From what I understand after thinking about this, delta epsilon really seems to formalize the notion of an infinitesimal. The constraint $0<|\delta-c|$ combined with the fact that there is no real number that doesn't satisfy the delta-epsilon definition, then don't these two combined facts mean that an infinitesimal exists just not within the reals? In other words, there is a value that is greater than zero that is smaller than all possible real numbers (from my understanding), and the only value this could be is an infinitesimal. But I know that infinitesimals don't exist within the realm of real analysis. So where is my thinking wrong?
Asked
Active
Viewed 1,206 times
2
-
See any textbook on Nonstandard Analysis. Abe Robinson devised this to make rigorous Leibniz's notion of infinitesimals. – Bill Dubuque Jun 23 '16 at 05:09
-
1@Bill, I think Abe had broader goals in mind :-) – Mikhail Katz Jun 23 '16 at 12:20
-
rb612, you may want to consult my answer here. – Mikhail Katz Jun 23 '16 at 12:22
-
@BillDubuque thanks Bill. I'm trying to ask how the conventional way of calculus and real analysis with epsilon-delta isn't just a formalized version of the infinitesimal. – rb612 Jun 23 '16 at 17:40
1 Answers
5
The opposition "limit versus infinitesimal" is a bit of a false opposition because limits are present both in the A-track approach working with an Archimedean continuum (the real numbers), and in the B-track approach working with a Bernoullian continuum (i.e., an infinitesimal-enriched one).
In the A-track, limit is defined via epsilon-delta definitions.
In the B-track, limit is defined in a more straightforward way using infinitesimals.
For example, $\lim_{x\to0}f(x)$ can be defined simply as the standard part of $f(\alpha)$ where $\alpha\not=0$ is infinitesimal.
Both epsilon-delta techniques and infinitesimals provide rigorous ways of handling the calculus. The relation between them can be stated as follows. The epsilon-delta techniques and definitions are a paraphrase of infinitesimal techniques and definitions.
Mikhail Katz
- 47,573
-
Thank you! Can you elaborate more on how track A does not use infinitesimals or something very similar? What I'm trying to get at is how epsilon-delta isn't just a rigorous version of infinitesimals. In other words, epsilon delta extends to all possible real numbers and the difference must be greater than zero. This seems like an infinitesimal rigorously defined to me. Am I wrong in thinking this? – rb612 Jun 23 '16 at 17:38
-
More precisely, epsilon-delta is not a "rigorous version of infinitesimals" but rather a paraphrase of arguments and definitions using infinitesimals. The definition of continuity is a good illustration of this. – Mikhail Katz Jun 24 '16 at 07:25