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The Gromov-Hausdorff metric makes the set of compact metric spaces into a metric space itself. I am wondering what some natural generalizations there are for arbitrary topological spaces. Namely, is there a natural topology on the set of (compact?) topological spaces?

Edit: I am not too concerned about set-theoretic issues, but perhaps part of the problem is to find a special collection of topological spaces that do form a set and have a natural topological structure. I am more interested in the topological structure.

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    $\begingroup$ The compact topological spaces do not form a set but a proper class, so you can not expect traditional topological structures on it. There are alternatives however, like Grothendieck topology. $\endgroup$
    – Zerox
    Commented yesterday
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    $\begingroup$ @Zerox you could add this as an answer. $\endgroup$
    – David Roberts ♦
    Commented yesterday
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    $\begingroup$ The following appears in Continuity and Baire functions by Edgar Raymond Lorch [American Mathematical Monthly 78 #7 (Aug-Sep 1971), pp. 748-762]: "This circumstance, that a collection of topologies is topologized, may seem a bit incestuous." Slightly more detail in this mathoverflow answer. $\endgroup$
    – Dave L Renfro
    Commented yesterday
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    $\begingroup$ I do not understand the question: Are you fixing a set $Y$ and looking for a topology on the set of topological spaces $(X,T)$, where $X\in 2^Y$? If so, what properties do you want this topology to have? Do you have a particular application in mind? $\endgroup$
    – Moishe Kohan
    Commented 19 hours ago
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    $\begingroup$ If you consider just the set of compact metric spaces and drop the isometry condition in the definition of the Gromov-Hausdorff metric all spaces have distance zero because you can make homeomorphic copies have arbitrarily small diameter in the Hilbert cube, and have these copies converge to the origin. You may want to try how you'd overcome this problem just in this particular case. $\endgroup$
    – KP Hart
    Commented 18 hours ago

2 Answers 2

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There is no topology on the set of all [compact] topological spaces, because there is no set of all [compact] topological spaces.

Given a set of topological spaces, consider the power set of its union. This power set with the indiscrete topology is a compact topological space. It is missing from the original set (even modulo homeomorphism) because it has larger cardinality than every member of the original set.

The reason we can topologize all compact metrizable spaces (up to homeomorphism) is because their cardinality is capped at $\mathfrak c$: π-Base, Search for Compact+Metrizable+~Cardinality $\leq\mathfrak c$. Note also that using the power set trick to produce a missing example fails, since the indiscrete topology is not metrizable for sets with multiple points.

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    $\begingroup$ If one came up with a decent notion of what it means for a sequence of topological spaces to converge to another topological space, then these set theoretic issues wouldn’t really be an issue; one could try to put a topology on any given set of spaces in which convergence behaves in the desired way. Maybe there would be some sequences (of spaces in the set) whose limit exists but is missing from the set, but that would be okay. On the other hand I don’t have a suggestion for such a notion of convergence… $\endgroup$
    – Dan Ramras
    Commented yesterday
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    $\begingroup$ @DanRamras To get a full description of the topological structure it is not enough to only look at convergent sequences, but convergent nets, which does involve set-theoretic issues I think. $\endgroup$
    – Zerox
    Commented 20 hours ago
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    $\begingroup$ One can avoid these set-theoretic issues either with grothendieck universes, or by looking at $\kappa$-small spaces for some cardinal: Fix a set $S$ of cardinality $\kappa$, and then just consider the set of all topologies on subsets of $S$. Every space of cardinality at most $\kappa$ is then homeomorphic to one of those, and they form a set. I believe the OP would be satisfied with an interesting topology on that set instead of "all topological spaces". $\endgroup$
    – Achim Krause
    Commented 16 hours ago
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    $\begingroup$ Well, we can partially order topologies by coarser/finer, and apply the partial order topology. $\endgroup$
    – Steven Clontz
    Commented 9 hours ago
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The set of all topologies on a given set $X$ admits a lattice structure under the refinement relation $\tau\leq\sigma$, whereby every $\tau$ open set is open with respect to $\sigma$, meaning that $\sigma$ refines $\tau$. See also this old MO question of mine.

This is a complete lattice, since the intersection of any set of topologies is a topology, and the union of topologies generates a unique smallest topology containing each of them.

Every lattice admits several natural topologies, and so indeed there are several natural topologies on the set of topological spaces on $X$. There is the lower-cone topology, for example, by which the basic open sets are all those topologies coarser than a given topology. The upper-cone topology, in contrast, would have basic open sets being all those topologies that refine a given topology. And the interval topology has basic open sets all those topologies not in a given closed interval $[\tau,\sigma]$.

It is interesting to consider how the compact and the Hausdorff topologies sit in the lattice of all topologies on a given set $X$, which is what my previous question had been about.

Meanwhile, there are also other topologies on the space of all topologies on a given set $X$. For example, we could consider the topology with basic open sets being determined by the set of topologies having any finite list of open sets as open. Or more generally, beyond finite, one could consider the set of topologies having any given allowed family of open sets as open. The world is truly open for natural topologies on the set of topologies on a given set $X$.

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    $\begingroup$ I had a similar thought in the comments of my answer. But I guess that really what we're observing is that there's a natural ordering of a set of topologies, but in general it isn't clear how you naturally topologize that ordering. $\endgroup$
    – Steven Clontz
    Commented 6 hours ago
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    $\begingroup$ Oh yes, I see that we hit upon the same idea. $\endgroup$
    – Joel David Hamkins
    Commented 6 hours ago

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