Could you give me an example of a complete metric space wiht covering dimension $> n$ all of which compact subsets have covering dimension $\le n$?
1 Answer
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 zerodimensional? This space (and thus any subset of it) is totally disconnected, and isn't it the case that for compact (metric) spaces, this implies zerodimensinal?

$\begingroup$ @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? $\endgroup$ Oct 26, 2010 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. $\endgroup$– εδOct 26, 2010 at 17:48

1$\begingroup$ @Willie, If your set $K$ is compact then there is projection $\pi:K\to\mathbb R^n$ to a coordinate nplane such that preimage of a set of small diameter has small diameter  from this you get that $\dim K=0$ $\endgroup$– εδOct 26, 2010 at 17:58

1$\begingroup$ Yes, complete Erdösspace (all sequences with only irrational coordinates) is complete, separable, dim= 1, and totally disconnected, so this example will do for the complete case. $\endgroup$ Nov 7, 2010 at 14:49

$\begingroup$ Why does the Erdös space or its irrational variation admit a complete metrics? $\endgroup$ Mar 1, 2013 at 4:29