## Awfully sophisticated proof for simple facts [closed]

It is sometimes the case that one can produce proofs of simple facts that are of disproportionate sophistication which, however, do not involve any circularity. For example, (I think) I gave an example in this M.SE answer (the title of this question comes from Pete's comment there) If I recall correctly, another example is proving Wedderburn's theorem on the commutativity of finite division rings by computing the Brauer group of their centers.

Do you know of other examples of nuking mosquitos like this?

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I once saw someone proving resolutions of singularities of curves by quoting Hironaka's theorem. – Richard Borcherds Oct 17 2010 at 15:23
rjlipton.wordpress.com/2010/03/31/april-fool – Steve Huntsman Oct 17 2010 at 15:42
Brauer groups and cohomology are certainly overkill for Wedderburn's theorem: if $D$ is a finite division algebra and $L$ is a maximal subfield, then the Noether-Skolem theorem shows that the multiplicative group of $D$ is a union of conjugates of that of $L$; hence $D$=$L$. – JS Milne Oct 17 2010 at 20:07
@Maxime: I have trouble believing that such a proof is actually non-circular. Surely such proofs form a step, however easy, in the classification. – Qiaochu Yuan Oct 17 2010 at 21:59
I once convinced myself the Cantor set is non empty because it is a descending intersection of non empty closed subsets of a compact set, before noticing it contains 0. – roy smith Jan 29 2011 at 6:48
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## closed as no longer relevant by Fernando Muro, Emil Jeřábek, Felipe Voloch, Mark Sapir, Andy PutmanMay 14 at 23:13

The density Hales-Jewett theorem implies that there cannot exist perfect magic hypercubes of fixed side length $k$ and arbitrarily high dimension $n$ whose cells are filled with the consecutive numbers $1,2,\dots,k^n$ and for which the numbers in cells along any geometric line sum to the magic constant $\frac{k(k^n+1)}{2}$.

For, take the cells with numbers $1,2,\dots,\left\lfloor\frac{k^n}{2}\right\rfloor$.

This always has density about $1/2$, and so by the density Hales-Jewett theorem, will contain a hyperline for sufficiently large $n$. But no $k$ numbers from this set of density about $1/2$ can ever sum to the magic constant.

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$Forest$ is in $P$. Given a finite undirected graph $G$ one can in polynomial time decide whether the input is a forest. The class of all finite forests is a minor-closed property and by the Robertson–Seymour theorem, there are finitely many forbidden minors. We can in $O(n^3)$ time test whether $G$ contains a forbidden minor and if not, output yes.

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Although I like the example, I'm not sure I follow your argument. For the case of forests we already know the finite set of forbidden minors: $\{C_3\}$. So Robertson-Seymour doesn't really enter the picture except via the $O(n^3)$ test, which is really a different theorem. – András Salamon Mar 28 at 23:33

Claim: $\sum\limits_{k=0}^n (-1)^k {n\choose k} = 0$ for all integers $n≥1$

Proof: Take the $n-1$-dimensional simplex $\Delta_{n-1}$. We can compute it's Euler characteristic by using simplicial homology. There are exactly $n \choose k+1$ many $k$-sub-simplexes of $\Delta_{n-1}$. Thus we get a simplicial chain complex of the form $\mathbb{Z}^{n\choose n} \to \mathbb{Z}^{n\choose n-1} \to \cdots \to \mathbb{Z}^{n\choose 2}\to\mathbb{Z}^{n\choose 1}$. So the Euler characteristic is $\chi(\Delta_{n-1}) = \sum\limits_{k=0}^{n-1} (-1)^k {n\choose k+1}=-\sum\limits_{k=1}^{n} (-1)^k {n\choose k}$
On the other hand $\Delta_{n-1}$ is contractible, and $\chi$ is homotopy-equivalence-invariant, so $\chi(\Delta_{n-1})=\chi(pt) =1$.
Putting those toghether we obtain: $0=\chi(\Delta_{n-1})-\chi(\Delta_{n-1})=1+\sum\limits_{k=1}^{n} (-1)^k {n\choose k}=\sum\limits_{k=0}^n (-1)^k {n\choose k}$

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If $0\le f_n \le 1$ is a sequence of continuous functions on $[0,1]$ that converges pointwise to $0$, then $\int_0^1 f_n(t) dt$ converges to $0$. Understandable by freshman, the statement is hard to prove using only the tools of calculus but is immediate from the dominated convergence theorem.

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I don't see this as a simple fact. To construct Lebesgue measure you usually have to prove such a statement (or something similar) anyway. – Mark Schwarzmann Jun 15 2011 at 15:10
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Kn is non-planar for n>4: it contradicts the four-color theorem.

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To qualify as a good answer, it has to be non-circular... Are we sure this passes that test? – Mariano Suárez-Alvarez Dec 30 at 2:42
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Around year 1970 a popular way to compute cohomology groups of the finite cyclic groups was by applying spectral sequences (which was quite an overkill).

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This was popular among whom? The book by Cartan and Eilenberg, the very first textbook on the subject, already has the computation done in terms of the usual very small periodic projective resolution: after that, using anything else to compute this seems pretty weird! – Mariano Suárez-Alvarez May 11 at 7:37
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As Helfgott uploaded a proof of the weak Goldbach conjecture it is now possible (but I guess circular) to proof that there are infinity many primes in this way.

Suppose there are only finite many primes, let $p_{\max}$ be the highest prime number, then $3 p_{\max}+2$ would be an odd number which is not the sum of 3 primes in contradiction to goldbachs weak conjecture.

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Vinogradov's theorem already suffices for this (but that theorem in turn relies on the prime number theorem, which certainly is stronger than Euclid's theorem). In any case Helfgott's argument uses effective estimates on the number of primes less than x which also gives Euclid's theorem. – Terry Tao May 14 at 20:30

The Jordan curve theorem. As far as I know, the "elementary" proof is quite involved, at least with respect to the intuitive plausibility of the statement.

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I think the idea of this question is to judge the simplicity of the fact by the length of the shortest possible elementary proof, not by the length of the statement. – HW Oct 17 2010 at 17:40
unknown - for suitable definitions of 'heuristic' and 'simple', yes, I do. But the key word in the question is 'disproportionate'. – HW Oct 17 2010 at 19:01
The Jordan curve theorem is not intuitive: it deals with continuous curves, and at that level of generality it is quite legitimate to expect the worst. The result is almost obvious for $C^1$ curves, of course, but there is a chasm between $C^0$ and $C^1$, and I can think of a couple of "intuitive" results like this which are not yet even proved in the $C^0$ case. See e.g. the square pegs & round holes problem quomodocumque.wordpress.com/2007/08/31/… which may be close to being solved, but has been open since 1911! – Thierry Zell Oct 17 2010 at 22:52
I am not sure it is so intuitive, even in the nice $C^1$ case. For example, could you explain to a child why the results holds in the plane and not in the torus ? – Hugh J Oct 18 2010 at 22:25
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In 1993 D.Christodoulou and S.Klainerman proved Global Nonlinear Stability of the Minkowski Space. It was one of the most important results of the mathematical General Relativity. Their proof was published in an over 500-pages book. It stated, that any initial data "sufficiently close" to that corresponding to the Minkowski space will remain so forever. Most of mathematical physicists belived this frankly simple fact but the proof was one of the most sophisticated in the field.

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"Simple" in the context of this topic means "simple to prove", not "easy to believe". – darij grinberg Mar 25 2011 at 15:02
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