# Unconventional (but instructive) proofs of basic theorems of calculus

Inspired by this questions asked on MathOverflow, I would like to ask if you know some "sophisticated" proofs of the basic theorems in a calculus course (that is, the ones that you can find, for instance, in Spivak's Calculus).

In this case, by "sophisticated", I do not mean awfully complicated, but unexpected, extremely clever and unconventional (and hopefully instructive), either because they use concepts from other areas of mathematics or because they enlighten a theorem by tackling it from a non-obvious and by no means standard(ized) perspective.

• The title says "unconventional" but the body says "sophisticated". Do you want unconventional but not necessarily sophisticated proofs? – JohnD Dec 15 '14 at 21:11
• @JohnD I wrote: "In this case, by "sophisticated", I do not mean awfully complicated, but unexpected, extremely clever and unconventional (and hopefully instructive)". So, yes. Do you have any such examples? – Dal Dec 15 '14 at 21:27
• "Extremely clever" is a high bar... ;-) – JohnD Dec 15 '14 at 21:28
• @JohnD Well, extremely clever for me, so it may as well feel slightly more than normal to you and other people here. – Dal Dec 15 '14 at 21:30

I always love to prove that:

If $\{a_n\}_{n\in\mathbb{N}}$ is a bounded real sequence, it has a converging subsequence.

with the Erdos-Szekeres', or Dilworth's, theorem:

(Erdos-Szekeres, finite version) Every sequence with $n^2+1$ terms admits a weakly monotonic subsequence with $n+1$ terms.

(Dilworth, infinite version) Every infinite POset contains an infinite chain or antichain.

To prove Erdos-Szekeres, we send $n^2+1$ people to a post office with $n$ employees, $n$ queues. When a person arrives, he takes place in the first queue such that he is taller than the last person in the queue. If at some point someone ($A$) is not able to take place, then the people in the last position of every queue and $A$ give a decreasing sequence. On the other hand, if everyone is able to take place, there is a queue with at least $n+1$ people in it, giving an increasing sequence.

So we can use Erdos-Szekeres' or Dilworth's theorem to extract a (weakly) monotonic and bounded subsequence from $\{a_n\}_{n\in\mathbb{N}}$. Such a subsequence is clearly converging to its $\sup$ or $\inf$, and we are done.

• This seems really good to me. Thank you very much for sharing it :). I will really appreciate also other examples. – Dal Dec 8 '14 at 11:46

I think the classic example of this is the whole field of non-standard analysis. It took 300 years to make infinitesimals rigorous (finally realized in the 1960's), but once equipped with such a toolkit you can derive all the basic calculus results (and much more) in just a couple of lines of infinitesimal algebra.

It's interesting to try to prove the basic results on derivatives without using the mean value theorem.

See for instance this previous MO discussion of the role of MVT in first-year calculus, a related post I made on the Expii "Mean Value Theorem" page, and (also on Expii) a MVT-free proof (inspired by the proof (given e.g. in Stein & Shakarchi) of Cauchy/Goursat from complex analysis, following a blog post of Gowers) of the fact that $f'=0$ on an interval implies $f$ is constant.

Here is an unconventional but instructive proof of the extreme value theorem for a continuous function $f$ on the interval $[0,1]$. Let $H$ be (gasp) an infinite hypernatural. Partition the interval into $H$ equal subintervals, each of infinitesimal length. Among the partition points $p_i$, choose the one, say $p_{i_0}$, with the maximal value of $f$. Now round off $p_{i_0}$ to the nearest real number $c$, so that $p_{i_0}$ is infinitely close to $c\in\mathbb{R}$. Then $c$ is a required maximum of $f$. This is because, by definition of continuity, the composed function $\text{st}\circ f$ is constant on the halo of $c$, where "st" is the standard part function.