Re: Is continuum completely filled up?

David Marcus <DavidMarcus@xxxxxxxxxxxxxx> writes

Let me be even more specific. I will define "the line segment
from P1 to P2" as the set of points P
{P : P = a*P1 + (1-a)*P2, 0 <= a <= 1}

So everything in this set is defined by a real number a
between 0 and 1. There's no such thing as a point which
is not associated with such an a, since that's my membership
test. Do you see that it's obvious there is no member of
my set not corresponding to some a?

No, but in the context of all of this, how do you adjust a such that
successive points in P (there can be no such thing) are continuous ?

Is this a serious question? Or, are you trolling? How can you ask a
question about "successive points" and then add "(there can be no such
thing)"? And, what does "continuous" mean in this context?

Well, yes it was a serious question. This thread is in the context of
"is continuum completely filled up". So I had previously suggested that
you could possibly have holes, depending on how you "construct" the
reals e.g. the 2D fractal curve filling all space, but with all points
both defined and a hole. That may be and probably is bollocks, but I
thought it deserved an answer.

Randy then said, words to the effect of, can't I construct a uniform
mapping of points from P1 to P2 via linear interpolation, and my
response was that, if we are concerned about holes amongst the reals,
and given that no two real points are adjacent, but that continuity is
assured by recursive procrastination (see my previous post), then it is
perfectly reasonable to ask how much one can increase the parameter "a"
by for the linear interpolation. Clearly you can't increase it by "just
enough" to ensure that successive points touch - you can't have adjacent
reals. So the linear interpolation does not ensure a continuous coverage
of the line by real numbers.

Anyway, even if from your perspective it is a dumb/incomprehensible
post, no I'm not trolling.

Regards
--
Andy Smith
.

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