Could you give me an example of a complete metric space wiht covering dimension $> n$ all of which compact subsets have covering dimension $\le n$?
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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?
answered Oct 26, 2010 at 16:43
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$\begingroup$ @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? $\endgroup$ Commented Oct 26, 2010 at 17:44
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$\begingroup$ 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-δCommented Oct 26, 2010 at 17:48
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1$\begingroup$ @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-δCommented Oct 26, 2010 at 17:58
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1$\begingroup$ 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. $\endgroup$ Commented Nov 7, 2010 at 14:49
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$\begingroup$ Why does the Erdös space or its irrational variation admit a complete metrics? $\endgroup$ Commented Mar 1, 2013 at 4:29