Re: abundance of irrationals!)

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. Or, are you saying there are only
finitely many binary numbers?

--
Smiles,

Tony
.

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