Comment by saithound

> 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.