# Are there more general spaces than Euclidean spaces to have the Heine–Borel property?

From Wikipedia

A metric space (or topological vector space) is said to have the Heine–Borel property if every closed and bounded subset is compact.

Any subset of a Euclidean space, including itself, has the Heine–Borel property. I was wondering if there are more general types of metric spaces, topological vector spaces, or whatever space where boundedness and closedness can make sense, such that they also have the Heine–Borel property?

Or does the Heine–Borel property characterize subsets of Euclidean spaces?

Thanks and regards!

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Very closely related: math.stackexchange.com/questions/33917/… –  Jonas Meyer Feb 2 '12 at 1:04
@JonasMeyer: Thanks! Yes. related. Maybe the Q&A there help this question in some way. But I haven't figured it out yet. –  Tim Feb 2 '12 at 1:14
I'm not sure what you mean when you say that every subset of Euclidean space has this property. For example: the open, unit ball $B \subseteq \mathbb{R}^n$ can be viewed as a metric space with the induced Euclidean metric. However, it does not possess the HB property ($B$ is closed in itself and bounded, but not compact). –  student Feb 2 '12 at 2:39
@Tim a subset of Euclidean space has the HB property if and only if it is closed as a subset of $\mathbb{R}^n$. The unit ball is not compact since the open cover $\{B(1 - 1/n)\}_{n \in \mathbb{N}}$ does not admit a finite subcover. –  student Feb 2 '12 at 2:57
The concept of a space withthis property is so important in functional analysis is so important that locally convex spaces which enjoy it have their own name---Montel space, so-called because Montel's theorem implies that the Fréchet space of holomorphic functions on an open subset of the complex plane has it. Many of the important spaces of test functions and distributions have it, as do complete nuclear space as mentioned in a resonse below. –  jbc Mar 19 '13 at 11:24

0) Note that in general a metric space has this property iff it is ball compact, i.e., closed balls of finite radius are compact.

Ball compact spaces are locally compact, but the converse does not hold: e.g. an infinite set endowed with the discrete metric $d(x,y) = \delta_{x,y}$ is locally compact, bounded and not compact, hence not ball compact.

1) A topological field is ball compact if and only if it is locally compact and not discrete. Thus the topological fields with this property are $\mathbb{R}$ and $\mathbb{C}$, $\mathbb{Q}_p$ and its finite extensions and $\mathbb{F}_q((t))$.

2) A finite product of ball compact metric spaces, endowed with (say) the product metric $d = \max_{i=1}^n d_i$ is ball compact.

Combining these, we find that any finite dimensional vector space over a nondiscrete locally compact field has this property. This directly generalizes the spaces $\mathbb{R}^n$ and there are branches of mathematics (number theory, representation theory, harmonic analysis) in which this generalization is very natural.

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+1, thanks! About "any finite dimensional vector space over a nondiscrete locally compact field", I was wondering if a topology can be induced on any vector space over a topological field, and how? –  Tim Feb 2 '12 at 8:38
@Tim: certainly topologies can be put on vector spaces over a topological field. If the topologies are derived from compatible norms (first on the field itself and then on the vectr space), the field norm is complete, and the vector space is finite-dimensional, then all such norms are (uniformly) equivalent. This is a familiar fact over $\mathbb{R}$ and the proof in the general case is much the same. –  Pete L. Clark Feb 2 '12 at 14:05
Thanks! (1) "This is a familiar fact over R and the proof in the general case is much the same." May I know where (books or links) I can become familiar with it? (2) I was wondering if there is some naturally unique topology on any vector space over a topological field? Naturally unique in the similar sense as the single topology induced by the metric in a metric space? –  Tim Feb 2 '12 at 15:17
@Tim: let $V$ be a finite-dimensional vector space over a ball-compact topological field $F$. Assume that $V$ is Hausdorff and that scalar multiplication and vector addition are continuous. If $e_1, ... e_n$ is a basis of $V$, then $c_1 e_1 + ... + c_n e_n$ is a continuous bijection $\phi : F^n \to V$. It induces continuous bijections on balls of any radius, and since continuous bijections from a compact space to a Hausdorff space are homeomorphisms, $\phi$ is a homeomorphism on all balls, hence a homeomorphism. So the topology on $V$ must be the product topology on $F^n$. –  Qiaochu Yuan Feb 4 '12 at 18:52
How wide-spread is this "ball compact" terminology? I never heard that (although it makes sense). I'm used to the term proper metric space (very wide-spread in metric geometry). Two explanations I've heard for that terminology: 1) because the distance functions $d(x,\cdot)$ are proper maps 2) because the (locally compact) isometry group of $X$ acts properly on $X$. –  t.b. Aug 8 '12 at 6:55

You have to choose how you want to generalize the Heine-Borel property. There is no difficulty in generalizing "closed", because it already makes sense for any topological space, but "bounded" does not; it's specific to spaces with extra structure. So you have to choose how exactly you want to generalize that structure in interpreting the Heine-Borel property.

Since there are tempting generalizations of "bounded" where the Heine-Borel property need not hold (e.g. arbitrary metric spaces, with "$E$ is bounded" defined to mean something like "$E$ is contained in some ball of finite radius", need not have the Heine-Borel property), some care must taken in doing this.

One generalization is to work with complete metric spaces, and to replace the naive notion of boundedness just described with that of total boundedness. In this situation it is true that a subset $E$ of a complete metric space is compact if and only if it is closed and totally bounded. (If you want a Heine-Borel-type theorem for metric spaces that might not be complete, you can do this too--- at the cost of replacing "closed" with something relating to completeness. Wikipedia's page on this is informative; the details are also in most analysis books that discuss metric spaces in general. I'm not certain, but I think Rudin's Principles of Mathematical Analysis contains at least some of the details--- or maybe he sticks them in exercises.)

You can go slightly more general than metric spaces by considering uniform spaces (which have enough extra structure beyond just the topology to be able to talk about things like "uniform continuity", but do not necessarily have topologies induced by metrics). Total boundedness can be formulated for uniform spaces, too, and it's what you need (along with a technical condition related to completeness) to characterize compact sets in this situation.

Going in a slightly different direction, in functional analysis at least, it is quite common to have subsets of interest in infinite dimensional vector spaces that are simply not going to be compact or totally bounded in any natural structure that you want them to have these properties in, but you still want to exploit the idea of some kind of "boundedness". The generalized notion of a bornological space (roughly speaking, a topological space endowed with a collection of subsets that one regards as being "bounded") is occasionally useful in this connection, although I can't think of any nice Heine-Borel-type theorem at this level of generality.

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+1, thanks! I am talking about boundedness as it is, no matter how it is defined in different spaces, not totally boundedness. I am also talking about closedness as wrt the topology. –  Tim Feb 2 '12 at 1:38
Thanks for mentioning bornological spaces. I was wondering if bounded metric spaces and bounded topological vector spaces are bornological spaces? –  Tim Feb 2 '12 at 8:41
@Tim re the first comment: one reason generalizations go beyond "boundedness as it is" is that compactness is a purely topological concept (ie, invariant under homeomorphism) and closedness is also--- but boundedness generally isn't, so "bounded and closed" isn't, so it can't be equivalent to compactness in general. This is why more general Heine-Borel theorems get more complicated and involve other structure. And once other structure is involved, "closed" tends to get more subtle, so that the combination of hypotheses remains invariant under topology-preserving changes in that structure. –  leslie townes Feb 2 '12 at 21:09
@Tim: The bounded sets in a metric space do give a bornology. I don't know what you mean by "bounded topological vector space"; the fact that a vector space contains arbitrarily large scalar multiples of its elements tends to prevent entire vector spaces from meeting "bounded-type" conditions. There is a natural bornology on any topological vector space, and in the finite dimensional normed case, its sets are the sets that are bounded in the usual sense; see that Wikipedia page. (In applications, bornologies are often used as replacements for compactness, not as a step toward it.) –  leslie townes Feb 2 '12 at 21:17
Thanks! By bounded subset in a topological vector space, I mean the same as this Wiki article –  Tim Feb 2 '12 at 21:26

Nuclear Fréchet spaces have this property.

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+1, thanks! Is an Euclidean space a nuclear Fréchet space? –  Tim Feb 2 '12 at 1:11
Yes. See e.g. the Wikipedia article on "nuclear space". –  Robert Israel Feb 2 '12 at 2:41

A metric space such that the compact sets are exactly the closed and bounded ones is called "Borel compact" (by definition). A metric space is Borel compact iff it is cofinally complete and regularly bounded:

A sequence $(x_n)$ in a metric space is cofinally Cauchy iff for every $\epsilon > 0$ there exists some ball $B(x, \epsilon)$ that contains $x_n$ for infinitely many $n$. A metric space $(X,d)$ is cofinally complete iff every cofinally Cauchy sequence has a convergent subsequence (or equivalently, has a cluster point).

A space is regularly bounded iff every closed and bounded set is totally bounded.

Also, a metric space is Borel compact iff every bounded sequence has a convergent subsequence.

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