Re: Is continuum completely filled up?

Andy Smith wrote:
Randy Poe <poespam-trap@xxxxxxxxx> writes
Andy Smith wrote:

Speaking intuitively, I would have said that the line in[0,1] is a
1-dimensional space and it is up to somebody to demonstrate that you can
cover it with (an uncountable infinity of) points of zero dimension.

The point of my post was that you can't even ask the
question "do the reals cover it" until "it" is defined. Any
attempt I might make to define the line (for instance
in terms of distance from 0) depends on the real numbers.

So I don't know what "it" is. If I think of it as a representatioin
of [0,1], then of course [0,1] covers it because [0,1] and
"it" are essentially the same thing. "It" has no existence (in
my scheme) outside of [0,1].

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?

--
David Marcus
.

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