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Fix a dimension $n$ (for me, I am particularly interested in $n =4$ but I am not sure if that is relevant). Is there a total ordering on the set of all smooth closed compact manifolds of dimension $n$?

I suppose that by ordering 3-manifolds using triangulations and then removing any repetition (using the fact that there is an algorithm to tell if two triangulated 3-manifolds are the same), 3-manifolds can be totally ordered.

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    That is a set (after modding about by diffeomorphism) with the same cardinality as $\Bbb N$ so it has a total order coming from a bijection with $\Bbb N$. For this question to be reasonably interesting, one should hope some interesting compatibility condiiton with the structure and the order. I am not convinced there is any interesting such condition. – 2017-01-12
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    @PVAL-inactive What is the argument (or a reference) for there only being a countable number of smooth manifolds up to diffeomorphism? – 2017-01-12
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    @user101010 In dimension 4 that's equivalent to PL structures, which are determined by their triangulations, which are finite combinatorial objects. In dimension greater than 4 there are countably many PL structures and Hirsch-Mazur obstruction theory says there's only finitely many smooth manifolds with a given PL structure. (I have no idea why you want to order manifolds.) – 2017-01-12
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    One should be able to construct a bijection directly from handle decompositions (it can surely be done in dimensions 4 or less). The higher dimensional case requires proving that there are only countable isotopy classes of higher dimensional links, which I do not know a proof of offhand. – 2017-01-12
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    It is a theorem, I think due to Cheeger, that there are only countably many compact topological manifolds up to homeomorphism. The proof used Edwards' theorem on $\epsilon$-homeomorphisms (they are homotopic to homeomorphisms if $\epsilon$ is small enough). – 2017-01-15

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