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Motivation: I am interested in computational representations of topological spaces which are particularly "explicit", in the somewhat vague sense that we can specify everything we care about using only binary strings. I'll try to illustrate what I'm getting at with examples, and hope my subsequent use of terminology will be clear (if it isn't, please let me know where):

  • the primary example of such a space is $2^\omega$ itself;
  • an open set (of $2^\omega$) can be specified by which basis sets it contains, and the set $\{(x,U) | x \in U\}$ is itself open;
  • a closed set can be specified via its complement, and is the image of the whole space under an explicit map which "has bounded complexity" (is 1-Lipschitz);
  • an $F_\sigma$ set is the image of an open set under an explicit map of "bounded complexity" (($1-\varepsilon$)-Hölder?);
  • but for subsets of $2^\omega$ that's as far as we can go, because $F_\sigma \Leftrightarrow \sigma-$compact, which is preserved under continuous maps. We could still specify e.g. a $G_\delta$ set, but there's no uniform way to represent its points. The best we could do is identify the points which provably live in the set given some axioms; such points form a $F_\sigma$ set.

We can also consider "explicit" coarsenings of the topologies on these sets; in particular, every explicit quotient will induce such a coarsening.

Start with an $F_\sigma$ subset of the lattice of open sets of $2^\omega$ (this lattice carries a natural topology, induced by the representation of open sets mentioned above). The $F_\sigma$ set will be used as a subbase for a new, coarser topology on $2^\omega$ or one of its explicit subsets.

Closing our subbase under finite intersections gives an $F_\sigma$ base. The question then is, which open sets of $2^\omega$ are generated by this base? Take any open set; its downward closure intersects the base in some $F_\sigma$ subset.

Now - the part I'm least sure about - taking unions gives an explicit map from $F_\sigma$ sets of opens to opens; the lattice is Hausdorff, so if our chosen open set differs from the union of the sets it contains from the base, we should find out after a finite amount of time.

In other words, the topology generated by the (sub)base is actually closed as a subset of the lattice, even though the base was merely $F_\sigma$.

Is this right? Am I right in thinking we can identify all the coarsenings of $2^\omega$: that is, they form a closed set of closed subsets of the lattice of open sets?

In other words, is there a single, explicit low-complexity map which gives us all explicit represented spaces?

Robin Saunders
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  • I'm beginning to suspect that the "union" map from $F_\sigma$ (or even closed) sets of opens to opens isn't continuous - or, perhaps, is continuous but not computable. If true, that's pretty unpleasant and also a little surprising... I'm trying to learn more about this area, but a lot of the material either isn't publicly available or is somewhat condensed. Any advice would be welcome! – Robin Saunders Jul 04 '20 at 16:30
  • Are you aware of the Effros-Borel structure on the space of closed subsets of $2^\omega$ (or any Polish space)? – Pedro Sánchez Terraf Jul 05 '20 at 12:34
  • Thanks for the pointer. I was not, though I can see from https://math.stackexchange.com/questions/1547337/relation-between-the-effros-structure-and-vietoris-topology that it's related to the downward closure I mentioned since for a space $X$ with $F$ closed and $U$ open, $F \cap U \ne \emptyset \Leftrightarrow U \nsubseteq X \setminus F$. – Robin Saunders Jul 06 '20 at 01:40
  • I've also been reconsidering my first comment. On the one hand, closed subsets of $2^\omega$ are images of the whole space so have a countable dense subset, and countable unions are computable. On the other hand, I only today realized that open sets live in $\Sigma^\omega$ rather than $2^\omega$ itself, and have been too foggy to check if cleaning this up is merely a matter of "book-keeping" (as the above implies, my intuition for this stuff is still pretty weak). – Robin Saunders Jul 06 '20 at 01:50
  • Beware of “countable unions are computable”... I'm not sure in what sense you say this. A recent question might be related to that. – Pedro Sánchez Terraf Jul 06 '20 at 02:34
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    Noted. I mean it in pretty much the sense of proposition 6(4.) here: https://arxiv.org/abs/1204.3763 where I represent maps $2^\omega \rightarrow \Sigma$ in terms of the induced maps from basis elements to $\Sigma$; such maps form an explicit closed subset of $\Sigma^\omega$ (not of $2^\omega$, as I'd assumed until today). – Robin Saunders Jul 06 '20 at 03:46

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