Re: An uncountable countable set

David Marcus wrote:
Mike Kelly wrote:
Tony Orlow wrote:
Where are the iterations mentioned there? You're missing the crucial
part of the experiment. By your logic, you could put them in in any
order and remove them in any order, and when you say both processes are
done, nothing's left, but that's BS. It ignores the sequence specified.
This is just a distraction.

Yes, if you insert and remove exactly the same balls then you get the
same result when you're done, no matter what order you did it all in.
Why is that BS? It seems blindingly obvious.

But I forgot, you think that if you shift all the insertions 1 minute
further back in time, you DO get an empty vase at noon, right? I really
don't understand how your mind works.

Try the mental picture with the water. We fill it up, then we start
letting it run out. No reason all the water shouldn't empty out of the
vase by noon.

--
David Marcus

Hi,

There are only sets in ZFC. When people talk about the real numbers in
ZFC, it is as a construction of sets in ZFC has been found to be
isomorphic in a relatively strong sense to the real numbers.
Obviously, mathematicians want to find the best representation of
everything the real numbers are or must be by their nature.

Then, that gets into that some feel that mathematics can't assume what
it sets out to prove. That's reasonable. By the same token, the real
numbers have many, many roles to fill, and some of them have, for
example, in the projectively extended real numbers, points at
plus/minus infinity.

It is relatively standard to define a real number as a Dedekind cut or
Cauchy sequence, which are basically defined in terms of sequences of
rationals, which resolve to generally the familiar decimal
representation which is adequate in finitely expressing rationals, and
with radicals, algebraics.

If a Dedekind cut is as was recently stated some "initial segment" of
the rationals, I wonder to what ordering that pertains.

Consider how that is to describe an irrational number. Basically the
sequence of elements is to converge towards the number. There's an
irrational less than one and greater than .9, in decimal, less than one
and greater than .99, less than one and greater than .999, etcetera.
The general consensus here is that .999... = 1, yet for each
..999...999, there is an irrational between it and one. So, does that
not seem that there are irrationals unrepresentable via
Dedekind/Cauchy? It would seem that certainly as the irrational is
some finite distance from 1 that it would be between .999...998 and
..999...999, and between that irrational and one are infinitely many
more numbers forever, there always exist irrationals between .999...999
and 1, and, for Dedekind/Cauchy to represent them, they must have a
unique representation.

For no finite number of 9's or rep-units in binary can these
irrationals in the diminishing remaining interval be represented, and
for any infinite number of rep-units the result is said to be one. So,
either between the finite and infinite those values are represented, or
they're not, and due to the completeness of the reals, if they're not,
then Dedekind/Cauchy, the standard set-theoretic method to construct
real numbers, is insufficient to construct some real numbers.

For any it's so, for all it's not, or vice versa. Don't worry I've
heard of the transfer principle.

Consider the representation of rational numbers, for example 9/10.
That would be .9, .90, .900, ..., .9(0): 9/10's. There is no last
element of that list, .9 could be an initial segment of a sequence for
9/10's or any irrational between .9 and 1.0, as above. The initial
sequence .9, .90 could be an initial sequence for any irrational
between .900 and .91. The initial segment .900 could be an initial
sequence for any number between .9000 and .901.

.90000 <= x <= .9001
.900000 <= x <= .90001
.9000000 <= x <= .900001
.90000000 <= x <= .9000001
.900000000 <= x <= .90000001
.9000000000 <= x <= .900000001
.90000000000 <= x <= .9000000001
.900000000000 <= x <= .90000000001
.9000000000000 <= x <= .900000000001
.90000000000000 <= x <= .9000000000001
..900000000000000 <= x <= .90000000000001
...

As the number of zeros diverges, the diminishing interval goes to zero,
where the lower and upper bounds are a and b, lim n->oo b-a = 0. For
any finite iteration there are obviously a continuum of elements that x
could be, so for a value, x, to not obviously be among a continuum of
possible values there must be infinitely many iterations.

Keep in mind that there are printed counterexamples to standard real
analysis with a least positive real.

Obviously the ground around .999... vis-a-vis 1 is very well turned,
that's the point, to some extent we're talking about significant
ephemera.

Look at the 1 on the right side above. Where does it go?

.90001 <= x <= .9002
.900001 <= x <= .90002
.9000001 <= x <= .900002
.90000001 <= x <= .9000002

Standardly, equal.

.90001 <= x <= .901
.900001 <= x <= .9001
.9000001 <= x <= .90001
.90000001 <= x <= .900001

.90001 <= x <= .91
.900001 <= x <= .901
.9000001 <= x <= .9001
.90000001 <= x <= .90001

.90001 <= x <= .91
.900001 <= x <= .901
.9000001 <= x <= .9001
.90000001 <= x <= .90001

.90001 <= x <= .91
.900001 <= x <= .91
.9000001 <= x <= .901
.90000001 <= x <= .9001

.90001 <= x <= .91
.900001 <= x <= .91
.9000001 <= x <= .91
.90000001 <= x <= .901

.90001 <= x <= .91
.900001 <= x <= .91
.9000001 <= x <= .91
.90000001 <= x <= .91

On the left and right side each converges to 9/10, but as this
continues the lhs is .90 and the rhs is .91.

So, is the vase empty at noon?

Ross

.

Relevant Pages

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  • Re: An uncountable countable set
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