Two definitions of Lebesgue covering dimension

As you refer to Engelking's "Dimension theory" book, I suppose you know the following two statements, but anyway, here they are:

  • The notions agree for separable metric spaces, by Exercise 1.7.E of Engelking's book.

  • The notions agree for paracompact spaces by Proposition 3.2.2 of Engelking's book. (In particular, if both dimensions are finite, they agree.)

An example where the two notions differ was suggested in the comment of Benoît Kloeckner: the long ray $[0,\omega_1)\times[0,1)$ with the lexicographic order topology. The long ray is locally compact but not paracompact. If we use the definition of covering dimension with arbitrary open covers, the long ray has infinite covering dimension (because finite covering dimension implies paracompactness). However, for the finitary covering dimension we get 1 as follows. Any finite open cover is essentially a couple of initial segments $((\lambda_1,a_i),(\lambda_2,b_i))$ with $\lambda_i$ countable ordinals and at least one long end $((\lambda,a),(\omega_1,1))$. The segments with the countable ordinals in the upper bound are homeomorphic to intervals $(a,b)\subseteq\mathbb{R}$. For the initial segments, we get a refinement such that at most two-fold intersections are nonempty (as one would do in $\mathbb{R}$), and only the last of the initial segments intersects the long end $((\lambda,a),(\omega_1,1))$.

Addendum: I just noted that $[0,\omega_1)$ with the order topology is an even simpler (and slightly more puzzling) example. It is locally compact, Hausdorff and not paracompact, but its finitary covering dimension is $0$.


Yes, the long ray $R$ works. If $\mathcal{U}$ is a finite open cover then $\bigcap\{R\setminus U:U\in\mathcal{U}\}=\emptyset$ and at least one of these closed sets must be bounded as in $R$ any two closed ubbounded sets have closed unbounded intersection. Fix one $V\in\mathcal{U}$ and an ordinal $\alpha$ such that $[\alpha,\omega_1)\subset V$ and refine the restriction of $\mathcal{U}$ to $[0,\alpha+1]$ to a cover of order $1$.