Basic results with three or more hypotheses

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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 $A$

Answer 6

Purely because I didn't see any combinatorial example:

The Blow-up Lemma says that if you have a regular partition of a graph $G$, and a bounded-degree subgraph $H$ on the same number of vertices, then you can embed $H$ in $G$ if you 'should be able to', that is you can find a homomorphism from $H$ to the reduced graph of $G$ (put edges between parts of the partition corresponding to dense regular pairs) which maps the right number of vertices of $H$ to each part of $G$. Provided that you have 'super-regularity', which more or less means no cheating by having isolated vertices. If you unpack the vagueness here you get a reasonable list of conditions.

If you want to go further, try repeating this in sparse graphs. Now $G$ will have to be a subgraph of some random or pseudorandom graph, otherwise (as the OP knows well) the whole theory falls apart, but in addition there are two other ways to 'cheat' (two quite distinct ways of interfering with vertex neighbourhoods) and you have to exclude these to have a Blow-up Lemma.

I think this kind of thing is pretty ubiquitous in mathematics, actually: you have a general idea that in all nice situations X will be true. But when you try to make it precise, you discover a bunch of distinct nasty situations in which X fails, so you exclude them and then you have a theorem with many hypotheses.