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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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closed as no longer relevant by Fernando Muro, Emil Jeřábek, Felipe Voloch, Mark Sapir, Andy Putman May 14 '13 at 23:13

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I once saw someone proving resolutions of singularities of curves by quoting Hironaka's theorem. – Richard Borcherds Oct 17 '10 at 15:23
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 '10 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 '10 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 '11 at 6:48

67 Answers 67

The case of Fatou's theorem for H^2 can be proven as follows:

By Carleson's theorem the series $ \sum a_n e^{i \theta n} $ converges for almost all $\theta$ if $ \sum |a_n|^2 < \infty$. Now we can appeal to Abel's theorem to conclude that the function $ f(z)= \sum a_n z^n$ has radial limits almost everywhere on the unit circle. (I am not sure if we can get non-tangential limits this way.)

But Carleson's theorem is a much more difficult theorem than what we have proved here. (I got this example from a Hardy space course I am taking right now.)

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I don't understand the details of Carleson's Theorem (who does? Genius required!), but I thought one of the main standard techniques for both results was to use maximal functions; so the two results definitely have strongly related proofs (with the Carleson one being much more difficult, of course). Although that doesn't mean it's actually circular, of course! – Zen Harper Dec 9 '10 at 8:42

One can use the continuous functional calculus of a C$^*$-algebra (namely $M_N(\mathbb{C})$) to prove that a normal matrix is diagonalizable.

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ok, but this is not really an awful sophisticated proof. this is the standard way of proofing the spectral theorem for normal operators. if one considers the special case of normal operators on finite dimensional spaces - viz, matrices - you get this. – Delio Mugnolo Feb 25 '13 at 5:35

$Forest$ 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 '13 at 23:33

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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Is that a "simple fact"? – Robin Chapman Oct 17 '10 at 16:58
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. – HJRW Oct 17 '10 at 17:40
unknown - for suitable definitions of 'heuristic' and 'simple', yes, I do. But the key word in the question is 'disproportionate'. – HJRW Oct 17 '10 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… which may be close to being solved, but has been open since 1911! – Thierry Zell Oct 17 '10 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 '10 at 22:25

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 Jun 15 '11 at 15:10
Yes, like Mark, I don't think this is in the intended spirit of the question, which is about sophisticated proofs for facts that have much easier proofs. – Todd Trimble Dec 22 '12 at 7:22

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 '12 at 2:42
@Mariano Suárez-Alvarez: I thought it's noncircular for sufficiently large n, that's why I phrased the example the way I did. It is probably circular for n=5. I am aware that this can be proved using "any subgraph of a planar graph is planar", and "K5 is nonplanar" or "Euler's theorem", all of which are preliminary results to 4-color-theorem, but it was not clear to me that this consistutes a circularity, as this is a statement with a quantifier, just a ridiculously easy one to prove. I was testing the limits of the question, in a sense. I agree it's not 100% in the spirit. – Ron Maimon Jan 5 '13 at 16:12
Hi @Ron, you don't know me but I was hoping to buy you a beer sometime in NYC. Shoot me an email if you'd like to say hello. (Couldn't find a better way to contact you!) – Jess Riedel Apr 17 '13 at 16:17
I gave once as an easy exercise on an exam the following easy problem: If we remove $2$ edges from $K_7$, the resulting graph is not planar...One student solved using the 4 colors theorem :) – Nick S Jan 8 '14 at 23:42

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 '13 at 7:37
@Mariano, I didn't say among whom to be discreet about it. At the time I rediscovered (it sounds funny) the periodic resolutions due to the free actions of the cyclic groups on the spheres. Later I saw a paper on periodic projective resolutions by Swan (it covered more advanced material of course). – Włodzimierz Holsztyński May 12 '13 at 4:12

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