Comment by xyzzyz
11 hours ago
There's no alternative that's significantly easier to understand and to use. The so-called "nonstandard analysis" hasn't caught on, because it's mostly the exact same arguments wrapped in slightly different language, not making them any simpler or shorter.
The language used by mathematicians is subject to constant evolution. 18th and 19th century results in analysis are not expressed and taught in the same way their original authors did it. Newton, Leibniz, Euler, Lagrange, Fourier, Riemann -- none of them expressed their results in terms of epsilons and deltas. These only caught on in the second half of 19th century, and they did so, because they were a better tool to rigorously prove the ideas.
New terminology inventions that make the subjects easier to understand take the field by storm. Some of the relatively recent examples are category theory, homological algebra, or, for that matter, the notion of sheafs, popularized by J.P. Serre himself. Mathematicians are very open to innovation, and intransigence is not the reason why we're stuck with epsilon-delta.
The reason is that nobody has yet come up with a better way of talking about these concepts. I repeatedly observe many people who seem to believe that their difficulty in understanding math stems from mathematicians gatekeeping their results. I think that this belief is just a coping mechanism. Mathematics is genuinely hard, and when people have trouble understanding something, it's easier to think that it's someone else's fault, rather than accepting one's own deficiencies.
> The so-called "nonstandard analysis" hasn't caught on, because it's mostly the exact same arguments wrapped in slightly different language
No. Let's take a nonstandard proof of the intermediate value theorem on [0,1] by Nelson.
By the transfer principle it is enough to prove this for a standard continuous function f on [0,1] with f(0)<0<f(1).
Take a finite subset of [0,1] containing every standard point. Colour its points blue, green, or red according to whether f is negative, zero, or positive at tha point.
The first point of the interval is blue and the last red. Hence either awe can find some green point, or we can find two neighbouring points that have different colours, the first blue and the second red.
In the first case there is a zero, so we are done. In the second, let the neighbouring points be p and q. By the completeness of the real numbers, every nonstandard real in [0,1] is infinitesimally close to exactly one standard real. So p and q are infinitesimally close to some standard real number, let's call it z.
Standard continuous functions send infinitesimally close points to infinitesimally close points. So f(p) and f(q) are both infinitesimally close to f(z). But f(p) is negative and f(q) is positive. The only standard number infinitesimally close to both positive and negative numbers is zero. Thus f(z) is zero. This proves the theorem.
You tell me, which standard proof is this? It's certainly not the nested interval proof. Not the supremum proof. Not the bisection proof in disguise. Which argument does it wrap in slightly different language? Can you point to a single textbook, course note or lecture that gives such an argument?
No. One could of course argue that this is not simpler/shorter than the usual arguments. But it is very different from them. Saying that it's the same arguments repackaged in a different language is just wrong, and detracts from an otherwise valid point.