Compactness is indeed a central theme in analysis and general topology.

Since the first courses in these subjects, we are exposed to criteria for compactness, such as the Heine-Borel Theorem for subspaces of $\mathbb{R}^n$.
Then we grow up to discover, in our first course of functional analysis, that things do not work so desirably in the case of infinite dimensional normed spaces, as through Riesz Lemma one shows that the unit ball is not compact in an infinite dimensional vector space. Much in the same line of thought, studying topological vector spaces, hence vector spaces endowed with a weaker notion of distance, we have another theorem by Riesz stating that a topological vector space is finite dimensional if and only if it is locally compact. To recover compactness in infinite dimensional spaces we introduce weak topologies and weak compactness, defined through the topological dual of a space.
Finally in the general context of topological spaces one can proove finite product of compacta compact without the axiom of choice , while the famous Tychonoff theorem relative to the infinite case is actually equivalent to it (or better to a weaker version). Moreover, when one increases the dimension of a space, most naturally does it through product we have $\mathbb{R};\ \mathbb{R}^n=\mathbb{R}^{n-1}\times\mathbb{R}...$

I have highlighted words much in a brainstorming fashion, to show the relations I know between compactness and dimensionality, intended in a broad sense. Compactness do recalls, in its definition, a notion of dimensionality, in the sense that if a space is compact, we are always able to extract a finite object, a cover, from an arbitrary one. What I would like to know is:

  • Is there more to this intuition? Can we formalize it?
  • Can we build a notion of dimensionality directly from compactness, in a way which is somehow consistent with the usual one, in the case of vector spaces?
  • Is there a more general relation between compactness and dimensionality, with compactness being a 'signature' for a notion of finite dimensionality?

Thanks in advance

  • 2
    $\begingroup$ Note that a Banach space can have compact sets which are not contained in any finite dimensional subspace, so compactness is not necessarily an indicator of finite dimension $\endgroup$ Jul 15, 2019 at 6:35
  • 4
    $\begingroup$ Look at $\prod_{n\in\Bbb N} [0,1/n]$ as a subspace of $\ell^2(\Bbb N)$ @Francesco $\endgroup$ Jul 15, 2019 at 7:47
  • 2
    $\begingroup$ It all depends on how general you want the definition of dimensionality to be. If you want to deal with all topological spaces, then what you want is hopeless. But if you restrict, say, to manifolds modeled on some Banach spaces, then a manifold is locally compact if and only if it is finite-dimensional, where dimension is simply the dimension of the Banach space. (This notion of dimension is the one one normally uses when dealing with manifolds.) Historically, this is how the notion of dimension got started. But then Lebesgue asked if this notion is a topological invariant. $\endgroup$ Jul 17, 2019 at 3:56
  • 2
    $\begingroup$ Brouwer proved that this is indeed the case (this is his famous invariance of dimension theorem) which eventually has led to an abstract notions (not all equivalent) of dimension which make sense for general topological spaces. These notions have very little to do with compactness. $\endgroup$ Jul 17, 2019 at 4:01
  • 1
    $\begingroup$ @AlessandroCodenotti: The circle $S^1$ is a compact topological group and of dimension $1$. The abstract product $\prod _{i \in I} S^1$ over the arbitrary index set $I$ is a compact topological group again (by Tychonoff), so locally-compact in particular, yet clearly of infinite dimension. The situation is different in topological vector spaces, because their structure is much more rigid than that of only a topological group. $\endgroup$
    – Alex M.
    Jul 19, 2019 at 14:28

1 Answer 1


It is not quite clear what you are asking, so I will answer your questions the way I understood them.

  1. "Is there more to this intuition? Can we formalize it?"

Sort of. The usual notion of compactness deals with finite open covers and their subcovers. One of the abstract notions of dimension (applied to all topological spaces) is due to Chech (based on ideas due to Lebesgue), it is called the covering dimension. It is formulated in terms of refinements of open covers. For $E^n$, covering dimension gives you the expected number, namely $n$. (Which is not at all obvious.) Thus, if you use the notion of covering dimension, then indeed, you get some formalization of your intuition of relation between compactness and dimensionality (both are defined in terms of certain procedures related to open covers, although the procedures are quite different if you look at them closely).

  1. "Can we build a notion of dimensionality directly from compactness, in a way which is somehow consistent with the usual one, in the case of vector spaces?"

It depends on what class of topological spaces you want to "cover". If you are satisfied with manifolds (defined as spaces locally homeomorphic to a certain, say, Banach, vector space) then yes: You define dimension of this manifold to be the dimension of the Banach space. This dimension will be a topological invariant and a manifold will be finite-dimensional if and only if it is locally compact. If you are satisfied with this class of topological spaces then you are done. Originally, going back to the 19th century, topology was developed primarily in the context of manifolds. However, eventually, people realized that this class is way too narrow and it took quite a bit of effort to develop notions of dimension for general topological spaces. In this degree of generality (even if you restrict to, say, metric spaces) finite dimensionality has nothing to do with compactness and your "vector space intuition" breaks down completely. The best thing to do is to abandon it and deal with dimensionality separately from compactness.

  1. "Is there a more general relation between compactness and dimensionality, with compactness being a 'signature' for a notion of finite dimensionality?"

No. A good example is the Hilbert cube: It is compact but infinite dimensional in any reasonable sense.

One last thing: Different parts of mathematics have their own notions of dimension. For instance, dimension in the sense of algebraic geometry is (mostly) different from the one used by topologists. If you study fractals then you get yet another notions of dimension, which are geometric rather than topological invariants. Why is this is another question: Everybody wants to have a single numerical invariant of their favorite class of spaces.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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