Re: Well Ordering the Reals

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 contesxt 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.

>
> > > 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.

>
> > 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. If there are no 0's and the string
is all 1's, invert the entire string.
>
> > 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.
>
> > and invert that portion of the string. This is the
> > general rule for increment in binary: locate the rightmost 0, and
> > invert all bits up to and including that bit.
>
> 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?
>
> Or does TO have some finite upper bound on the number of digits that one
> "limit point" can cover?
Nope. I am saying the rightmost 0 to the left of the 0 point.
>
> > > > >
> > > > > > What is it you hate so much about me or my ideas? Is it just
> > > > > > because I smell like a cat?
> > > > >
> > > > > I am quite fond of most cats. But ideas that are
> > > > > self-contradictory and people who try to sell them I am not
> > > > > fond of.
> > >
>

--
Smiles,

Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.

Relevant Pages

  • Re: Well Ordering the Reals
    ... > Dense can have more than one meaning, which you aptly demonstrate every day. ... rightmost zero in an unending sequence of ever more rightward zeros, ... of TO's T-errible two-ended infinite string of digits. ...
    (sci.math)
  • Re: Well Ordering the Reals
    ... An ordered set is dense if and ... the integers look dense on that scale. ... "rightmost" zero. ... point) there is an unending sequence of digits each finitely far from ...
    (sci.math)