Could you give me an example of a complete metric space wiht covering dimension $> n$ all of which compact subsets have covering dimension $\le n$?


A guess

In $l^2$ Hilbert space, consider the set $E$ of points with all coordinates rational. Erdös (reference) showed that $E$ has topological dimension $1$. (In separable metric space, all notions of topological dimension coincide.)

Does this $E$ have the property that every compact subset is zero-dimensional? This space (and thus any subset of it) is totally disconnected, and isn't it the case that for compact (metric) spaces, this implies zero-dimensinal?

  • $\begingroup$ @Gerald: I can't quite figure out how compactness can be used to rule out the element on the boundary that Erdos constructed. Help please? $\endgroup$ – Willie Wong Oct 26 '10 at 17:44
  • $\begingroup$ Thank you, I add "completeness" but Erdös constructs also a complete set ($R_1$). It seems to work, but I need to think a bit. $\endgroup$ – ε-δ Oct 26 '10 at 17:48
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    $\begingroup$ @Willie, If your set $K$ is compact then there is projection $\pi:K\to\mathbb R^n$ to a coordinate n-plane such that preimage of a set of small diameter has small diameter --- from this you get that $\dim K=0$ $\endgroup$ – ε-δ Oct 26 '10 at 17:58
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    $\begingroup$ Yes, complete Erdös-space (all sequences with only irrational coordinates) is complete, separable, dim= 1, and totally disconnected, so this example will do for the complete case. $\endgroup$ – Henno Brandsma Nov 7 '10 at 14:49
  • $\begingroup$ Why does the Erdös space or its irrational variation admit a complete metrics? $\endgroup$ – Włodzimierz Holsztyński Mar 1 '13 at 4:29

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