Two definitions of Lebesgue covering dimension

Question

Maybe this question has already been considered here, but after a quick search I didn't find what I was looking for.

As I see, in the literature there are two different definitions of the topological/Lebegue covering dimension.

$\textbf{Definition 1.}$ (e.g. Munkres, General topology) A topological space $X$ is said to have the Lebesgue covering dimension $d<\infty$ if $d$ is the smallest non-negative integer with the property that each open cover of $X$ has a refinement in which no point of $X$ is included in more than $d+1$ elements.

$\textbf{Definition 2.}$ (e.g. Engelking, Ryszard, "Dimension theory" or Pears, "Dimension theory of general spaces") A topological space $X$ is said to have the Lebesgue covering dimension $d<\infty$ if $d$ is the smallest non-negative integer with the property that each $\textbf{finite}$ open cover of $X$ has a refinement in which no point of $X$ is included in more than $d+1$ elements.

Obviously these two definitions agree for compact spaces, but I suppose the first one is stronger in general. For which other class(es) of spaces these two definitions coincide (and for which they don't?

Also, I didn't find much results in the literature about "dimension theory" with respect to Definition 1. (Reference?) For example, is the analogue of the Subset theorem (Pears, Theorem 3.6.4) true when dealing with Definition 1 (i.e. If $S$ is a subset of a totally normal space $X$, then $\dim S \leq \dim X$)?

Answer 1

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 $0$.

Answer 2

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 $1$.