Re: Is continuum completely filled up?

I

Distinct points on the line cannot have a separation of 0. If a and b
are distinct real numbers, then | a - b | > 0.

However, there is a way to represent the concept of things being "next to
each other" on the line. That is to consider sets of points, rather than
individual points. For example, let

A = { x in Q : x < 0 } (the negative rationals),
B = { x in Q : x > 0 } (the positive rationals).

The sets A and B are distinct, but they approach each other more closely
than any positive real number. In fact, we can say that distance(A,B) =
0, where the distance between sets is defined by

distance(X,Y) = inf{ |x-y| : x in X and y in Y }.

Here, "inf" stands for "infimum", which means the greatest lower bound.

But notice, even though distance(A,B) = 0, there is nevertheless a gap
between the sets. The number 0 lies strictly between A and B and does
not belong to either.

The point is, requiring a distance to be zero is not sufficient to
guarantee the absence of a gap. After all, points have zero width and
therefore they can fit into a gap without causing any actual separation.

This is why the notion of completeness does not rely on distance.
Instead, completeness uses the least upper bound property. The set A
does not have a maximum element, but it does have a least upper bound,
and the LUB is a real number. The same is true of any nonempty set of
reals that is bounded above.

Thank you, understood that, and I can see that things are inevitably more tricky than they might seem. But I don't see that that answers my simplistic argument about gaps between the numbers?
--
Andy Smith
.

Relevant Pages

  • Re: Is continuum completely filled up?
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  • Re: Is continuum completely filled up?
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