Re: Multiple infinities - one more look

On Dec 7, 9:35 pm, "FredJeffr...@xxxxxxxxx" <FredJeffr...@xxxxxxxxx>
wrote:
On Dec 6, 7:02 pm, Venkat Reddy <vred...@xxxxxxxxx> wrote:

multiple infinities - lets check them "really"

I was reading about cardinalities and the diagonal argument.

For me, here is how things are. For integer sequences such as 1,2,3,..
and 2,3,4,... there is a possibility that one can always insert a new
number at the end.

No, then you no longer have a sequence. You can insert a new number at
the BEGINNING of a sequence and the result is a sequence: insert 1 at
the beginning of
2, 3, 4, ... and get
1, 2, 3, 4, ...

But 2, 3, 4, ..., 1 is not a sequence. Sequences have a first member
but no last member.
2, 3, 4, ... has no last member.
2, 3, 4, ..., 1 has a last member, namely 1.

Seehttp://en.wikipedia.org/wiki/Ordinal_number

The order type of a sequence is omega (the greek letter that looks
like a w). The order type when you "insert a new number at the end" of
a sequence is omega + 1

That's one of the differences between finite and infinite sets: Any
way you order a finite set, the order structure is the same (order
isomorphic). An infinite set has many different possible orderings
with different order structures.

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.

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.

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.

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 2-dimensional order.
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?

- venkat
.

Relevant Pages

  • Re: An uncountable countable set
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  • Re: An uncountable countable set
    ... It ignores the sequence specified. ... mathematicians want to find the best representation of ... there always exist irrationals between .999...999 ... they're not, and due to the completeness of the reals, if they're not, ...
    (sci.math)
  • Re: Multiple infinities - one more look
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  • Re: Galileos Paradox and the Project of the Reals
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  • Re: Cantor Confusion
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