sci.math
[Top] [All Lists]

Re: An uncountable countable set

Subject: Re: An uncountable countable set
From: Tony Orlow
Date: Tue, 31 Oct 2006 12:57:01 -0500
Newsgroups: sci.math
stephen@xxxxxxxxxx wrote:
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
stephen@xxxxxxxxxx wrote:
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
stephen@xxxxxxxxxx wrote:
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
stephen@xxxxxxxxxx wrote:
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
stephen@xxxxxxxxxx wrote:
<snip>

What does that have to do with the sets IN and OUT? IN and OUT are the same set. You claimed I was losing the "formulaic relationship" between the sets. So I still do not know what you meant by that statement. Once again
         IN = { n | -1/(2^(floor(n/10))) < 0 }
        OUT = { n | -1/(2^n) < 0 }

I mean the formula relating the number In to the number OUT for any n. That is given by out(in) = in/10.
What number IN?  There is one set named IN, and one set named OUT.
There is no number IN.  I have no idea what you think out(in) is
supposed to be.  OUT and IN are sets, not functions.

OH. So, sets don't have sizes which are numbers, at least at particular moments. I see....
If that is what you meant, then you should have said that.
And technically speaking, sets do not have sizes which are numbers,
unless by "size" you mean cardinality, and by "number" you include
transfinite cardinals.
So, cardinality is the only definition of set size which you will consider.....your loss.
If somebody presents another definition of set size, I will
consider it.  You have not presented such a definition.



I have presented an approach that works for the majority of infinite bijections, and explained some of the exceptions.

Given that you cannot answer how many Turing machines are, or
how many sets of prime numbers there are, or pretty much
any set that is not a member of P(N), I do not see how you
can claim it works for sets in general.

<snip>


In any case, it still does not make any sense. I am not sure what |IN| is for any n. IN is a single set. There is only
one set, and it does not depend on n.  In fact, there isn't
an n specified in the problem.   Yes I used the letter n in
the set description, but that does not define an entity named 'n'.

There most certainly is an 'n'. The problem describes a repeating process, each repetition of which is indexed with a successive n in n, and during each repetition of which ball n is removed. What do you mean there's no n???
I am talking about two sets that I have explicitly defined.
There is no 'n'.  I have been very clear about what I am
talking about.  I cannot help it if you cannot get over
your fixation with balls and vases.


Yes, from that perspective, they are the same set. But, there is an n in the original problem,

I am not talking about the original problem.  How many times
do you have to be told that?

<snip>


So here is one last attempt.  The balls and vase problem
is not a physical problem.

No kidding. And I was going to set up a test bed in the basement....

Who are the one applying physical reasoning to it.

Pretending that it is a physical
problem is pointless.  It is phrased in such a way to suggest
a physical problem, so that it might confuse your physical
intuitions, but it is not a physical problem.

Show me where your formulation relates t and n. Where is the schedule stated in that formulation? It's not. Try again.

I never said that was a formulation of the problem.  It is however
an accurate description of the set of balls that are added to
the vase, and the set of balls removed from the vase.  But that
was not the point of the question.

Even 'finite approximations' of it are not physical problems.
Consider the case where you start adding balls at 11pm.
Suppose you had 2000 balls.  The last set of 10 balls will
be added at time at 10^-56 seconds before noon.  But that
is less than Planck time, and is therefore impossible according
to known physics.  How big are these balls?  Lets suppose
they are 1cm in diameter, and moving them into the vase requires
the ball be moved 1cm.  Once you get up to ball 350 or so,
you will have to move the balls faster than the speed of light.

Blah blah blah. Thanks for the reality lesson. Now take a lesson in realism.

More pointless insults.

This is simply not a physical problem.   It is an abstract
problem.  It is a mathematical problem.  The "physical" part
is just a distraction.

So is all your blabbering....

More pointless insults.  You seem to be incapable of discussing
this rationally.

So if instead, someone had just posed this problem
Let
         IN = { n | -1/2^(floor(n/10)) < 0 }
        OUT = { n | -1/2^n < 0 }

What is | IN - OUT | there would be controversy.
Note, there are not balls or vases, or times or anything
in this problem.  Just two sets.  It would help if you
bother to stop and think for a second before responding.



Where are the iterations mentioned there?

What iterations?  I am not talking about iterations.  I am
just posing a problem about sets that should be totally uncontroversial.

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.

No, the words "balls" and "vase" are the distraction.  If you want
to avoid distractions, phrase the question strictly in mathematical
terms.

Given that for every positive integer -1/(2^(floor(n/10))) < 0
and -1/(2^n) < 0, both sets are in fact the same set, namely N.

Do you agree, or not?  Or is it the case that the
         "formulaic between the sets is lost."
?

Stephen
The formulaic relationship is lost in that statement. When you state the relationship given any n, then the answer is obvious.
What relationship?  For a given n, -1/(2^(floor(n/10))) < 0
if and only if -1/(2^n) < 0.  The two conditions are logically
equivalent for positive integers.  If n is a member of IN,
n is a member of OUT, and vice versa.

What other relationship do you think there is between
        -1/(2^(floor(n/10))) < 0
and
        -1/(2^n) < 0
??
Like, wow, Man, at, like, each moment, there's, like, 10 going in, and, like, Man, only 1 coming out. Seems kinda weird. There's, like, a rate thing going on.... :D
What rate?  There is no rate.  There are just two sets
         IN = { n | -1/2^(floor(n/10)) < 0 }
        OUT = { n | -1/2^n < 0 }
9 balls/iteration.
What balls?   What iterations?  The two sets are defined
as
         IN = { n | -1/2^(floor(n/10)) < 0 }
        OUT = { n | -1/2^n < 0 }

Why do you keep talking about balls and vases?

Because that's the original problem, and your formulation states nothing about iterations n or the relation between n (ball number) and t, from which the iteration information could also be deduced.

But as everybody keeps pointing out to you, this is not
necessarily a formulation of the original problem.  It
is just a question about sets.

Anyway, this is pointless.  You are apparently incapable
of discussing the abstract question posed, and instead
just want to toss about lame insults and ignore whatever
else says.

Stephen

From a set theoretic perspective, the sets are the same by axiom of extensionality. That does not take into account any order, which the original sequence clearly has. So, what is your point?

<Prev in Thread] Current Thread [Next in Thread>
Privacy Policy