6
$\begingroup$

It is well known that every compact metric space is the continuous image (or quotient) of a Cantor space (i.e. a totally disconnected perfect compact metric space). Furthermore, Cantor sets are all homeomorphic. For this reason it is tempting to think of "the" Cantor set as the initial object in the category of compact metric spaces, but the issue with this is that quotient maps from the Cantor set to a compact metric space are not unique. So my question is: is there a category where the objects are compact metric spaces, and the maps are such that the Cantor set is actually the initial object?

Some ideas:

  • There is a canonical way to produce a totally disconnected space from any topological space $X$: consider the complete Boolean algebra of regular open sets $RO(X)$, and by Stone duality $RO(X)$ is the Boolean algebra of clopen sets for a totally disconnected compact Hausdorff space $S$ whose points are the ultrafilters in $RO(X)$. However, the space $S$ may not be a Cantor set since it may have isolated points, and it is not clear to me how we would put a restriction on continuous maps so that $S$ constructed this way has a unique map to $X$.

  • In topological dynamical systems there is the concept of a symbolic representation, which is a certain encoding of a dynamical system as a shift space. More precisely, for a topological dynamical system $(X,\varphi)$ with $\varphi:X \to X$ a homeomorphism, and $\mathcal{P}$ a finite topological partition of $X$, we get a map $\pi : \Sigma \to X$ where $\Sigma \subseteq |\mathcal{P}|^\mathbb{Z}$ is the subset such that $\pi(s) = \bigcap_{n \in \mathbb{Z}} \varphi^{-n}(P_{s_n})$ is a well-defined map. Note that $\Sigma$ is a subspace of the Cantor set $|\mathcal{P}|^\mathbb{Z}$. For some classes of dynamical systems, such as expansive ones, this map can be made to be more or less unique. However, I am only aware of these symbolic representations working (i.e. the map $\pi$ is onto and finite-to-one) for certain expansive dynamical systems, and there are compact metric spaces such as the Hilbert cube which do not admit an expansive homeomorphism. Also this highly depends on the choice of partition (although there are nice choices).

It seems like one way we might make such a category is if the objects of the category are compact metric spaces which have attached to them some canonical self-homeomorphism (along with a partition), and the maps in the category would be continuous surjections which intertwine these self-homeomorphisms. Then there would be a unique map from the Cantor set (with the shift homeomorphism) which is the object's "symbolic representation" as described above. But it is not clear to me if there is an immediate counter-example to this approach or if it is tractable in the first place.

$\endgroup$
2
  • 3
    $\begingroup$ What you're looking for is the Cantor space, rather than the Cantor set. Being an abstract topological space, we're rid of the fixation of one set or the other. It won't solve the uniqueness of the morphism, but it will improve your starting point. $\endgroup$
    – Asaf Karagila
    May 13 '20 at 7:50
  • $\begingroup$ @AsafKaragila you are right, I will edit it to say Cantor space. $\endgroup$ May 13 '20 at 7:53
4
$\begingroup$

Not sure whether this is exactly what you want, but one type of solution would be the following (which I will motivate first):

Why is $\mathbb{Z}$ the initial object in the category of rings? Well, because the category of rings is the category of $\mathbb{Z}$-algebras. That means that an object in this category is given by a ring $R$ together with a structure morphism $\mathbb{Z} \rightarrow R$ and morphisms are morphisms of rings that respect the structure morphisms.

Analogously, you can define a category whose objects are topological spaces/compact metric spaces equipped with a structure morphism from the cantor space etc. Since the structure morphisms have to be respected, this will make the cantor space initial.

These types of categories are called coslice categories in case that you want to read more about them.

Edit: Here is the stuff mentioned in the comments. Denote by $C$ the cantor space and let $X$ and $Y$ be compact metric spaces. Note that the cantor space as object of the above category is the identity $\text{id}_C \colon C \rightarrow C$.

A morphism $f \colon X \rightarrow Y$ in this category is a commutative diagram $$\require{AMScd} \begin{CD} C @>{\varphi_X}>> X \\ @VV{\text{id}_C}V @VV{f}V \\ C@>{\varphi_Y}>> Y \end{CD} $$ where $\varphi_X$ and $\varphi_Y$ are the structure morphisms. If we now choose $X = C$, i.e. the identity on $C$, we get

$$\require{AMScd} \begin{CD} C @>{\text{id}_C}>> C \\ @VV{\text{id}_C}V @VV{f}V \\ C@>{\varphi_Y}>> Y \end{CD} $$

which forces $f$ to be the structure morphism $\varphi_Y$. Hence the identity $\text{id}_C \colon C \rightarrow C$ is initial.

This proof does of course not depend on our specific category, but rather work for any coslice category and the dual proof for any slice category.

$\endgroup$
4
  • $\begingroup$ Thanks for the response! I've been looking up coslice categories and I think this would definitely be a useful perspective to approach this problem. Now I have to think about in what sense a compact metric space has a unique structure morphism from the cantor space. $\endgroup$ May 14 '20 at 19:01
  • $\begingroup$ You do not need this. It will be initial no matter how many structure morphisms you can choose. Let me edit to make this clear. Every choice of structure morphism defines a different object. Think about $k$-algebras for example. There is not necessarily only one morphisms from $k$ to your algebra, but you choose a morphism to give your object the structure of a $k$-algebra. $\endgroup$
    – Con
    May 14 '20 at 19:33
  • $\begingroup$ That makes a lot of sense, thanks for the clarification! $\endgroup$ May 14 '20 at 19:43
  • $\begingroup$ Always glad to help. $\endgroup$
    – Con
    May 14 '20 at 19:44

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.