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What are the most interesting results in mathematics that say there are only finitely many of something?

Examples:

  • If it's ever shown that there are only finitely many twin primes, that would fit here.
  • The compactness theorem for first-order logic is an example. Suppose $S$ is a set of first-order sentences. Suppose $\varphi$ is a first-order sentence that is true in every structure in which every sentence in $S$ is true. Then for some finite $S_0\subseteq S$, $\varphi$ is true in every structure in which every sentence in $S_0$ is true.
  • The Robertson–Seymour theorem is an example. If a class of finite graphs is closed under taking minors, then it is equal to the class of graphs that have no minor isomorphic to any of the members of a set of "forbidden minors". The theorem says that that set is always finite. (Among such classes of graphs are planar graphs, outer-planar graphs (embeddable in a plane with all vertices on a circle), graphs knotlessly embedable in $\mathbb R^3$, and graphs linklessly embedable in $\mathbb R^3$. In some cases, finding the finite set of forbidden minors is challenging.
  • The Borel–Cantelli lemma considers an infinite sequence $E_1, E_2, E_3, \ldots$ of events in a probability space. It says that if $\Pr(E_1)+\Pr(E_2)+\Pr(E_3)+\cdots<\infty$ then only finitely many of $E_1, E_2, E_3, \ldots$ occur. (Or maybe more precisely,: the probability that only finitely many occur is $1$. The set of outcomes for which infnitely many of the events occur may be non-empty, but its measure is $0$.)

PS: Maybe two different sorts of results can fit here:

  • Results that say, for example, exactly eleven of something exist, and maybe lists them. For example, there are only five regular polyhedra, and we all know which five. If only finitely many twin primes exist, then perhaps a theorem would say exactly which ones they are.
  • Results that say that something must in every instance be finite, but do not and cannot give a finite upper bound. The The Borel–Cantelli lemma is of that sort. Even specifying which events are in the sequence and what their probabilities are does not make it possible to give a finite upper bound. The compactness theorems of logic are also of that sort. And maybe I should have mentioned compactness theorems of topology? Tychonoff's theorem, maybe? The Robertson–Seymour theorem in general is of that sort, but some of its concrete instances are substantial theorems in their own right. The one that says planar graphs are those that exclude two specified forbidden minors. I seem to recall a simple example in which their were something like 30 forbidden minors – maybe it was triangulations of a torus or something like that.
  • Maybe in some cases the question of whether an example is of the first kind or the second might itself be a hard problem.

Despite my inclusion of the first bullet point above, somehow I hesitate to include uniqueness theorems in general. There are zillions of those, and I don't think anyone would want to compile a comprehensive list of uniqueness theorems in mathematics unless maybe they're creating a reference book.

Michael Hardy
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    There are only finitely many sporadic simple groups? – Hagen von Eitzen Jul 05 '13 at 13:35
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    Picard's theorem: If $z_0$ is an essential singularity of $f$, then all complex values with possibly one exception are assumed infinitely often by $f$ in every punctured neighbourhood of $z_0$. – Daniel Fischer Jul 05 '13 at 13:38
  • @HagenvonEitzen : I think that fits in this list. You could post it as an answer. If this question is considered worthy of attention, I suspect each answer will be only one example or only a few. – Michael Hardy Jul 05 '13 at 13:43
  • Just which tags this question should bear might be a bit of a challenge. – Michael Hardy Jul 05 '13 at 13:56
  • There's finitely many integer solutions to $x^n + y^n = z^n, n \ge 3$. :) Love the question! – Daniel R Jul 05 '13 at 14:00
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    There are exactly three (up to isomorphism) infinite dimensional Banach spaces that have a unique normalized unconditional (Schauder) basis: $c_0$, $\ell_1$, and $\ell_2$. – David Mitra Jul 05 '13 at 14:04
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    [Frobenius theorem]: there are, up to isomorphism, precisely three finite-dimensional real associative division algebras, namely $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$. – wildildildlife Jul 05 '13 at 14:09
  • Artinian rings only have finitely many prime ideals. – Ragib Zaman Jul 05 '13 at 14:18
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    Almost by definition, if there are only finitely many twin primes then there is a theorem that says 'the only twin prime pairs are $p_0, p_0+2; p_1, p_1+2; \ldots; p_n, p_n+2$'. 'Most interesting' is terribly subjective; maybe it would be better to ask for collections of objects that would naively be expected to be infinite but turn out to be finite? – Steven Stadnicki Jul 05 '13 at 17:02
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    Without some additional constraint on answers, this seems perilously close to 'list your favorite theorem that involves finite quantities in some way', and so hopelessly broad. – Steven Stadnicki Jul 05 '13 at 17:05
  • @StevenStadnicki : There is a constraint, not altogether explicit (but readers are presumed to have some common sense) that what is sought are not sets that are trivially seen to be finite. – Michael Hardy Jul 05 '13 at 18:41
  • The number of solutions to $n! + 1 = m^2$ (Brocard's problem) is finite provided that the abc conjecture is true. Since it has allegedly been proven recently, I guess this would fit in your list. – Daniel R Jul 05 '13 at 18:46
  • @StevenStadnicki I'm not sure your claim is true - it is a different thing entirely to say "there are finitely many twin primes" as compared to "the twin primes are $p_1,\dots,p_n$". Knowing only that there are finitely many, you can never be sure your list is complete (there could be another pair not in your list and it would still be finite). In order to get such a result, you need an upper bound proof like "there are no twin primes greater than $10^{10^{10}}$". – Mario Carneiro Jul 05 '13 at 22:42
  • @MarioCarneiro If the number of twin primes is finite, then there is a true theorem that states that 'the twin primes are $p_1, \ldots, p_n$'. That theorem may not be provable from a given axiom system, but it is none the less true. – Steven Stadnicki Jul 06 '13 at 03:35
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    @StevenStadnicki Call me a formalist, but for exactly this reason I find "true" and "false" to be much less useful concepts in formal logic as compared to "provable" and "provably false" (and provably unprovable, etc). Sure, if your axiom system is consistent relative to some model and proves finitely many twin primes, you can say that there is some set $p_1,\dots,p_n$ that is the set of all twin primes in the model, and so "the twin primes are $p_1,\dots,p_n$" is true in this model. But the whole process is wholly nonconstructive, and you still have no idea what $p_1,\dots,p_n$ are. – Mario Carneiro Jul 06 '13 at 07:17
  • There are thousands of results like this in mathematics, so your question seems almost limitlessly broad. Also, what is your motivation for asking it? You don't say... – Pete L. Clark Jul 18 '13 at 22:57
  • @PeteL.Clark : It may be fairly broad, but maybe "most interesting", if construed grudgingly enough, can narrow it down. I seem to recall that there was a specific reason why I thought of this. Probably I'll remember what it was at some point...... – Michael Hardy Jul 18 '13 at 23:09
  • @Michael: Isn't "most interesting" subjective and argumentative? – Pete L. Clark Jul 18 '13 at 23:10

3 Answers3

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A beautiful one, from the theory of quadratic forms:

A quadratic form is simply a homogeneous polynomial of degree two in several variables, so it has the form $$f(x_1,\dots,x_n)=\sum_{i=1}^n \sum_{j=1}^i a_{ij}x_ix_j, $$ and we associate to it the symmetric matrix $A=(b_{ij})_{i,j=1}^n$ where $b_{ii}=a_{ii}$ for all $i$, and $b_{ij}=b_{ji}=a_{ij}/2$ for $j<i$. The matrix is integral iff all the $b_{ij}$ are integers, and the form is positive definite iff $f(\mathbf x)>0$ for all vectors $\mathbf x\ne0$.

That the form represents an integer $k$ means that there are integers $x_1,\dots,x_n$ such that $f(x_1,\dots,x_n)=k$.

The $\mathbf {15}$ theorem of J. H. Conway and W. A. Schneeberger, from $1993$, says that if a positive definite quadratic form (in any number of variables) with integral matrix represents the numbers $$ 1, 2, 3, 5, 6, 7, 10, 14, 15, $$ then it represents every positive integer.

An example is the form $x^2+y^2+z^2+w^2$. (That it represents every positive integer is Lagrange's four squares theorem.)

When we replace the requirement that the matrix is integer valued with the more general requirement that $f(x_1,\dots,x_n)$ is an integer whenever the $x_i$ are integers, the corresponding result is the $\mathbf{290}$ theorem of M. Bhargava and J. P. Hanke, from $2005$, that states that such a form represents all positive integers iff it represents $$ 1, 2, 3, 5, 6, 7, 10, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 34, 35, 37, 42, 58, 93, 110, 145, 203, 290. $$

Unfortunately, the paper by Bhargava and Hanke never appeared, though detailed sketches are available. Wikipedia has additional details on these results. For a recent extension, see Quadratic forms representing all odd positive integers, by J. Rouse.

This question has further references.

Community
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Andrés E. Caicedo
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Gitik and Shelah have shown the following: If $\kappa \leq 2^{\omega}$ is real valued measurable (or even quasi measurable), then the set of infinite cardinals below $\kappa$ can be partitioned into finitely many intervals on which the function $\lambda \rightarrow 2^{\lambda}$ is constant.

For each $n \geq 1$, they also constructed model where the number of these intervals is $n$. The reference is: More on simple forcing notions and forcing with ideals, Annals of Pure and Applied Logic 59 (1993) 219-238, DOI: 10.1016/0168-0072(93)90094-T.

Martin Sleziak
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hot_queen
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This result in Universal Algebra is an obvious one but I find it very useful (and often used) in many branches of Algebra:

Let $A,B,C$ be algebras over a signature $\Omega$. Let $f:A\rightarrow C$ and $g:A\rightarrow B$ be epimorphisms such that $\ker g\subset\ker f$. Then there exist exactly one homomorphism $\bar{f}:B\rightarrow C$ such that $\bar{f}g=f$.

A related results in Algebra (theory of groups, rings, modules...) are better known as Isomorphism Theorems.

One more, rather obvious but useful, construction:

Let $X$ be a set and $f:X\rightarrow A$ a (set theoretic) function into the $\Omega$-algebra $A$. Then there exists exactly one homomorphic extension $\bar{f}:F_\Omega(X)\rightarrow A$ of the function $f$.

Again, the uniqueness of the homomorphic extension is rather obvious, but the whole construction is of great importance.

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    Is this result trivial, or is only its proof trivial? – Michael Hardy Jul 18 '13 at 23:16
  • @Michael: good point. I changed "trivial" to "obvious". All those theorems with commuting triangles that I mentioned have trivial proofs but are far from being trivial. –  Jul 21 '13 at 17:18