MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the following statement of the Arzela-Ascoli theorem.

Theorem. Let K be a compact topological space and let S be a subset of C(K). Then S is relatively compact if and only if S is uniformly bounded and equicontinuous.

There are various hypotheses needed here, but they divide up naturally into two classes: some, such as the compactness of K, are setting the scene, whereas others, such as the equicontinuity of S, are the "real" hypotheses that we assume. This is reflected in the way we state the theorem, putting the scene-setting assumptions in a sentence that begins "Let" so that the meat of the theorem can appear uncluttered in a second sentence that begins "Then".

What interests me is that nearly always when we do this we seem to have either one or two hypotheses. For example, a compact Hausdorff space is normal, or a metric space is compact if and only if it is complete and totally bounded. In this question I am asking for good exceptions to this rule. A truly good exception would be a statement of an undergraduate-level theorem that sets the scene and then talks about an object X, concluding, in the main sentence, that if X is A, B and C, then X is D, where A, B and C are adjectives or short adjectival phrases. (Thus, a technical lemma that needs five complicated conditions in order to hold does not count as a good exception.) It doesn't have to be from general topology -- it's just that there seem to be a lot of adjectives floating around in that area. At the time of writing, I don't have a single good example, though I fully expect them to exist.

Note that this is really a question about mathematical language, and in particular what prompts us to make definitions. After all, if we have a theorem that X is A, B and C implies that X is D, we can always define an X to be E if it is A and B, in which case we will have split the statement up into two parts, one saying that A and B imply E (a definition) and the other that E and C imply D (a theorem). It seems to me that we have a tendency to do this kind of thing because we like two-hypothesis statements.

I'm not going to use the big-list tag though, because I secretly hope that the result will be only a rather small list.

Edit: Some of the examples below are excellent. But I think I don't really want to count examples where we say something about a function between two different objects, where it is obviously quite natural to want information about the function and both objects. (For example, the statement that a continuous bijection from a compact topological space to a Hausdorff topological space is a homeomorphism needs at least three hypotheses for this reason.) Also, the distinction between scene setting and genuine meaty hypotheses is essential (even if slightly vague) if this question is to make any sense at all.

I would of course be happy with an example where we have a function between two objects, we regard all properties about the objects as scene setting, and we claim that three conditions about the function imply a fourth.

share|cite|improve this question
Whitehead: If X and Y are CW-complexes, both simply connected, and f: X ---> Y is a quasi-isomorphism, then f is a homotopy equivalence. If I remembered the theorem correctly, then that should be an example of three hypotheses leading to one conclusion. – Dylan Wilson Oct 22 '10 at 8:49
Dear Kevin, The same class of examples came to my mind immediately! – Emerton Oct 22 '10 at 15:32
Many modern modularity lifting theorems nowadays have a gazillion hypotheses, and the conclusion "...then rho comes from a modular form". Look at the new preprint on potential modularity and change of weight, by Barnet-Lamb, Gee, Geraghty and Taylor. Theorem A has three hypotheses and theorem B has six. Of course this isn't an undergraduate-level example, but it is certainly not a "technical lemma"---these are the main results of the paper, and the paper is expecting to sell based on these theorems with many hypotheses. – Kevin Buzzard Oct 22 '10 at 16:01
If a group is finitely generated, abelian and torsion free, then it is isomorphic to Z^n. – Ben Linowitz Oct 22 '10 at 16:36
A topological group is a profinite group (that is, an inverse limit of finite groups) if and only if it is compact, Hausdorff, and totally disconnected. – KConrad Oct 22 '10 at 17:39

36 Answers 36

A non-empty, perfect, compact, totally disconnected, Hausdorff, second countable topological space is homeomorphic to the Cantor set.

share|cite|improve this answer
Also exactly the kind of thing I was after! – gowers Oct 22 '10 at 15:11
In fact there is the notion of a “Stone space” (compact, totally disconnected, Hausdorff). Thus it can be reduced to “every non-empty, second countable, perfect Stone space is homeomorphic to the Cantor space”. Or using Stone duality: There exists one and only one (up to isomorphism) countable, atomless Boolean algebra. Thus you could argue that “non-empty, compact, totally disconnected, Hausdorff” is just the setting related to Boolean algebras. – The User May 15 '13 at 19:11

The real line $\langle\mathbb{R},\lt\rangle$ is (up to isomorphism) the unique nonempty, separable, complete, dense, endless total order.

(All conditions are needed: Without separable we have for example $[0,1]\times\Bbb R$ with lexicographic order, without complete we have $\Bbb Q$, without dense we have $\Bbb Z$, without endless we have $[0,1]$, all with standard order)

share|cite|improve this answer
Exactly the kind of thing I was after! – gowers Oct 22 '10 at 15:10
I'm glad you like it. Let me mention that Souslin's Hypothesis ('s_problem) is the assertion that the statement remains correct even when the word separable is weakened to the assertion that every family of pairwise disjoint intervals is countable. This hypothesis is independent of ZFC. – Joel David Hamkins Oct 23 '10 at 2:32
I assume you forgot dense as otherwise $\langle \Bbb Z,{<}\rangle$ spoils the show. – Hagen von Eitzen Nov 19 '15 at 22:03
Yes, thanks! Of course I had meant to include that. – Joel David Hamkins Nov 19 '15 at 22:41

An integral domain is called a Dedekind domain if it's not a field and every nonzero proper ideal admits a unique factorization into prime ideals. This is the most concrete way to say what a Dedekind domain is. But how do you check if a ring is a Dedekind domain? Emmy Noether found three conditions: if a domain is Noetherian, integrally closed, and one-dimensional then it's a Dedekind domain. Moreover the converse holds, so you can't make the number of hypotheses smaller in a non-artificial way. (In some references you will find those three conditions used as a definition of Dedekind domains.)

share|cite|improve this answer

If two quadratic forms over $\mathbb{Q}_p$ have the same rank, discriminant and Hasse invariant, then they are equivalent.

share|cite|improve this answer

The central limit theorem: if random variables $\{X_n\}_{n \in \mathbb{N}}$ are (A) Independent, (B) Identically distributed, and (C) have finite variance then (D) $(\sum_1^n X_i - n\mu)/\sqrt{\sigma^2 n} \to N(0,1)$.

share|cite|improve this answer
An interesting phenomenon occurs here: we introduce the abbreviation "i.i.d." and almost think of it as a single condition (meaning "nice" in an appropriate sense). – gowers Oct 22 '10 at 15:22
For generalizations it's useful to be able to take either perspective: (A) and (B) as a single "niceness" condition; or, separate them and try to relax them separately. – Louigi Addario-Berry Oct 22 '10 at 18:01
The "identically distributed" part can be replaced by much weaker conditions that can be combined with conditions on the variance. See the Lindeberg or Lyapunov conditions. That would leave only two ingredients, independence and, say, the Lindeberg condition. – Douglas Zare Nov 19 '15 at 23:37

A compact convex subset of $\mathbb{R}^n$ with nonempty interior is homeomorphic to the $n$-dimensional ball.

share|cite|improve this answer

A basic example from undergraduate topology that comes to my mind is the theorem on the existence of universal covers.

Theorem. Let $X$ be a topological space. Then, there exists a universal covering space $\pi\colon \tilde{X}\rightarrow X$ if $X$ is connected, locally path connected and semi-locally simply connected.

share|cite|improve this answer

In representation theory, one oftens sees long lists of adjectives: "If $V$ is an irreducible, admissible, smooth representation, then ... ".

In the theory of group schemes, similarly long lists can appear: "If $G$ is a reduced commutative finite flat group scheme then ... " or "If $G$ is a connected commutative finite flat group scheme then ...". (Here "group scheme" is one term --- it is the basic object --- but the other three adjectives are applied independently, although "finite" and "flat" come together so often that maybe you can argue they should be treated as a single property.)

In the theory of automorphic forms and Galois representations one has "If $\pi$ is a regular, algebraic, essentially conjugate self-dual, cuspidal automorphic representation, then ...". (In this case people introduced the pleasing acronym RAECSDC in order to simplify statements.)

None of these examples are from undergraduate mathematics, of course, and ideed they are taken from areas with some reputation for technical complexity. The examples of modularity theorems that Kevin mentions in his comment above are from the same field as my RAECSDC example. I think that the long lists of adjectives in the statements of results from these fields is certainly related to their reputation for being technical.

share|cite|improve this answer

How about the Stone-Weierstrass theorem? If $\mathcal{A}$ is a collection of real-valued continuous functions on the compact Hausdorff space $X$ which (1) is an algebra, (2) separates points, and (3) contains the constants, then it is dense in $C(X)$. (For complex-valued functions, add (4) closed under conjugation.)

Measure theory has the monotone class and $\pi$-$\lambda$ theorems that are of a similar nature, but there we usually assign names to the hypotheses (e.g. a $\lambda$-system, which is short for three different properties).

share|cite|improve this answer

Dear Tim: perhaps I misunderstood your question, but I think any reasonably technical field has a few examples. I heard algebraic geometry is such a field, so I looked up a couple of sources and found these two:

1) Let $f: R\to S$ be a local homomorphism of local rings. Suppose $R$ is regular, $S$ is Cohen-Macaulay, $f$ is finite and $\text{dim} R = \text{dim} S$. Then $f$ is flat.

This is a basic result that is used quite frequently. I don't think you can drop any of the hypotheses, they are all basic definitions and independent of each other.

Or how about:

2) A morphism of schemes $f: X \to S$ is quasi-finite if it is locally of finite type, quasi compact and has finite fibres.

Again, I don't think you can drop any of the hypotheses. I am reasonably sure EGA has a few more results like this (-:

share|cite|improve this answer

Any field that is algebraically closed, characteristic zero, and of continuum cardinality is ring-theoretically isomorphic to the complex numbers.

share|cite|improve this answer
I think it's better to think of this theorem as: "Any two algebraically closed fields of the same characteristic and transcendence degree are isomorphic." Then "algebraically closed" is part of the setting, not the sort of hypothesis Gowers is looking for. – Noah Snyder Oct 22 '10 at 22:33
@Scott: since we're talking about language: I'm curious what the "ring-theoretically" is doing in your sentence. If you took it out, then to me the meaning and the clarity are completely unchanged. What's your take on it? – Pete L. Clark Oct 23 '10 at 2:15
I guess that Scott wanted to underline the fact that such fields need not be topologically isomorphic. – Alex B. Oct 23 '10 at 6:35
@Noah: I concede that the theorem you state is more general, and that the statement I gave is typically found as a corollary of it. However, I think there is independent merit in characterizing a field that we already know and love, and in that case, "algebraically closed" is no longer part of the setting. – S. Carnahan Oct 23 '10 at 8:54
@Pete: I prefer to specify the category in which we take isomorphisms if it is not too inconvenient and if there is some risk of ambiguity. I decided it was a habit worth cultivating to reduce the chance of error in my own research. – S. Carnahan Oct 23 '10 at 9:04

Here is another one: a finite irreducible aperiodic Markov chain is ergodic.

share|cite|improve this answer

A countable, dense linear ordering without first or last element is isomorphic to $\mathbb Q$.

I once heard someone use the acronym DLOWFOLE. That reduces the number of hypotheses but I think it's sort of cheating.

share|cite|improve this answer
I'll never remember the acronym DLOWFOLE. I'd rather just say "countable dense linear ordering without endpoints" (DLOWE). – Todd Trimble Mar 8 '13 at 14:45

A nice example from recent work in set theory.

Theorem (Viale). Assume Martin's maximum, and that every limit cardinal is a strong limit. Suppose that $N$ is an inner model, that $N$ has the same cardinals as $V$, and that $V$ is a forcing extension of $N$. Then every $\omega_1$-sequence of ordinals is in $N$.

We actually expect that the assumptions that limit cardinals are strong limit, and that $V$ is a forcing extension of $N$, can be removed.

share|cite|improve this answer

What about the textbook general version of the original Gödel incompleteness theorem: if $T$ is recursively axiomatized, sufficiently strong, and $\omega$-consistent, it is incomplete (where sufficient strength means representing every recursive function)?

share|cite|improve this answer

Krein-Milman Theorem. In a Hausdorff, locally convex topological vector space (my one), a compact (my two) convex (my three) subset is the closed convex hull of its extreme points.

It has wonderful applications. For instance, that $L^1({\mathbb R}^n)$ is not the dual of a Banach space.

Baire Theorem. In a complete metric (one) space, a denumerable (two) intersection of dense open (three) subsets in dense.

It is used in the proof of

Banach Theorem. Let $E$ be a Banach space (one), $F$ be a Banach space (two), $f:E\rightarrow F$ be linear, bounded (three). Then $f$ is open (the image of the unit ball is a neighborhood of $0_F$).

share|cite|improve this answer
These ones don't quite count, because the hypotheses are referring to different things (e.g. see my additional remark above). For instance, I regard your first hypothesis of the Krein-Milman theorem as setting the scene, and there are only two hypotheses concerning the subset. – gowers Oct 22 '10 at 15:10
Tim, I must be dumb, but I don't see any difference between the structures of your example (Ascli-Arzela) and Krein-Milman. I give up. – Denis Serre Oct 22 '10 at 15:27
@Andreas. Thanks for the correction. – Denis Serre Oct 22 '10 at 15:40
Just to be clear, the Arzela-Ascoli theorem is meant as an example with TWO hypotheses. I count everything in the first sentence as background. So it was not supposed to be an example of what I was looking for. – gowers Oct 22 '10 at 15:41
I just wanted to point out that the Banach (or Open Mapping) theorem should have a surjectivity hypothesis. Preferring to remain agnostic about what constitutes a single hypothesis for the sake of this thread, I don't want to add it myself, but I thought I should at least mention it lest someone be mislead about the technical content. – Noah Stein Mar 25 '11 at 18:23

Let $G$ be a discrete group. If there exists a subgroup of $G$ which is (1) infinite, (2) normal, and (3) amenable, then the first $l^2$-Betti number of $G$ vanishes.

share|cite|improve this answer

The HSP-theorem from universal algebra: If a class of algebraic structures (over a given signature) is closed under homomorphic images, substructures and products, then it is defined by a set of equations.

share|cite|improve this answer

Lindstrom's theorem: if $L$ is a regular logic which is compact, has the Lowenheim-Skolem property, and extends first-order logic, then $L$ is (equivalent to) first-order logic. Compactness and the Lowenheim-Skolem property are both very important notions, which are (in abstract model theory) often studied independently of each other; regularity and extending first-order logic are slightly more minor, but I still think they are substantial enough to count as individual hypotheses. ("Regular" means that given a formula $\phi$ and a predicate symbol $U$, there is a single formula $\phi^U$ such that for all structures $M$ in a language containing $U$ and all symbols used in $\phi$, we have $M\models\phi^U\iff M^U\models\phi$.)

share|cite|improve this answer

Poincaré Conjecture. If you are (A) a 3-manifold, (B) closed, and (C) simply-connected, then you are (D) the 3-sphere.

share|cite|improve this answer
I would think A and B would both be considered part of the "background" here... – Harry Altman Oct 22 '10 at 17:14
Fair point. Although, I also think that manifold is conveniently interpreted here to include compact and connected. – Tony Huynh Oct 22 '10 at 18:39

Every compact, connected, locally connected metric space is the continuous image of the unit interval

share|cite|improve this answer

The principle of transfinite induction is often stated as the following theorem.

Theorem. Suppose that $A$ is a class of ordinals. If

  • (zero) $0$ is in $A$,
  • (successor) whenever an ordinal $\alpha$ is in $A$, then $\alpha+1$ is also in $A$, and
  • (limit) if $\lambda$ is a limit ordinal and $\lambda\subset A$, then $\lambda\in A$,

then $A$ contains all ordinals.

There are other accounts of transfinite induction that unify the hypotheses into the single statement that whenever all smaller ordinals than an ordinal $\alpha$ are in $A$, then $\alpha$ is in A, and it is considered more elegant to use that formulation when it is possible, but nevertheless many uses of transfinite induction consist in verifying the three properties above.

share|cite|improve this answer
I agree that that's three hypotheses, but they aren't given by adjectives or adjectival phrases. In theory they could be: one could say something like this (my definitions are made up but their meanings should be obvious). Let A be a set of ordinals. If A is properly rooted, successor-closed and limit-closed, then A contains all ordinals. – gowers Oct 22 '10 at 15:15
I've always felt that zero should be a limit ordinal. The statement of transfinite induction benefits from this point of view: we may omit the first hypothesis! By the way, is there a good reason to exclude zero from the definition of limit ordinal? It seems artificial to me. – John Wiltshire-Gordon Oct 22 '10 at 17:53
It makes sense to call 0 a limit ordinal, and some authors do so. But often you have to distinguish between 0 and nonzero limit ordinals in inductive definitions/proofs. For example: $\aleph_\delta$ is defined as $\bigcup_{\alpha < \delta} \aleph_\alpha$ for nonzero limit ordinals,whereas $\aleph_0$ is defined as $\omega$. – Goldstern Jan 25 '13 at 11:39
@Goldstern: This simply shows that we have chosen the wrong indexing for the alephs. If we use that definition at all limit ordinals (and cardinal successor at successor ordinals as usual), then the resulting hierarchy starts by enumerating the finite cardinals, and than at stage $\omega$ it reaches the usual $\aleph_0$ and continues with the normal alephs from there. Very natural, no? It also removes the need to give special treatment to $n=0$ in various basic propositions about cofinalities, etc. – Peter LeFanu Lumsdaine Nov 20 '15 at 0:16
@PeterLeFanuLumsdaine: First: Even if you would use a different indexing for the alephs, this would not invalidate my point. There still are many other examples where you have to distinguish between 0 and nonzero limit ordinals in inductive definitions/proofs, such as the inductive definition of ordinal arithmetic. – Goldstern Nov 20 '15 at 14:30

Whyburn's Theorem: Let $S$ be a planar set that is compact, connected, locally connected, nowhere dense, and such that any two components of the complement are bounded by disjoint simple closed curves. Then $S$ is homeomorphic to the Sierpinski carpet.

share|cite|improve this answer

The examples I've seen so far are not undergraduate-level, at least, not anywhere I've taught. The Fundamental Theorem of Galois Theory is undergraduate-level, and can be stated, in part, as follows: if $K$ is separable (that's one), normal (that's two), and finite (that's three!) over $F$, then the number of elements in the Galois group of $K$ over $F$ equals the degree of $K$ over $F$.

share|cite|improve this answer
Gerry, one usually sets the stage for this by starting with a finite extension K/F. Then if (and only if) K is normal and separable over F, #Aut(K/F) = [K:F]. That's just my two (not three) cents... – KConrad Oct 22 '10 at 10:21
The distinction OP makes between "setting the scene" and the "real" hypotheses is, I confess, not clear to me. For what it's worth, I was more-or-less quoting from Adamson, Introduction To Field Theory: "Let $K$ be a separable normal extension of finite degree $n$ over a subfield $F$, with Galois group $G$. Then $G$ has order $n$." I even took the "Let ... then" construction from Adamson. – Gerry Myerson Oct 22 '10 at 10:37
I had changed the "Let... then" to "If... then" because it was used in a single sentence, which is not the same way it is used in the quote from Adamson's book (where it's "Let.... Then"). – KConrad Oct 22 '10 at 10:52
Oh. OK. Say, is someone going through this discussion downvoting answers? I see three answers at minus 1. – Gerry Myerson Oct 22 '10 at 12:57

Tim: here's one from a course I am giving now (I think you know which): let $B\subset \mathbb{R}^n$ be a nonempty subset. Then there is a norm on $\mathbb{R}^n$ whose open unit ball is $B$ iff $B$ is open, convex, symmetric and bounded. (I think it would be poor style to move "nonempty" into the 2nd sentence, since that is such an obvious condition, but that would make 5...)

share|cite|improve this answer
That's a nice example, but in a way it illustrates exactly the point that we don't like too many hypotheses, since we go on to define a convex body in R^n to be a compact convex set with nonempty interior. Then an equivalent statement to yours is that a set is the closed unit ball of a norm if and only if it is a symmetric convex body. – gowers Oct 22 '10 at 15:05
That is still 3 hypotheses: "symmetric", "convex", "body" :-) – Tony Scholl Oct 22 '10 at 15:49
That's logically true, but psychologically something happens with the new terminology. A convex body isn't a body that's convex (what's a body?) but an indivisible term that happens to be made of two words. And it becomes the object studied rather than hypotheses applied to a more general object. – gowers Oct 22 '10 at 20:02

Two answers.I'm thinking about trees because of the two out of three property: For a simple (no multiple edges) undirected graph G, any 2 of the three conditions

  1. cycle-free (better acyclic)
  2. connected
  3. #edges=#vertices-1

Means G is a tree.

Of course that isn't what you asked. Condition three is not one word although we could coin uni-deficicient

so simple+undirected+acyclic+connected defines tree.

Certain sets are relations (basically any set of ordered pairs). Not counting that as a condition

For a relation, reflexive+symmetric+transitive defines Equivalence relation

similarly reflexive+antisymmetric+transitive defines partial order

Actually there are $k$-ary relations for other $k$ so one could quantify over all relations and restrict to binary relations.

share|cite|improve this answer

Here is a really basic example. Suppose that $\phi:M_n(K)\rightarrow K$ satisfies $(a)$ $\phi$ is multilinear, $(b)$ $\phi$ is alternating and $(c)$ $\phi(E_n)=1$. Then it follows $(d))$ $\phi(A)=\det (A)$.

share|cite|improve this answer

Let $K$ be an ordered field, and $v$ a valuation on $K$ with convex valuation ring. If $(K,v)$ is henselian, the value group is divisible, and the residue field is real-closed, then $K$ itself is a real-closed field.

There is also an analogous statement for algebraically closed fields of characteristic $0$.

share|cite|improve this answer

Let the set $S\subset R^d$ be (a) nonempty, (b) closed, and (c) with no isolated points. Then, $S$ is uncountable.

share|cite|improve this answer

Your Answer


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.