What are the most overpowered theorems in mathematics?

By "overpowered," I mean theorems that allow disproportionately strong conclusions to be drawn from minimal / relatively simple assumptions. I'm looking for the biggest guns a research mathematician can wield.

This is different from "proof nukes" (that is, applying advanced results to solve much simpler problems). It is also not the same as requesting very important theorems, since while those are very beautiful and high level and motivate the creation of entire new disciplines of mathematics, they aren't always commonly used to prove other things (e.g. FLT), and if they are they tend to have more elaborate conditions that are more proportional to the conclusions (e.g. classification of finite simple groups).

Answers should contain the name and statement of the theorem (if it is named), the discipline(s) it comes from, and a brief discussion of why the theorem is so good. I'll start with an example.

The Feit-Thompson Theorem. All finite groups of odd order are solvable.

Solvability is an incredibly strong condition in that it immediately eliminates all the bizarre, chaotic things that can happen in finite nonabelian simple groups. Solvable groups have terminating central and derived series, a composition series made of cyclic groups of prime order, a full set of Hall subgroups, and so on. The fact that we can read off all that structure simply by looking at the parity of a group's order is amazing.

closed as primarily opinion-based by Najib Idrissi, user26857, user91500, user147263, yoknapatawpha Oct 25 '15 at 14:45

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    I heard Feit-Thompson is getting nerfed in the next version. – Federico Poloni Nov 7 '13 at 9:13
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    disproportionately strong conclusions, minimal / relatively simple assumptions, Fermat's little theorem? – Paul S. Nov 7 '13 at 19:03
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    @FedericoPoloni: nerfed? next version? can you explain what that means? I conjecture it relates to video game patches. – Nikolaj-K Nov 8 '13 at 9:20
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    @NickKidman It's a joke, you guessed well. "Overpowered" is (also) common computer game jargon for some powers/features in a game that are much stronger than the others and should be reduced in power ("nerfed") to preserve the balance of the game. Feel free to remove the comment if you think it's not funny or it shouldn't be allowed here because it's gratuitously off-topic. – Federico Poloni Nov 8 '13 at 9:32
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    @FedericoPoloni: I will sue you in federal court. – Nikolaj-K Nov 8 '13 at 9:40

33 Answers 33

I would include the pigeonhole principle in such a list. This principle is one that is easy to explain to a 4 year old, and at the same time it is used throughout all areas of mathematical research. As an example see Proof of Van der Waerden's theorem (in a special case).

  • So simple, but leads to interesting results! – Cruncher Nov 7 '13 at 14:08
  • Isn't this a little like including the pythagorean theorem? It's a fundamental idea that is used everywhere-hence it is fundamental. Perhaps the original question needs revision. – abnry Nov 7 '13 at 18:55
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    But it's also extremely "flexible". The Pythagorean theorem applies to right triangles, but that's a very particular case. The pigeonhole principle shows up in mealy every field of mathematics, and many of its applications are "clever", rather than straightforward. – Henry Swanson Nov 7 '13 at 21:18

Look into complex analysis. You have insanely powerful theorems all stemming from a function being analytic :).

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    I always though of the results in complex as being a bit like black magic (that is, lots of results seemingly from nowhere). I've since decided the real culprit is in the subtle definition of differentiability, which is actually quite restrictive in maps from $\mathbb R^2$ to $\mathbb R^2$. For instance, on an intuitive level, the maps are conformal almost everywhere. – Brady Trainor Nov 7 '13 at 7:06
  • +1 Such as: (Simple) differentiability implies analytic; Liouville, Picard ... – Hagen von Eitzen Nov 7 '13 at 7:22
  • @BradyTrainor, black magic indeed. – Mr.Fry Nov 7 '13 at 7:33
  • I remember one of my professors telling us that a mathematical proof is just a succession of trivial equivalent statements to get you from a statement A to another statement B. In that sense, no theorem is overpowered as long as it is an equivalence statement. It has as much power as you put in it. – Raskolnikov Nov 7 '13 at 8:14
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    The OP justifies his meaning of "overpowered" in the post. He means a simple assumption that yields a multitude of results. :) But yes you are correct. – Mr.Fry Nov 7 '13 at 8:52

Orbit-stabilizer. There's so many things that can be described as group actions, and tons of counting arguments reduce to it.

The contraction Mapping Theorem. It simply states if $X$ is a complete metric space and $T:X\rightarrow X$ is a contraction mapping then there is a unique fixed point. This theorem is used a lot in studying solutions in numerical analysis and ordinary and partial differential equations.

Stokes' theorem - The integral of a differential form $\omega$ over the boundary of some orientable manifold $\Omega$ is equal to the integral of its exterior derivative $d\omega$ over the whole of $\Omega$, i.e. $$\int_{\partial\Omega}\omega=\int_\Omega d\omega$$

It's amazing and beautiful not only because of its brevity and elegance but also because it unifies from fundamental theorem of calculus, Green's theorem to divergence theorem, even in some generalization, Cauchy's theorem in complex analysis. And we are very glad to accept it from geometric view since it embodies some sort of self-cancellation and symmetry. Moreover, Stokes' theorem is applied to varieties of areas, especially in physics like electromagnetics, hydrodynamics and so on.

  • For comparison, written in vector notation for $\mathbb{R}^3$, Stokes' theorem is: $\oint_{\partial\Omega} \vec{F} \cdot d\vec{r} = \iint_\Omega (\nabla \times \vec{F}) \cdot \hat{n} \, dS$ – Mateen Ulhaq Apr 10 '17 at 4:00

The CENTRAL LIMIT THEOREM. Ever since Laplace proved it, it forms the basis of Probability Theory and Statistics. We can make sense of the world with it, it is simply indispensable and all that thanks to the geniuses of Laplace and Gauss.

Open any probability theory book and you will immediately stumble upon CLT. Under very general conditions we can approximate any distribution with the nice Bell-curve even in not so large samples. In applied research this is a blessing unparalleled with any other theorem.

It is no exaggeration to claim that modern Statistics have been built on these premises and CLT should make it to the top of your list, top 5 minimum ;)

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    ...the basis of Probability and Statistics applicable only when the assumptions are satisfied. In particular when independence is violated, CLT can fail badly in the real world - as opposed to what Savage called small world, and what Taleb called Ludic, ie games or toy systems. – alancalvitti Dec 17 '14 at 1:29

Cantor's Theorem.

For every set $A$, $|A|<|\mathcal P(A)|$.

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    Whoops, just went through another round in my head. 0 is indeed less than 1. – Caleb Jares Nov 7 '13 at 15:37
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    And it's a good thing too! Although $0=1$ would be an overpowered theorem indeed! :) – Asaf Karagila Nov 7 '13 at 15:38

The field of complex numbers $\mathbb{C}$ is algebraically closed. The fundamental theorem of algebra is a result which we use in lot of places. It has its own importance in Mathematics.

Same we can see with any fundamental theorem in Mathematics like Fundamental Theorem of Arithmetic, Fundamental Theorem of Calculus, $\ldots$

The Classification of Closed Surfaces - A closed surface is fully determined by its Euler characteristic (or genus) and whether it is orientable. Further, every closed surface is either

  • The sphere $S^2$,
  • A finite connect sum of tori $T_n=T^2\#T^2\#\cdots\#T^2$, or
  • A finite connect sum of real projective planes $P_n=\mathbb{R}P^2\#\mathbb{R}P^2\#\cdots \mathbb{R}P^2$.

The fact that the homeomorphism type of a closed surface is determined by just two very simple invariants is a very powerful fact indeed. The Euler characteristic is a very easy to calculate invariant and orientability is an especially easy concept to understand and determine for a surface. To think that all closed surfaces (the definition of which is, at first glace, seemingly quite broad) are determined by these criteria is quite remarkable.

It's also worth noting that the closely related statement for $2$-manifolds with boundary and marked points is just as simple to state and even more powerful.

  • But this theorem does simply not have hypotheses... – Mariano Suárez-Álvarez Nov 8 '13 at 9:09
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    I think being a closed surface is a hypothesis on a topological space. The next question to ask is 'can we classify $3$-manifolds?' which is only a slightly different hypothesis and is taking much longer to fully solve (we're nearly there now I think). – Dan Rust Nov 8 '13 at 10:07

The Stone–Weierstrass theorem.

Every continuous function $f: [0,1] \rightarrow \mathbb{R}$ can be uniformly approximated by polynomial functions.

It is 'overpowered' because one only needs to have that $f$ is continuous and we get that we have an approximation of $f$ with polynomials, which behave very nice in many regards.

To give an example, you can use it to prove Brouwers Fixed Point Theorem in the general case from the special case where $f$ is twice differentiable.

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    How can you prove BFPT with that? I am curious! – Pedro Tamaroff Nov 7 '13 at 19:37
  • @PedroTamaroff: In short - When you know that it is true for all $C^2$ functions, you can look at an arbitrary continuous $f$. Then you take a convergent subsequence of the fixed points of the approximating polynomials. A small calculation shows that the limit will be a fixed point of $f$ too. – Listing Nov 7 '13 at 19:49
  • Do you have some reference to read it in detail? – Pedro Tamaroff Nov 7 '13 at 19:50
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    Yes, theorem 1.15 here: mate.polimi.it/viste/pagina_personale/pp/121/fp.pdf but it is also very short. – Listing Nov 7 '13 at 19:53
  • Does this theorem work for all finite intervals? (Scale and shift an arbitrary finite interval around until it fits inside of $[0, 1]$.) – Mateen Ulhaq Apr 10 '17 at 4:06

Gödel's Completeness Theorem.

If $T$ is a first order theory, then $T\vdash\varphi$ if and only if $T\implies\varphi$.

We use this theorem so much, we don't even notice anymore.

The theorem comes from the field of logic. Whenever we want to prove that the theory of groups prove a statement, we begin by considering an arbitrary group, and we show that the group satisfies that statement. Bam, completeness!

If mathematicians consider their work within the context of $\sf ZFC$ (even implicitly), then proving something about $L_2[0,1]$ is in fact proving that $\sf ZFC$ proves that the object which has this and that properties also satisfies that thing. In essence, this is another very implicit use of completeness.

  • What do you mean by T => phi ? – Serge Seredenko Nov 9 '13 at 23:06
  • Every model of $T$ is a model of $\varphi$. – Asaf Karagila Nov 9 '13 at 23:07
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    @AsafKaragila: isn’t it more standard to denote the second relation in the statement by $T \vDash \varphi$? (See e.g. wikipedia.) – Peter LeFanu Lumsdaine Nov 14 '13 at 22:53
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    @Donkey_2009: For finite objects we don't need choice; we also don't need choice to prove the completeness theorem when the language is well-orderable (e.g. finite or countable). Suppose that Voevodsky manages to overthrow conventional mathematics and 20 years from now we all use proof assistants to verify our proofs. The move from arbitrary group to an actual proof is important, and we know that we can do it due to the completeness theorem. – Asaf Karagila Feb 11 '15 at 21:01
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    @Donkey: I will spend two weeks or so in the Cambridge area, doing math and whatnot. Who knows, you might run into me somewhere. – Asaf Karagila Feb 17 '15 at 15:44

The inverse function theorem (for $C^1$ multivariable maps between open sets of euclidean space). You just check that at some point, some determinant is non zero (which is easily computed) and you get that the maps is locally a diffeomorphism at that point!

Given a short exact sequence $0\to A\to B\to C\to 0$ of chain complexes in an abelian category, there exists a long exact sequence $\cdots\to H_{i+1}(C)\to H_i(A)\to H_i(B)\to H_i(C)\to\cdots$ of their homology groups. This gives the long exact sequence for derived functors, and thus allows the production of a myriad of long exact sequences that give immense computational power based on the usually easy to check condition of short exact sequence.

Zorn's lemma.

True, it is equivalent to the axiom of choice. However it is also a theorem of $\sf ZFC$. The uses of this lemma are uncanny and can be found anywhere in mathematics where infinite objects (of unlimited cardinality) are involved.

If $(P,\leq)$ is a partial order in which every chain has an upper bound, then there exists a maximal element in $P$.

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    "The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's lemma?" — Jerry Bona. I like how that joke sums up intuitions (rather than logic) regarding the Axiom of Choice and Zorn's Lemma. – minopret Nov 9 '13 at 20:46
  • I used to like it. Then I learned so many other twisted equivalents to the axiom of choice. Zorn's lemma seems simple and intuitive enough now. – Asaf Karagila Nov 9 '13 at 20:52

Lagrange's Theorem makes proving some aspects of group theory insanely simple. Like, for instance, the proof that any group of prime order is cyclic.

Sylow's first, second and third theorem which are about finite groups are a powerful result in proving results about finite groups. As a nice application of this we can prove that, if $G$ is a finite group of order $pq$ where $p$ and $q$ are distinct primes and $p \lt q$, then if $p \mid q-1$ then there exists a unique non abelion group of order $pq$ in the other case the group is actually turns to be cyclic.

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    Those are important, but they're not terribly powerful in my estimation. Most arguments with Sylow's theorems require a lot of additional legwork. – zibadawa timmy Nov 7 '13 at 6:52

Donaldson's Theorem A: The intersection form of a closed simply-connected definite smooth 4-manifold is diagonalizable.


The intersection form of a 4-manifold contains a large amount of topological information about a manifold, particularly when it is simply-connected. Donaldson's theorem thus tells us that topologically a closed definite smooth 4-manifold can't be too wild. For example, when this is combined with Freedman's work (which would also be a good candidate for an answer to this question), one gets the existence of non-smoothable manifolds.

The changing order of integration in multiple integrals.It is the key of almost all applied mathematics to science.

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    Fubini's theorem is the name. – IAmNoOne May 17 '14 at 6:20

Rice's Theorem is extremely powerful one. It basically says that you cannot invent an algorithm which, taking programming source code as an argument, would discover any property of the function computed by that algorithm.

Example: I really need to make a analyzer, which would take any program and tell me whether that program prints $0$ being given $0$ for input or not. Rice's theorem says I will never have such analyzer.

  • Do you mean "Discover every property of the function computed by that algorithm". How is this related to the Halting Problem? – Daron Nov 10 '13 at 0:40
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    Halting problem is a specific case. You cannot discover whether given algorithm stops or not in general. In other words, you can't discover if a given function is defined in given point. Rise's theorem says that you can't discover any property of a function at all. A little addition: any non-trivial property. – Serge Seredenko Nov 10 '13 at 0:45
  • I think you can trivially prove it given the Halting Problem, though. I might be wrong about this, but just construct a function where whatever property one is trying to discover about it depends on the Halting Problem. – bushdid911 Nov 25 '17 at 21:33
  • @bushdid911 Yes, the theorem is easy to prove, which is nice. – Serge Seredenko Nov 25 '17 at 22:25

Brouwer's fixed point theorem is ridiculously useful.

Perhaps even better is the mean value theorem. We continue to mystify undergraduates by teaching them something in calculus which is not really for solving problems, but it is an awesome tool for actually proving things.

The existence of a primitive root in $(\mathbb{Z}/p\mathbb{Z})^{\times}$, and the Well-Ordering Principle, if you're talking about number theory. The former gives a lot of structure to the prime multiplicative groups, while well-ordering can prove linear diophantine equations, $(a,b)=1 \land a|bc\Rightarrow a|c$, and unique prime factorization in the natural numbers. Also this video by the author of Brown Sharpie is relevant.

Oh! Don't forget about AM-GM, probably the most used inequality theorem ever.

Row-rank equals column-rank

If you have a matrix, then the dimension of the column-space is the same as the dimension of the row-space. Doesn't it sound amazing? I think, because of its applications, it should be called the fundamental theorem of linear algebra.

  • Same answer as that of user1747134 – example Jun 5 '14 at 22:45
  • Has this theorem ever been applied to anything outside of academic mathematics? – DanielV Feb 17 '15 at 16:36

Convex implies continuous.

The existence of non (Lebesgue) measurable sets.

The halting problem.

Yoneda's lemma.

  • Aside from the satisfaction of knowing, what powerful consequences come from the existence of non-measureable sets? – LSpice May 10 at 16:38

In my opinion the Lebesgue convergence theorem (dominated and monotonic) are very powerful tools and have important consequences =)

Convergence citerium for geometric series: For any $r\geq 0$, the geometric series $\sum_{n=0}^\infty r^n$ converges if and only if $r<1$. Much of complex analysis is, in some sense, based off of this fact, as is much of the theory of Banach algebras.

Convergence of monotone sequences: If $a_n\in\mathbb R$ is a non-decreasing sequence, then $a_n$ converges to its least upper bound. I don't think the importance of this result can by overstated. Generalizations (refinements?) to increasing nets of operators and the all important Beppo Levi Theorem come to mind.

I hope powerful Mathematics Techniques will also be accepted here.

I am thinking about Fourier Transform and Fourier Series. For me it seems powerful tool which used in analysis very much.

Similarly Laplace transform kind of transforms are used well in solving differential equations efficiently.

Here are a couple from graph theory:

Hall's Marriage Theorem. A bipartite graph $G$ with partite sets $A$ and $B$ has a matching saturating $A$ if and only if for any $S \subseteq A$, there are at least as many neighbors of vertices in $S$ as there are vertices of $S$.

Not only does the result have many applications itself, but it is also equivalent to several other important and far-reaching results, including:

König's Theorem

Menger's Theorem

Dilworth's Theorem

Max Flow / Min Cut Theorem

Robertson-Seymour Theorem. Among any infinite list of graphs, one graph is a minor of another.

Simple to state, yet very difficult to prove, with so many applications. Particularly, any minor-closed set of graphs can be characterized by a finite list of excluded minors. The proof of the result also developed many new tools for proving other graph minor results.

Gödel's incompleteness theorems

Hypotheses : you have a (recursively enumerable) consistent theory $T$ that contains integers, addition and multiplication (not big hypotheses in math !)

Conclusions :

  1. there is a formula $F$ that is true in some model of $T$ but there is no proof of $F$ in $T$.
  2. The consistency of $T$ is not provable in $T$

All hypothesis are required : counter-example are well know without multiplication for example. I'm amazed that basic arithmetic operations can have such strong consequences.

  • I got two downvotes for this answer. Did I write something wrong ? – Xoff Nov 8 '13 at 13:28
  • I'm not sure, but maybe because the hypotheses misses a very important statement, that T is consistent. Otherwise both conclusions are wrong. – Serge Seredenko Nov 9 '13 at 23:10
  • Sometimes, we miss the most obvious things. :) – Xoff Nov 10 '13 at 11:47
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    You also need to add recursively enumerable to he requirements from $T$. – Asaf Karagila Nov 10 '13 at 11:59

Any $L_p$-integrable function can be $L_p$-approximated by step functions.

This is proved using the same methods as the monotone class theorems and its variants, and can be viewed as the strongest and most powerful version of the monotone class theorem, particularly when working with functions defined on $\mathbb R^n$. The reason is that step functions are finite sums of functions which are essentially trivial, so this theorem allows us to convert any complicated question about integrals into one about those finite sums.

Example (Fubini's Theorem on $\mathbb R^2$): Suppose $f:\mathbb R^2\to\mathbb R$ is such that:

$$ \int_{\mathbb R^2} |f(x,y)| d(x,y) <\infty $$


$$ \int_\mathbb R\left(\int_\mathbb R f(x,y) dy \right)dx = \int_\mathbb R\left(\int_\mathbb R f(x,y) dx \right)dy = \int_{\mathbb R^2} f(x,y) d(x,y) $$

Proof: $f$ is $L_1$-integrable, so we can replace $f$ by a sequence $(f_n : n\in\mathbb N)$ of functions converging to $f$ in the $L_1$-norm such that each $f_n$ takes the form $$ f_n(x,y)=\sum_{j=1}^n a_j 1_{I_j\times J_j} $$ where $I_j,J_j$ are intervals in $\mathbb R$ and $a_j$ is a real number.

Then certainly we have $$ \int_\mathbb R\left(\int_\mathbb R f_n(x,y) dy \right)dx = \int_\mathbb R\left(\int_\mathbb R f_n(x,y) dx \right)dy = \int_{\mathbb R^2} f_n(x,y) d(x,y) $$

So, by dominated convergence we must get the stated result.

Of course, Fubini's theorem is not hard and there are much slicker proofs. Indeed, the theorem we are using is a lot more concrete than other forms of the monotone class theorem, and doesn't give you particularly nice proofs. But you can't beat it for proofs that don't require you to think!

My lingebra professor remarked, that the well known identity $Rank(A) = Rank(A^T)$ for any matrix $A$ can be derived with no special assumptions. (But actually, $A$ musn't have elements from finite field.)

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    What is the problem in finite fields? – Martín-Blas Pérez Pinilla Feb 13 '14 at 7:35
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    It's easy to find a counterexample to the statement. Consider a matrix over $GF(4)$ : $A = \begin{pmatrix} 0 & 2 \\ 0 & 3\end{pmatrix}$. It has $Rank(A) = 2$, indeed, in $GF(4)$, there is: $0\cdot 2=0, 1\cdot 2=2, 2\cdot 2=4=0, 2\cdot 3=6=2$ so the second row is linearly independent of the first one. On the other side, obviously $Rank(A^T)=1$. – user1747134 Mar 2 '14 at 12:41
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    Ehm, isn't $2=0$ in $GF(4)=\mathbb{Z}_2[x]/(x^2+x+1)$? – Leon Apr 26 '14 at 22:34
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    GF(4) is not Z4 like you're treating it; Z4 is not a finite field and hence does not apply. The theorem does hold for finite fields. – Geoff Naylor Jun 5 '14 at 23:46
  • Also, by the 'obvious' definition of linear (in)dependence, the rows of $A$ are linearly dependent: namely, $3\begin{pmatrix} 0 & 2 \end{pmatrix} - 2\begin{pmatrix} 0 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 0 \end{pmatrix}$, and the coefficients are non-$0$ in $\mathbb Z/4\mathbb Z$. – LSpice May 12 at 23:14

protected by MJD Jul 30 '14 at 14:11

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