Re: Well Ordering the Reals
- From: Virgil <ITSnetNOTcom#virgil@xxxxxxxxxxx>
- Date: Tue, 06 Dec 2005 17:00:38 -0700
In article <MPG.1dff9bce5e943ea898a82a@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
> Virgil said:
> > In article <MPG.1dfe65ec152d14598a819@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
> >
> > > Virgil said:
> >
> > > > Density is not a matter of scale. An ordered set is dense if and
> > > > only if between any two members there is another member.
> > >
> > > It really is. If you stand infinitely back and look at the number
> > > line, the integers look dense on that scale. It's a matter of
> > > relatively infinitesimal distance between elements of the set, is
> > > all.
> >
> > "Dense" has a mathematical meaning in the context of ordered sets. In
> > mathematical discussions such meanings take priority over any
> > non-mathmatical meanings. And I would like to see TO try to stand
> > infinitely far back from any line.
>
> Dense can have more than one meaning, which you aptly demonstrate every day.
"Dense" has only one meaning in the context of ordered sets, though TO
seems to be too dense to comprehend that fact..
>
> >
> > > > Here is one more question: Imagine a "TO-number" with zeros from
> > > > its left end rightward to the furthest extent covered by some known
> > > > internal "limit point', and all 1's from there on rightward. What
> > > > is its successor?
> > > A 1 where the rightmost 0 is, and 0's from there rightward,
> >
> >
> > But according to TO's own descriptions, there cannot be any such
> > "rightmost" zero. To the right of any limit point (except the right end
> > point) there is an unending sequence of digits each finitely far from
> > that limit point, so there cannot be a "last" one.
>
> Since we are talking about whole numbers, bits to the right of the 0 point
> are
> all zeroes and ignored. So, I am referring to the rightmost 0 that is left of
> the 0 point.
TO is relying on the non-existent again. He is referring to the
rightmost zero in an unending sequence of ever more rightward zeros,
those which are a finite number of places to the right of the left end
of TO's T-errible two-ended infinite string of digits.
>
> >
> > > just like
> > > regular binary naturals would be.
> >
> >
> > Not at all like regular binaries (with only finitely many digits in
> > each) would be.
> Exactly as they would be. To increment any positive integer, find the
> rightmost
> 0 and invert it and every bit to its right.
But, as TO constructs them, there can be TO-strings staring at the left
with 0, and having an endless string of 0's followed by infintiely many
1's.
> > > Carry the increment across all 1's to the first 0,
> >
> >
> > Which can not exist in TO's construction!
> Of course it can. Please give an example where you think it doesn't work, so
> I
> can analyze your misconception.
I have: A string starting at the left with a zero at every finite
natural distance from the left end, and 1's everywhere else.
There cannot be a rightmost 0 in such a number because there cannot be
a largest finite natural.
> >
> > How can one locate the rightmost digit among a set of digits for which
> > there is no rightmost?
> To the left of the 0 point. These are whole numbers we're incrementing right?
These are TO-numbers, which makes them wholly illusional. But as TO
descxribed them, they have a leftmost digit, a rightmost digit,
uncountably many digits in between in a sequential order. Have I
misreprestened these TO-numbers in any way?
> Nope. I am saying the rightmost 0 to the left of the 0 point.
Which does not exist (see above), because each 0 is followed by another
0, but all of them are followed by uncountalby many 1's.
It is TO's system, not mine, so he is the one responsible for this
anomaly.
.
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