TITLE: Two definitions of Lebesgue covering dimension
QUESTION [15 upvotes]: 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 $\dim S \leq \dim X$)?

REPLY [3 votes]: 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 $1$.