Ceiling function of an infinitesimal

Question

I was working with infinitesimals and I came across the problem: what is the ceiling function of an infinitesimal? Wolfram Alpha says an infinitesimal equals zero, so therefore the ceiling function of it should equal zero. But what I've seen is that an infinitesimal is not equal to zero but less than any positive real value, so I thought it would be one. What is the right answer?

What's your definition of an infinitesimal? It's not rigorous but wikipedia says "an infinitesimal object is an object that is smaller than any feasible measurement, but not zero in size" — graydad, Jan 20 '16 at 04:55
Hmm, I'd be inclined to B.S. my way and say that as an infintismal has no measurable value it isn't measurable greater than 0 and the ceiling should be 0. I don't actually know, but I don't think the ceiling function is meaningful and I'm guessing that anyting other than 0 will lead to inconsistant results. What is the floor of 1 - an infintismal. subtracting by an infintisimal doesn't make it measurably smaller so 0 seems wrong. — fleablood, Jan 20 '16 at 06:46
Can you give a plausible example of why and when you would want to consider applying floor or ceilings to infinitesimals? — mrf, Jan 20 '16 at 09:14
If you're working with smooth infinitesimal analysis (which rejects the law of the excluded middle and has infinitesimal numbers $\epsilon\in\Bbb R$ such that neither $\epsilon=0$ nor $\epsilon\ne0$), the ceiling function isn't defined at all. (You can't say $\epsilon>0$, because that would imply $\epsilon\ne0$, which is false; similarly for $\epsilon<0$.) In fact, without the law of the excluded middle, you can't prove that any discontinuous functions exist. Indeed, in smooth infinitesimal analysis, all functions are continuous. — Akiva Weinberger, Jan 20 '16 at 13:54

score 7 · Answer 1 · answered Jan 20 '16 at 09:10

If you’re working in the hyperreals, the question is meaningful, and the answer follows immediately from the definitions:

if $\alpha$ is a positive infinitesimal, then $0<\alpha<1$, so $\lceil\alpha\rceil=1$, and
if $\alpha$ is a negative infinitesimal, then $-1<\alpha<0$, so $\lceil\alpha\rceil=0$.

score 0 · Answer 2 · answered Jan 21 '16 at 16:24

To respond to a closely related issue raised in a comment following the question, note that applying the floor function to hyperreals plays an important role when one wishes to define the hyperintegers. Namely, a hyperinteger can be defined as a hyperreal number $H$ satisfying $H=\lfloor H\rfloor$. Hyperintegers play a very basic role in the theory. For example, the convergence of a sequence $(u_n)$ to $L$ can be characterized as follows: $u_H\approx L$ for all infinite positive hyperintegers $H$, where $\approx$ is the relation of infinite proximity (i.e., the difference of the two sides is infinitesimal).

Ceiling function of an infinitesimal

2 Answers2