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Let $X$ be a (not necessarily metrizable) Hausdorff compact space of covering dimension = 1.

Is it possible to express $X$ as a filtering projective limit of finite graphs?

Here finite graphs means topological realizations of finite graphs, with (surjective?) continuous maps between them.

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  • $\begingroup$ I guess you mean at the end "with some vertices"? Also I guess you just mean "finite graphs", i.e., also with finitely many vertices. Also I guess you want a filtering projective limit (a non-filtering projective limit will most likely not be 1-dimensional). $\endgroup$
    – YCor
    Commented Oct 18, 2019 at 21:04
  • $\begingroup$ Once the question gets cleaned up, it is a standard Yes. (The same for the Hausdorff compact n-dimensional case; the covering dimension is understood here). $\endgroup$
    – Wlod AA
    Commented Oct 18, 2019 at 22:33
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    $\begingroup$ @WlodAA I cleaned up accordingly. It would be useful if you post an answer, possibly with a reference. $\endgroup$
    – YCor
    Commented Oct 19, 2019 at 10:09
  • $\begingroup$ This kind of classical research was to a large part a domain of the old Soviet topology (P.S.Aleksandrov school). I am poor when it comes to references, I have limited means. Basically, it is about finite open nerves, the ordering of the finite open covers (every element of the finer cover has to be a subset of an element of the coarser cover), and the maps into the nerve simplicial implementation. (I may write it as an answer but I was scolded in the past when I did similar things). An easier way to write it down would be without nerves. The projections would not be surjective. $\endgroup$
    – Wlod AA
    Commented Oct 19, 2019 at 19:56
  • $\begingroup$ I was reacting fast and careless. The fact that one must have projections which are not necessarily surjections indicates that the result is not easy (possibly 1-dim case is simpler but I don't know). No wonder that the non-countable (i.e. non-separable) case is harder than the countable sequences (the separable compacta) where projections can be all onto. $\endgroup$
    – Wlod AA
    Commented Oct 19, 2019 at 23:41

1 Answer 1

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References to Engelking's Dimension Theory (1978) ISBN 0-444-85176-3. The answer is yes for compact metrizable spaces, see Section 1.13. In general it is no in general, see Example 3.3.8 (Lokucievskii's example of a compact space $X$ with $\dim X=1$ and $\mathop{\mathrm{ind}}X=2$).

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