading zeros, each of the
second half of the sequence has a leading nonzero constant, an
infinitarily unique constant. Then, the alphabet of values is
infinitary, to wellorder the integers in a manner thus that there are
infinitely many elements less than a particular element. Then, in an
infinite base, that is to say where there are infinitely many
constants, similarly to base one, base two, and base three, the
antidiagonal argument doesn't apply, instead simply confirming the
successor.
For any finite list of constants in successive order, the antidiagonal
is the succesor, and order type, and in ubiquitous ordinals,
powerset. (That's again with the notion of successor ordinals formed
that are simply the powerset, the succesor ordinal's mechanistic form
is the powerset.)
That three space dimensions are particularly concise is not obviously
related, yet some people can find a physical parallel to any
mathematical statement. (V = L).
Infinite sets are equivalent. That is so in a theory where all
infinite sets are axiomatized as irregular, in moreso that the are
their own rootsets and powersets.
The powerset has lots use in terms of counting things in the finite
and small. The value 2^n appears in many forms. That's where 2^n
appears in many composite forms. That's how many elements are in the
powerset of the nset, with size n: 2^n. It might seem interesting
to make tables and then sum the rows of various counts and then assign
them to geometrical figures. Anyways a wide variety of other numbers
are very useful in terms of the cardinalities of combinatorics of
finite sets. (I pronounce finite finit, with no emphasis, not fi
night, as well, infinite infinit. ) Consider for example n!, n
factorial, which is the product 1 through n. That's the number of
ways n things can be put in an order, a permutation, that's how many
transitive orders there are of those things. Then using two
variables, the count n of objects and subset's count k of objects,
there are such useful notions as the choice: function, n choose k,
subset function: x subset number k, and cycle function: n cycle number
k, also know as the binomial coefficients and Stirling cycle and
subset numbers, Stirling numbers of the variously first and second
kind.
n!: count of permutations of an nset
s(n,m): count of permutations of an nset containing mmany
permutation "cycles",
In Knuth's notation for n cycle k the binomial coefficients is the
stacked numbers in the parentheses, and n cycle k is the stacked
numbers in the square brackets.
http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html (cycle
number)
A nice thing about them is that they can be computed in many ways.
I wonder how many there are of various partial permutation cycles, in
that there aren't complete transitive cycles. Maybe I should
investigate that for more than a minute, not the associated Stirling
numbers.
The Stirling subset number, which is the count of partitions of an n
set into mmany disjoint subsets, is written in the Knuth notation the
stacked elements in the curly parentheses. Consider for example
pigeonhole problems, where there are n objects and m cubbyholes, how
many ways there are to put a given number of objects into bins so none
of the bins is empty. That's exactly n subset m. Then, there are
more ways than that to put a given number of objects into bins, with
possible some of the bins being empty. Knowing these exact values is
very useful in formulating counting arguments so that then
probabilities are easy to determine, about what happens when a set of
objects is randomly distributed to separate bins.
http://links.jstor.org/sici?sici=00029890(199410)101%3A8%3C771%3AATNON%3E2.0.CO%3B2U
I could easily go on about the tremendous utility of counting
arguments of finite combinatorics. It would really be a reasonable
thing to do to define the counts of the objects of nsets and
generally multisets, or vice versa, as expressions of primary
interest and concise notation in definitions of convenient terms. The
point is that in the consideration of algorithms that would expect to
be useful on finite sets, they often are in the application to systems
modeled by iteratively larger finite sets, infinite sets.
To be putting the points on a line, put them in a line.
Ross

Finlayson Consulting
.
 References:
 Multiple infinities  one more look
 From: Venkat Reddy
 Re: Multiple infinities  one more look
 From: FredJeffries@xxxxxxxxx
 Re: Multiple infinities  one more look
 From: Venkat Reddy
 Re: Multiple infinities  one more look
 From: Mike Terry
 Re: Multiple infinities  one more look
 From: Venkat Reddy
 Multiple infinities  one more look
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