## Basic results with three or more hypotheses

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.

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If a group is finitely generated, abelian and torsion free, then it is isomorphic to Z^n. – Ben Linowitz Oct 22 2010 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 2010 at 17:39

Let $A\subset \mathbb{R}^d$ be (a)closed, (b) convex, and (c) contains the origin. Then $\left( A^{o} \right)^{o} = A$ where $o$ denotes the polar of $A$

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This is P. Lévy's characterisation of Brownian motion: let $X=(X_t : t\geq 0)$ be a continuous martingale with quadratic variation equal to $t$; then $X$ is a Brownian motion.

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

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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 $0$.

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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 $M\models\phi^U\iff M^U\models\phi$.)

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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.

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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)?

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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.

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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 $(d))$ $\phi(A)=\det (A)$.

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Assume CH. Let X be a compact zero-dimensional space without isolated points such that:

1) X is an $F$-space (that is, disjoint open $F_\sigma$ sets have disjoint closures).

2) $X$ is an almost $P$-space (that is, every non-empty $G_\delta$ set has non-empty interior).

3) $X$ has a base of cardinality continuum.

Then $X$ is homeomorphic to $\beta \mathbb{N} \setminus \mathbb{N}$

Parovicenko's characterization of $\beta \mathbb{N} \setminus \mathbb{N}$ is actually equivalent to CH by a result of van Douwen and van Mill.

It's interesting that Parovicenko's theorem is equivalent to a boolean-algebraic statement which uses only two essential hypotheses...

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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.

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

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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.

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I'll never remember the acronym DLOWFOLE. I'd rather just say "countable dense linear ordering without endpoints" (DLOWE). – Todd Trimble Mar 8 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.

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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.

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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.

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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).

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

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A very simple example (maybe too simple) which seems to be fit this cathegory is the Rolle's Theorem.

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

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

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I would think A and B would both be considered part of the "background" here... – Harry Altman Oct 22 2010 at 17:14
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Any field that is algebraically closed, characteristic zero, and of continuum cardinality is ring-theoretically isomorphic to the complex numbers.

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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 2010 at 22:33
I guess that Scott wanted to underline the fact that such fields need not be topologically isomorphic. – Alex Bartel Oct 23 2010 at 6:35

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

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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.

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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.

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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 2010 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 2010 at 17:53
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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...)

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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 2010 at 15:05
That is still 3 hypotheses: "symmetric", "convex", "body" :-) – Tony Scholl Oct 22 2010 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 2010 at 20:02

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) $(\sum_1^n X_i - n\mu)/\sqrt{\sigma^2 n} \to N(0,1)$.

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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 2010 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 2010 at 18:01

The following is classical in numerical linear algebra.

Let $A\in M_n({\mathbb R})$ be tridiagonal (one) with an invertible diagonal D (two). Assume that the eigenvalues of $J:=I_n-D^{-1}A$ belong to $(-1,1)$ (my three). Then the relaxation method converges for every choice of the relaxation parameter $\omega$ in the interval $(0,2)$; the optimal parameter is unique and equal to $$\omega^*=\frac{2}{1+\sqrt{1-\rho(J)^2}}.$$

If you do not like my two, you can take $\omega\in(0,2)$ as an hypothesis.

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

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Exactly the kind of thing I was after! – gowers Oct 22 2010 at 15:10
I'm glad you like it. Let me mention that Souslin's Hypothesis (en.wikipedia.org/wiki/Suslin'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 2010 at 2:32