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It is well-known (1, 2) that the cardinality of a separable Hausdorff topological space is at most $2^{2^{\aleph_0}}$. Therefore, the collection $\mathcal{A}$ of all (homeomorphism classes of) separable Hausdorff topological spaces is a set. Can the set $\mathcal{A}$ be equipped with a natural topology?

Smiley1000
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  • Much narrower, but you might be amused by the existence of a natural metric on the space of (isometry types of) compact metric spaces, namely the Gromov-Hausdorff metric. – Noah Schweber Feb 11 '24 at 20:14
  • I'm not sure if this helps, but an example where a topology is put on a certain collection of topologies on a set is cited in this mathoverflow answer. Regarding that answer, see my 3 June 2002 sci.math post Incestuous Mathematics. – Dave L. Renfro Feb 11 '24 at 20:21
  • Every partial order has a natural topology (subbasis of upward and downward sets). Is there a natural partial order on these spaces? – Steven Clontz Feb 11 '24 at 20:22
  • Define a relation $A\leq B$ if $A$ is homeomorphic to some subspace of $B.$ Can we show this is a partial order on $\mathcal A?$ This essentially means, if $A,B$ are separable and Hausdorff, then $A\leq B$ and $B\leq A$ implies $A$ and $B$ are homeomorphic. But I'mnot sure this is true. – Thomas Andrews Feb 11 '24 at 20:28
  • Doesn't work - we tried that in another post a while back. Think about [0,1] and (0,1). – Steven Clontz Feb 11 '24 at 20:42
  • @Thomas Andrews: FYI, I think a version of your relation was introduced by Fréchet in 1910. See the page references to Sierpiński's topology book cited in this MSE answer. For example, on p. 130 (lines $-18$ to $-15)$ Sierpiński says: "If $P$ is homeomorphic with a subset of $Q$ but $Q$ is not homeomorphic with any subset of $P,$ then $P$ is said to have a smaller dimensional type than $Q$ and we write $dP<dQ$ (or $dQ>dP).$ Clearly if $dP<dQ$ and $dQ \leq dR$ then $dP < dR.$" – Dave L. Renfro Feb 11 '24 at 20:50
  • All this interested commentary and not any voting that would actually help the question receive an answer – FShrike Feb 11 '24 at 20:51
  • There is no collection, let alone a set, of homeomorphism classes of separable Hausdorff topological spaces. What you can do is this: Fix any set $S$ of cardinality $2^{2^{\aleph_0}}$. Then you can define the set $\mathfrak H$ of all separable Hausdorff topological spaces whose underlying sets are subsets of $S$. Each separable Hausdorff topological spaces is homeomorphic to an element of $\mathfrak H$. – Paul Frost Feb 12 '24 at 11:08
  • @PaulFrost Thank you, this is what I meant – Smiley1000 Feb 12 '24 at 11:46
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    For anyone thinking about @StevenClontz idea regarding a natural partial order on these spaces, may want to have a look at https://math.stackexchange.com/q/4818460/1210477 to see a list of failing candidates. [To clarify, the idea itself is a good one, just wanted to reference some natural approaches to it that won't work] – M W Feb 12 '24 at 20:40
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    https://mathoverflow.net/q/496630/155881 – Smiley1000 Jun 24 '25 at 05:06

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