Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
There are only finitely many sporadic simple groups? –  Hagen von Eitzen Jul 5 '13 at 13:35
1  
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 5 '13 at 13:38
3  
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 5 '13 at 14:04
4  
[Frobenius theorem]: there are, up to isomorphism, precisely three finite-dimensional real associative division algebras, namely $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$. –  wildildildlife Jul 5 '13 at 14:09
1  
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 5 '13 at 17:02
show 13 more comments

3 Answers

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.

share|improve this answer
    
"Unfortunately, the paper by Bhargava and Hanke never appeared, though detailed sketches are available." The paper hasn't appeared yet, but what was holding it up has now been resolved so it should appear relatively soon. –  Pete L. Clark Jul 18 '13 at 22:58
    
Oh, that's great to hear. Thanks, @PeteL.Clark. –  Andres Caicedo Jul 19 '13 at 1:24
    
(An update.) –  Andres Caicedo Jul 20 '13 at 1:27
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
1  
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. –  Godot Jul 21 '13 at 17:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.