Monday, November 28, 2022

Precise lengths

As usual, write [a,b] for the interval of the real line from a to b including both a and b, (a,b) for the interval of the real line from a to b excluding a and b, and [a, b) and (a, b] respectively for the intervals that include a and exclude b and vice versa.

Suppose that you want to measure the size m(I) of an interval I, but you have the conviction that single points matter, so [a,b] is bigger than (a,b), and you want to use infinitesimals to model that difference. Thus, m([a,b]) will be infinitesimally bigger than m((a,b)).

Thus at least some intervals will have lengths that aren’t real numbers: their length will be a real number plus or minus a (non-zero) infinitesimal.

At the same time, intuitively, some intervals from a to b should have length exactly b − a, which is a real number (assuming a and b are real). Which ones? The choices are [a,b], (a,b), [a, b) are (a, b].

Let α be the non-zero infinitesimal length of a single point. Then [a,a] is a single point. Its length thus will be α, and not a − a = 0. So [a,b] can’t always have real-number length b − a. But maybe at least it can in the case where a < b? No. For suppose that m([a,b]) = b − a whenever a < b. Then m((a,b]) = b − a − α whenever a < b, since (a, b] is missing exactly one point of [a,b]. But then let c = (a+b)/2 be the midpoint of [a,b]. Then:

  1. m([a,b]) = m([a,c]) + m((c,b]) = (ca) + (bcα) = b − a − α,

rather than m([a,b]) as was claimed.

What about (a,b)? Can that always have real number length b − a if a < b? No. For if we had that, then we would absurdly have:

  1. m((a,b)) = m((a,c)) + α + m((c,b)) = c − a + α + b − c = b − a + α,

since (a,b) is equal to the disjoint union of (a,c), the point c and (c,b).

That leaves [a, b) and (a, b]. By symmetry if one has length b − a, surely so does the other. And in fact Milovich gave me a proof that there is no contradiction in supposing that m([a,b)) = m((b,a]) = b − a.

1 comment:

Alexander R Pruss said...

Yes, that's the proposal I suggest in the post.