Re: abundance of irrationals!)

Virgil said:
> In article <MPG.1ceaecbfd0fbb9a9989c0a@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
>
> > Russell said:
> > > aeo6 Tony Orlow wrote:
> > > > Virgil said:
> > > > > In article <MPG.1ceaad4d31013157989bf7@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > > > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> > > > >
> > > > > > Randy Poe said:
> > > > > > >
> > > > > > > mueck...@xxxxxxxxxxxxxxxxx wrote:
> > > > > > > > Randy Poe wrote:
> > > > > > > > > A number x being in your list means that there is m
> > > > > > > > > such that the m-th number in your list is x.
> > > > > > > >
> > > > > > > > No, these numbers exist in the same way as Cantor's
> > > antidiaonal.
> > > > > > >
> > > > > > > Then it's not a list of real numbers.
> > > > > > >
> > > > > > > First, Cantor's antidiagonal exists in the sense of NOT
> > > > > > > being on the list. So you have constructed a list by
> > > > > > > which all real numbers have the property of not being
> > > > > > > on the list?
> > > > > > >
> > > > > > > There is one and only one meaning of "a list of real
> > > > > > > numbers". That is that you have an association between
> > > > > > > every natural number m and every real number a_m on your
> > > > > > > list.
> > > > > > >
> > > > > > > If you respond to this remark with "no", as you did:
> > > > > > > > > A number x being in your list means that there is m
> > > > > > > > > such that the m-th number in your list is x.
> > > > > > >
> > > > > > > then x is not on your list. Because that is the meaning
> > > > > > > of "on a list". There is no other.
> > > > > > >
> > > > > > > If you want to come up with some other construction
> > > > > > > (some sort of infinitely deep binary tree, according
> > > > > > > to comments from others), that's fine. But it's not
> > > > > > > a list, and it doesn't advance your argument that
> > > > > > > a list can contain all real numbers in (0,1).
> > > > > > >
> > > > > > > - Randy
> > > > > > >
> > > > > > >
> > > > > > Excuse me, but the nodes of an infinitely deep binary tree can
> > > > > > obviously be enumerated in a linear manner, corresponding to
> > > binary
> > > > > > integers and thus to the naturals.
> > > > >
> > > > > But how does one enumerate the paths (from root node downward
> > > forever)
> > > > > in such a tree? It is not at all clear that this can be done
> > > linearly
> > > > > with an ordering to match that of the binary integers, as each path
> > >
> > > > > would be infinitely long, and not representable by a binary
> > > integer.
> > > > >
> > > > > And it is the paths that turn out not to be countable, not the
> > > nodes.
> > > > Absolutely wrong. We start at the top, element 1. The two child nodes
> > > are 10
> > > > and 11. The two nodes off 10 are 100 and 101, and the two off 11 are
> > > 110 and
> > > > 111. For each node, we add a digit to the right for each cild node, a
> > > 0 for the
> > > > left and a 1 for the right. At each level n (starting with 1) we have
> > > 2^(n-1)
> > > > nodes, numbered from 2^(n-1) to (2^n)-1, which are all the binary
> > > numbers with
> > > > n significant digits. By normal tree traversal, an infinite binary
> > > tree cannot
> > > > be traversed one branch at a time, but like the rationals in a 2D
> > > array, they
> > > > can be traversed "diagonally".
> > >
> > > That is a traversal of the *nodes*. Everybody agrees that
> > > the nodes are countable. See above.
> > The nodes represent the elements. The paths are the means of derivation for
> > the
> > members.
> > >
> > > There is no difference in enumerability, despite
> > > > Cantor's diagonal proof which cimple proves there are more reals than
> > > naturals.
> > >
> > > But you don't have any infinite sequences of digits in your
> > > enumeration. None! So you haven't represented any irrationals
> > > in your enumeration.
> > This was supposed to be an infinite binary tree, with infinite length
> > branches.
> > You want to go down the middle of it? Use 0101010101...... Place a digital
> > point before it and you'll find 1/3 all the way at the bottom of it, after
> > infinite digits. You want irrationals? Go look up how to define any digit of
> > pi
> > in hexidecimal without knowing the previous digits, tanslate the system into
> > binary, and construct your infinite string. I haven't bothered, but I am
> > pretty
> > sure it can be done. Digital number systems are valid enumerations of
> > naturals
> > and reals.
> > >
> > > > It is a leap to claim that the reals are therefore not "countable",
> > > because it
> > > > rests on the unsound notion that all "countable" sets are equal.
> > >
> > > No it isn't, no it doesn't, and no they aren't. (But countably
> > > infinite sets do have equal cardinality.) And, as stated above,
> > > not relevant anyway here since you've left out almost all of
> > > the reals in your enumeration.
> > I meant countably infinite, sorry. I still disagree. That assumption is
> > wrong,
> > and the whole conclusion that reals are not countable rests on them being a
> > larger set than naturals, because Cantor decided they couldn't be enumerated.
> > It's swiss cheese theory.
> > >
> > > That's just
> > > > wrong, and mathematically insane.
> > > > >
> > > > >
> > > > >
> > > > > > If you want to complain that it would require infinite digits for
> > >
> > > > > > most values, that's irrelevant. it's still an enumeration of the
> > > reals.
> > > > >
> > > > > WRONG!
> > > > That's a pretty weak argument. perhaps you'd like to expand on that
> > > thought.
> > > > >
> > > > > > Enumeration is irrelevant to the level of infinity, except
> > > insofar as we must
> > > > > > rearrange the quantities in order to enumerate them linearly.
> > > Don't look in
> > > > > > your bag of axioms for proof of this. It's been conveniently
> > > excluded.
> > > > >
> > > > > Only by concentrating on counting the wrong things.
> > > > The "wrong" things? define what is right and wrong, oh purveyor of
> > > mathematical
> > > > morality.
> > >
> > > Not immoral. Merely insufficient to do the task that you
> > > claim.
> > I don't even know what you are referring to as "wrong" or why. You might want
> > to be more specific. Then again you might not.
> > >
> > > > >
> > > > > Each path of your infinite binary tree corresponds to a real
> > > number, so
> > > > > until you count paths, not merely nodes, you are not done.
> > > > >
> > > > I counted the nodes in a way that traverses the paths all together.
> > > Check it
> > > > out.
> > >
> > > No, only the finite paths. Sorry.
> > >
> > >
> > Not after an infinite number of iterations.
>
> But you have accomplishedat most countably many iterations, by some rule
> that you have yet to define, and that ain't enough.
>
> Before you can claim to have iterated your way through all of those
> infinitely long strings of binary digits, you will have to show us in
> detail how you get from any one of them to the next.
>
> Which one do you start with?
> Which one comes next? And next, and so on ad infinitum?
> And cn you show that your rule doesn't muss any of the strings at all?
>
> Your hand wving just isn't good enough to convinve anyone with the
> possible exception of yourself.
>
> Or, are you saying there are only
> > finitely many binary numbers?
>
> The other way! There are more than countably many of them.
Interesting, since they are in 1-1 correspondence with the naturals. Would you
like a helmet so the flopping doesn't hurt your head?
>

--
Smiles,

Tony
.

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