Products of spaces of covering dimension zero

Question

Recall that a space has covering dimension zero if every open cover of it can be refined to a disjoint open cover.

Question 1. Does the product of two spaces of covering dimension zero have covering dimension zero?

Question 2. Does the product of infinitely many spaces of covering dimension zero have covering dimension zero?

Question 3. Are the answers different for topological spaces and locales?

I think I have a proof that works for compact spaces of covering dimension zero, using arguments similar to the one to prove the tube lemma and the Tychonoff theorem. However, there are certainly non-compact spaces of covering dimension zero.

Answer 1

It should be well-known that a regular, T1 space has covering dimension zero, iff it is paracompact and strongly zero-dimensional. Hence, a Lindelof zero-dimensional T1 space has covering dimension zero (see Engelking, General topology 6.2.7).

In particular, the Sorgenfrey line has covering dimension zero, but its square is not normal, hence does not have covering dimension zero.

Update
It should be noted that the above definition of "covering dimension zero" is not the usual one. The usual one only requires that $X$ is Tychonoff and every finite cover of cozero sets has a finite open refinement, i.e. it is strongly zero-dimensional. Of course, my above answer is according to the OP's definition. Spaces with this property are frequently called "ultraparacompact".

According to Engelking, General topology, p.363, there also exist two strongly zero-dimensional spaces such that their product is not strongly zero-dimensional. However, these examples seem to be very difficult. (Therefore, they are not even described there.)

Answer 2

For the record, let me give a direct proof that the Sorgenfrey plane is a counterexample to the first question.

Lemma. A topological space or locale with a basis of clopens and with the property that every open cover has a countable subcover has covering dimension zero.

Proof. Suppose given an open cover. We need to find a disjoint open cover refining it. By hypothesis, the given open cover can be refined by a countable clopen cover. Given an enumerated clopen cover $U_0, U_1, U_2, \ldots$, define $V_n = U_n \setminus ( U_0 \cup \ldots \cup U_{n-1} )$. Then each $V_n \subseteq U_n$, is open, and is disjoint from $V_0, \ldots, V_{n-1}$ by construction, so we have the desired disjoint open cover.　◼

Proposition. The Sorgenfrey line has covering dimension zero.

Proof. The Sorgenfrey line has a basis of clopens: indeed, each of the intervals $[a, b)$ is clopen. Furthermore, every open cover has a countable subcover. Thus, we may apply the lemma.　◼

Proposition. The Sorgenfrey plane does not have covering dimension zero.

Proof. For each real number $z$, let $A_z = [-z, \infty) \times [z, \infty)$ and let $B_z = (-\infty, -z) \times (-\infty, z)$. $A_z$ and $B_z$ are open subsets of the Sorgenfrey plane, and given $(x, y)$, if $x \ge -y$ then $(x, y) \in A_y$, whereas if $x < -y$ then $(x, y) \in B_{y + 1}$, so we have an open cover. Only $A_z$ contains $(-z, z)$, so any refinement of this cover is uncountable. On the other hand, any disjoint open cover must be countable (after excluding empty subsets), because every non-empty open subset contains a point with rational coordinates. Therefore the open cover constructed here has no disjoint refinement.　◼