Re: Is continuum completely filled up?

Andy Smith wrote:
Thanks. I had a similar clarification a few posts back from Dave Marcus
who observed that defining all the reals is different from defining one
instance of the reals once the reals have been defined, and I need to
read up on "defining the reals".

Actually, that would be "constructing the reals".

Following a rather chaotic line of thought re the title of this thread,
what about space filling curves - covering the plane with a curve is the
same issue as covering the line with points.

Space filling curves e.g. Hilbert, Peano are fractals; there is a
generator function - a suitably line defined in e.g. a square. At each
iteration the generator is e.g. reduced in linear scale by a factor 2,
replicated by 4 (with some defined rotations for Hilbert) and the scheme
is then iterated. As I understand it, the gist of the argument that it
is space filling is that, as a function of iteration number n, one can
show that one can choose an n such that the distance from any point is
less than any eta>0 and that the distance reduces for increasing n.

The argument is more complicated than that. What you wrote just shows
that we can find a sequence of iterations that come arbitrarily close to
a given point. However, it turns out that the limit defines a curve and
this curve completely fills the space.

What about foreground and background - like Escher's array of white
birds flying one way with black gaps, but on inspection it is equally
black birds flying the other on a white background (Dave Marcus
mentioned Godel, Escher, Bach - a book I haven't looked at for 25 years,
but it has nice examples of all of that as illustrations). Anyway, what
I mean is, suppose that instead of the generator function being a
geometric line, make it a thick line, such that the area of the line is
the same as the area of its enclosing space - if you wish, you can
consider it as two geometric curves of zero width, delineating the
boundaries between white and black.

Then, when we iterate using the same argument as for a single line, we
can conclude that any point on the plane is covered by both curves.
However, by construction, there are as many white points (or holes) as
there are black points. And I think that points alternate between black
and white on subsequent iterations, so that their black/white status is
undefined in the limit? I am now very aware that I shouldn't apply
finite reasoning to infinite sets, but in this case there is a complete
symmetry between the status of black/white, so anything applicable to
black is equally so to white, finite or infinite.

One could maybe do something similar for defining the reals (I have
still to read up on that). But one could, following an analogy with the
2D case above do something like - define the reals as a subset of the
rationals in a recursive fashion, proceeding 1 bit at a time, so
starting with {.0,.1}, then {.00,.01,.10,.11} etc. However, at each
iteration n divide the set of rationals into two sets - black (exist) or
white(holes), according to : if n even, let black = all even bit strings
(last bit = 0), white = all odd (last bit = 1); for odd n, the reverse.
At the limit the black/white status of any real number is undefined
(the last bit, there isn't one, = either 0 or 1), but any real can still
be defined with the rationals in either set - from the point of view of
the Cauchy sequence defining a real, it doesn't matter if the last bit
(there isn't one) is a 0 or a 1? So maybe it is a matter of definition
as to whether the continuum is filled up?

Well, this is kind of backwards. We can do lots of things. Some of them
accomplish something and some don't. The real numbers are a specific
thing. If you want to define/construct something else, it may have
different properties. Why is that surprising?

--
David Marcus
.

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