totally disconnected and zero-dimensional spaces

Question

When do the notions of totally disconnected space and zero-dimensional space coincide? From what I gather, there are at least three common notions of topological dimension: covering dimension, small inductive dimension, and large inductive dimension. A secondary question, then, would be to what extent and under what assumptions the three different definitions of zero-dimensional coincide. For example, Wikipedia claims that a space has covering dimension zero if and only if it has large inductive dimension zero, and that a Hausdorff locally compact space is totally disconnected if and only if it is zero-dimensional, but I can't track down their source and would like to understand the proofs. I would appreciate any explanation, or a reference, since this is a pretty textbookish question.

Answer 1

Just to agree on notation: A space is zero-dimensional if it is $T_1$ and has a basis consisting of clopen sets, and totally disconnected if the quasicomponents of all points (intersections of all clopen neighborhoods) are singletons. A space is hereditarily disconnected if no subspace is connected, i.e., if the components of all points are singletons. (Edit: There seems to be disagreement about the names of these properties. Often what I call hereditarily disconnected is called totally disconnected and what I call totally disconnected is then called totally separated.)

Note that zero-dimensionality implies Hausdorffness. Zero-dimensional implies totally disconnected since every point can be separated from every other point by a clopen set.
Totally disconnected implies hereditarily disconnected: given a set $A$ with at least two points, one point is not in the quasi-component of the other and hence the two points can be separated by a clopen set. Hence the set $A$ is not connected. This shows that the space is hereditarily disconnected.

On the other hand, if $X$ is locally compact and hereditarily disconnected, take $x\in X$ and let $U$ be an open set containing $x$.
By local compactness, find an open neighborhood $V\subseteq U$ of $x$ whose closure $\overline V$ is compact.
In a compact space, components and quasi-components coincide and hence the quasi-component of $x$ in $\overline V$ is $\{x\}$ (you don't need this if you are not interested in hereditary disconnected spaces but just totally disconnected ones). Using compactness again, there are finitely many clopen subsets $C_1\dots,C_n$ of $\overline V$ such that $x\in C_1\cap\dots\cap C_n\subseteq V$. The intersection of the $C_i$ is closed in $X$ since this intersection is compact. It is open in $X$ since it is open in $V$ and $V$ is open in $X$.
This shows that the clopen subsets of $X$ form a basis.
Hence $X$ is zero-dimensional.

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Edit: As suggested by Joseph Van Name, I include a proof that in a compact space the components coincide with the quasi-components.

Let $X$ be a compact space and $x\in X$. The component $C$ of $x$ is the union of all connected subsets of $X$ containing $x$. If $A\subseteq X$ is clopen and $x\in A$, then the component of $x$ is contained in $A$. It follows that the component of $x$ is contained in the quasi-component $Q$ of $x$.

In order to show that the component $C$ and the quasi-component $Q$ coincide, it is now enough to show that $Q$ is connected. Observe that $Q$ is closed in $X$ and thus compact. Now suppose that $Q$ is not connected. Then there are nonempty, relatively open subsets $A$ and $B$ of $Q$ such that $A\cap B=\emptyset$ and $A\cup B=Q$. Note that $A$ and $B$ are relatively closed in $Q$ and hence compact. Hence $A$ and $B$ are closed in $X$.
Two disjoint closed sets in a compact space can be separated by open subsets, i.e., there are disjoint open sets $U,V\subseteq X$ such that $A\subseteq U$ and $B\subseteq V$.

We have $$Q=\bigcap\{F\subseteq X:F\mbox{ is clopen and }x\in F\}$$ and thus $$\bigcap\{F\subseteq X:F\mbox{ is clopen and }x\in F\}\cap(X\setminus(U\cup V))=\emptyset.$$ By compactness there are finitely many clopen sets $F_1,\dots,F_n$ containing $x$ such that $$F_1\cap\dots\cap F_n\cap(X\setminus(U\cup V))=\emptyset.$$ Let $F=F_1\cap\dots\cap F_n$.
$F$ is clopen and we have $Q\subseteq F\subseteq U\cup V$.

We have $$\overline{U\cap F}\subseteq\overline U\cap F=\overline U\cap(U\cup V)\cap F=U\cap F.$$ It follows that $U\cap F$ is clopen in $X$. We may assume $x\in A$. Since $B$ is nonempty, there is some $y\in B$. But now $y\not\in U\cap F$. It follows that $y$ is not in the quasi-component of $x$, a contradiction.

Answer 2

Following Victor Protsak's suggestion, I took the answer to this question and turned it into a paper found here Ultraparacompactness and Ultranormality, so it may be easier to read that paper than to read the answer here on MO.

The notions that you are looking for are ultraparacompactness (covering dimension depending on how you define it), ultranormality (large inductive dimension zero), and of course the notion of a zero-dimensional space (small inductive dimension zero). A Hausdorff space $X$ is said to be ultraparacompact if every open cover can be refined by a partition into clopen sets. It should be noted that there appears to be some disagreement on the definition of covering dimension since some people require your original cover to be finite and some people consider arbitrary covers, so the notion of ultraparacompactness may or may not coincide with the notion of covering dimension zero. It seems as if the standard but not universal practice is to define the covering dimension in terms of finite covers. The spaces with large inductive dimension zero are known as ultranormal spaces. In other words, a Hausdorff space is ultranormal if and only if whenever $R,S$ are disjoint closed sets, there is a clopen set $C$ with $R\subseteq C,S\subseteq C^{c}$. Ultraparacompactness, ultranormality, and zero-dimensionality are the zero-dimensional analogues of the notions of paracompactness, normality, and regularity. Many in the notions and results from general topology have analogous zero-dimensional notions and results. Clearly every ultraparacompact space is paracompact, and every ultranormal space is normal. It is easy to see that every ultraparacompact space is ultranormal and every ultranormal space is zero-dimensional. However, the converses fail to hold.

$\large\mathbf{Examples}$

Every compact totally disconnected space is ultraparacompact and hence ultranormal as well.

The space $\omega_{1}$ of all countable ordinals with the order topology is ultranormal. If $R,S$ are two disjoint closed subsets of $\omega_{1}$, then either $R$ or $S$ is bounded, so say $R$ is bounded by an ordinal $\alpha$. Then since $[0,\alpha]$ is compact and zero-dimensional, the set $[0,\alpha]$ is ultranormal. Therefore there is a clopen subset $C\subseteq[0,\alpha]$ with $R\subseteq C$ and $S\cap C=\emptyset$. However, the set $C$ is clopen in $\omega_{1}$ as well. Therefore $\omega_{1}$ is ultranormal. To the contrary, the space $\omega_{1}$ is not even paracompact. The cover $\{[0,\alpha)|\alpha<\omega_{1}\}$ however does not have a locally finite open refinement since if $\mathcal{U}$ is an open refinement of $[0,\alpha)$, then for each $\alpha<\omega_{1}$ there is some $x_{\alpha}<\alpha$ where $(x_{\alpha},\alpha]\subseteq U$ for some $U\in\mathcal{U}$. However, since the mapping $\alpha\mapsto x_{\alpha}$ is regressive, everyone who knows anything about set theory can tell you that there is an ordinal $\beta$ where $x_{\alpha}=\beta$ for uncountably many $\alpha$. Since $\mathcal{U}$ refines $\{[0,\alpha)|\alpha<\omega_{1}\}$, each $U\in\mathcal{U}$ is bounded, so the ordinal $\beta$ must be contained in uncountably many $U\in\mathcal{U}$.

Now consider the space $\mathbb{R}$ with the lower limit topology. In other words, with this topology, $\mathbb{R}$ is generated by the basis $\{[a,b)|a<b\}$. Then $[0,\infty)$ is an ultraparacompact space with this topology. Let $\mathcal{U}$ be an open cover of $[0,\infty)$. Let $x_{0}=0$. If $\alpha$ is an ordinal and $\sup\{x_{\beta}|\beta<\alpha\}<\infty$, then let $x_{\alpha}$ be a real number such that $x_{\alpha}>\sup\{x_{\beta}|\beta<\alpha\}$ and $[\sup\{x_{\beta}|\beta<\alpha\},x_{\alpha})\subseteq U$ for some $U\in\mathcal{U}$. Then $\{[x_{\alpha},x_{\alpha+1})|\alpha\}$ is a partition of $[0,\infty)$ into clopen sets that refines $\mathcal{U}$. Thus, $[0,\infty)$ is ultraparacompact. The space $\mathbb{R}$ with the lower limit topology is ultraparacompact as well since $\mathbb{R}$ is isomorphic to the countable sum of spaces $[0,\infty)$ with the lower limit topology. However, it is a well known counterexample that the product $\mathbb{R}\times\mathbb{R}$ is not even normal. We conclude that $\mathbb{R}\times\mathbb{R}$ is a zero-dimensional space which is not ultranormal.

In the paper Nonequality of Dimensions for Metric Spaces, Prabir Roy shows that certain space $\Delta$ is a complete metric space of cardinality continuum which is zero-dimensional, but not ultranormal, and hence not ultraparacompact. In fact, later in the paper Not every 0-dimensional realcompact space is $\mathbb{N}$-compact, Peter Nyikos shows that this space is not even $\mathbb{N}$-compact (a space is $\mathbb{N}$-compact if and only if it can be embedded as a closed subspace of a product $\mathbb{N}^{I}$ for some set $I$). This result strengthens Roy's result since every ultraparacompact space of non-measurable cardinality is $\mathbb{N}$-compact and the first measurable cardinal is terribly large if it even exists.

A metric $d$ on a set $X$ is said to be an ultrametric if $d$ satisfies the strong triangle inequality: $d(x,z)\leq Max(d(x,y),d(y,z))$, and a metric space $(X,d)$ is said to be an ultrametric space if $d$ is an ultrametric. Every ultrametric space is ultraparacompact.

There are zero-dimensional locally compact spaces that are not ultranormal. The Tychonoff plank $X:=((\omega_{1}+1)\times(\omega+1))\setminus\{(\omega_{1},\omega)\}$ is zero-dimensional (even strongly zero-dimensional; i.e. $\beta X$ is zero-dimensional) but not ultranormal.

$\large\mathbf{Results}$

A subset $Z$ of a space $X$ is said to be a zero set if there is a continuous function $f:X\rightarrow[0,1]$ such that $Z=f^{-1}[\{0\}]$. A completely regular space $X$ is said to be strongly zero-dimensional if whenever $Z_{1},Z_{2}$ are disjoint zero sets, there is a clopen set $C$ with $Z_{1}\subseteq C$ and $Z_{2}\subseteq C^{c}$. It is not too hard to show that a completely regular space $X$ is strongly zero-dimensional if and only if the Stone-Cech compactification $\beta X$ is zero-dimensional. We say that a cover $\mathcal{R} $ of a topological space $X$ is point-finite if $\{R\in\mathcal{R}|x\in R\}$ is finite for each $x\in X$, and we say that $\mathcal{R}$ is locally finite if each $x\in X$ has a neighborhood $U$ such that $\{R\in\mathcal{R}|U\cap R\neq\emptyset\}$ is finite. Recall that a Hausdorff space is paracompact iff every open cover has a locally finite open refinement.

A uniform space $(X,\mathcal{U})$ is said to be non-Archimedean if $\mathcal{U}$ is generated by equivalence relations. In other words, for each $R\in\mathcal{U}$ there is an equivalence relation $E\in\mathcal{U}$ with $E\subseteq R$. Clearly, every non-Archimedean uniform space is zero-dimensional. If $(X,\mathcal{U})$ is a uniform space, then let $H(X)$ denote the set of all closed subsets of $X$. If $E\in\mathcal{U}$, then let $\hat{E}$ be the relation on $H(X)$ where $(C,D)\in\hat{E}$ if and only if $C\subseteq E[D]=\{y\in X|(x,y)\in E\textrm{for some}x\in D\}$ and $D\subseteq E[C]$. Then the system $\{\hat{E}|E\in\mathcal{U}\}$ generates a uniformity $\hat{\mathcal{U}}$ on $H(X)$ called the hyperspace uniformity. A uniform space $(X,\mathcal{U})$ is said to be supercomplete if the hyperspace $H(X)$ is a complete uniform space.

$\mathbf{Proposition}$ Let $X$ be a locally compact zero-dimensional space. Then $X$ is ultraparacompact if and only if $X$ can be partitioned into a family of compact open sets.

$\mathbf{Proof}$ The direction $\leftarrow$ is fairly trivial. To prove $\rightarrow$ assume that $X$ is a locally compact zero-dimensional space. Then let $\mathcal{U}$ the collection of all open sets $U$ such that $\overline{U}$ is compact. Then since $X$ is locally compact, $\mathcal{U}$ is a cover for $X$. Therefore there is a partition $P$ of $X$ into clopen sets that refines $\mathcal{U}$. Clearly each $R\in P$ is a compact open subset of $X$. $\textrm{QED}$

$\textbf{Theorem}$ Let $X$ be a Hausdorff space. Then $X$ is normal if and only if whenever $(U_{\alpha})_{\alpha\in\mathcal{A}}$ is a point-finite open covering of $X$, there is an open covering $(V_{\alpha})_{\alpha\in\mathcal{A}}$ such that $\overline{V_{\alpha}}\subseteq U_{\alpha}$ for $\alpha\in\mathcal{A}$ and $V_{\alpha}\neq\emptyset$ whenever $U_{\alpha}\neq\emptyset$.///

To prove the above result, one first well orders the set $\mathcal{A}$, then one shrinks the sets $U_{\alpha}$ to sets $V_{\alpha}$ in such a way that you still cover your space $X$ at every point in the induction process. See the book Topology by James Dugundji for a proof of the above result.

$\textbf{Theorem}$ Let $X$ be a Hausdorff space. The following are equivalent.

1. $X$ is ultranormal.

2. Whenever $(U_{\alpha})_{\alpha\in\mathcal{A}}$ is a point-finite open cover of $X$, there is a clopen cover $(V_{\alpha})_{\alpha\in\mathcal{A}}$ such that $V_{\alpha}\subseteq U_{\alpha}$ for each $\alpha$ and $V_{\alpha}\neq\emptyset$ whenever $U_{\alpha}\neq\emptyset$.

3. If $(U_{\alpha})_{\alpha\in\mathcal{A}}$ is a locally-finite open cover of $X$, then there is system $(P_{\alpha})_{\alpha\in\mathcal{A}}$ of clopen sets such that $P_{\alpha}\subseteq U_{\alpha}$ for $\alpha\in\mathcal{A}$ and $P_{\alpha}\cap P_{\beta}=\emptyset$ whenever $\alpha,\beta\in\mathcal{A}$ and $\alpha\neq\beta$.

4. $X$ is normal and strongly zero-dimensional.

$\textbf{Proof}$ $1\rightarrow 2$. Since $X$ is normal, there is an open cover $(W_{\alpha})_{\alpha\in\mathcal{A}}$ such that $\overline{W_{\alpha}}\subseteq U_{\alpha}$ for each $\alpha\in\mathcal{A}$ and $W_{\alpha}\neq\emptyset$ whenever $U_{\alpha}\neq\emptyset$. Since $X$ is ultranormal, for each $\alpha\in\mathcal{A}$, there is some clopen set $V_{\alpha}$ with $\overline{W_{\alpha}}\subseteq V_{\alpha}\subseteq U_{\alpha}$.

$2\rightarrow 3$. Now assume that $(U_{\alpha})_{\alpha\in\mathcal{A}}$ is a locally-finite open over of $X$. Let $(V_{\alpha})_{\alpha\in\mathcal{A}}$ be a clopen cover of $X$ such that $V_{\alpha}\subseteq U_{\alpha}$ for each $\alpha\in\mathcal{A}$. Well order the set $\mathcal{A}$. The family $(V_{\alpha})_{\alpha\in\mathcal{A}}$ is locally-finite, so since each $V_{\alpha}$ is closed, for each $\alpha\in\mathcal{A}$, the union $\bigcup_{\beta<\alpha}V_{\beta}$ is closed. Clearly $\bigcup_{\beta<\alpha}V_{\beta}$ is open as well, so $\bigcup_{\beta<\alpha}V_{\beta}$ is clopen. Let $P_{\alpha}=V_{\alpha}\setminus(\bigcup_{\beta<\alpha}V_{\beta})$. Then $(P_{\alpha})_{\alpha\in\mathcal{A}}$ is the required partition of $X$ into clopen sets.

$3\rightarrow 1,1\rightarrow 4$. This is fairly obvious.

$4\rightarrow 1$. This is a trivial consequence of Urysohn's lemma.

$\textbf{QED}$

$\textbf{Theorem}$ Let $X$ be a Hausdorff space. The following are equivalent.

1. $X$ is ultraparacompact.

2. $X$ is ultranormal and paracompact.

3. $X$ is strongly zero-dimensional and paracompact.

4. Every open cover of $X$ has a locally finite clopen refinement.

5. $X$ is zero-dimensional and satisfies the following property: let $I$ be an ideal on the Boolean algebra $\mathfrak{B}(X)$ of clopen subsets of $X$ such that $\bigcup I=X$ and if $P$ is a partition of $X$ into clopen sets, then $\bigcup(P\cap I)\in I$. Then $I=\mathfrak{B}(X)$.

6. $X$ has a compatible supercomplete non-Archimedean uniformity. ///

The paper $\omega_{\mu}$-additive topological spaces by Giuliano Artico and Roberto Moresco gives several characterizations of when a $P_{\kappa}$-space (a space where the intersection of less than $\kappa$ many open sets is open) is ultraparacompact.

$\large\textbf{The Point-Free Context}$.

The following results on zero-dimensionality, ultranormality, and ultraparacompactness are part of my own research. If you think I have wrote too much already, you don't like me, or if you don't believe in pointless topology, then I suggest that you stop reading this answer here.

The notions of ultranormality, zero-dimensionality, and ultraparacompactness make sense in a point-free context, and by a generalization of Stone duality, the notions of ultranormality, zero-dimensionality, and ultraparacompactness translate nicely to certain kinds of Boolean algebras with extra structure.

A frame is a complete lattice that satisfies the following infinite distributivity law $$x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i}).$$ If $X$ is a topological space, then the collection of all open subsets of $X$ forms a frame. Frames generalize the notion of a topological space, and frames are the central object of study in point-free topology. For Hausdorff spaces, no information is lost simply by considering the lattice of open sets of the topological space. More specifically, if $X,Y$ are Hausdorff spaces and the lattices of open sets of $X$ and $Y$ respectively are isomorphic, then $X$ and $Y$ are themselves isomorphic. Furthermore, many notions and theorems from general topology can be generalized to the point-free context. Moreover, the notion of a frame is very interesting from a purely lattice theoretic perspective without even looking at the topological perspective. The reader is referred to the excellent new book Frames and Locales: Topology Without Points by Picado and Pultr for more information on point-free topology.

For notation, if $X$ is a poset and $R,S\subseteq X$, then $R$ refines $S$ (written $R\preceq S$) if for each $r\in R$ there is some $s\in S$ with $r\leq s$. If $x\in X$, then define $\downarrow x:=\{y\in X|y\leq x\}$.

A Boolean admissibility system is a pair $(B,\mathcal{A})$ such that $B$ is a Boolean algebra and $\mathcal{A}$ is a collection of subsets of $B$ with least upper bounds such that

i. $\mathcal{A}$ contains all finite subsets of $B$,

ii. if $R\in\mathcal{A},S\subseteq\downarrow\bigvee R,R\preceq S$, then $S\in\mathcal{A}$ as well,

iii. if $R\in\mathcal{A}$ and $R_{r}\in\mathcal{A},\bigvee R_{r}=r$ for $r\in R$, then $\bigcup_{r\in R}R_{r}\in\mathcal{A}$,

iv. if $R\in\mathcal{A}$, then $\{a\wedge r|r\in R\}\in\mathcal{A}$ as well.

Intuitively, a Boolean admissibility system is Boolean algebra along with a notion of which least upper bounds are important and which least upper bounds are not important.

For example, if $B$ is a Boolean algebra, and $\mathcal{A}$ is the collection of all subsets of $B$ with least upper bounds, then $(B,\mathcal{A})$ is a Boolean admissibility system.

If $A$ is a Boolean subalgebra of $B$, and $\mathcal{A}$ is the collection of all subsets of $R$ where the least upper bound $\bigvee^{B}R$ exists in $B$ and $\bigvee^{B}R\in A$. Then $(A,\mathcal{A})$ is a Boolean admissibility system.

A Boolean admissibility system $(B,\mathcal{A})$ is said to be subcomplete if whenever $R,S\subseteq B$ and $R\cup S\in\mathcal{A}$ and $r\wedge s=\emptyset$ whenever $r\in R,s\in S$, then $R\in\mathcal{A}$ and $S\in\mathcal{A}$.

If $L$ is a frame, then an element $x\in L$ is said to be complemented if there is an element $y$ such that $x\wedge y=0$ and $x\vee y=1$. The element $y$ is said to be the complement of $y$ and one can easily show that the element $y$ is unique. The notion of a complemented element is the point-free generalization of the notion of a clopen set. Let $\mathfrak{B}(L)$ denote the set of complemented elements in $L$. Then $\mathfrak{B}(L)$ is a sublattice of $L$. In fact, $\mathfrak{B}(L)$ is a Boolean lattice. A frame $L$ is said to be zero-dimensional if $x=\bigvee\{y\in\mathfrak{B}(L)|y\leq x\}$ for each $x\in L$.

A Boolean based frame is a pair $(L,B)$ where $L$ is a frame and $B$ is a Boolean subalgebra of $\mathfrak{B}(L)$ such that $x=\bigvee\{b\in\mathfrak{B}(L)|b\leq x\}$ for each $x\in L$. Clearly every Boolean based frame is zero-dimensional.

A zero-dimensional frame $L$ is said to be ultranormal if whenever $a\vee b=1$, there is a complemented element $c\in L$ such that $c\leq a$ and $c'\leq b$.

A cover of a frame $L$ is a subset $C\subseteq L$ with $\bigvee C=1$. A partition of a frame $L$ is a subset $p\subseteq L\setminus\{0\}$ with $\bigvee p=1$ and where $a\wedge b=0$ whenever $a,b\in p,a\neq b$.

A zero-dimensional frame $L$ is said to be ultraparacompact if whenever $C$ is a cover of $L$ there is a partition $p$ that refines $C$.

There is a duality between the category of Boolean admissibility systems and Boolean based frames. If $(B,\mathcal{A})$ is a Boolean admissibility system, then let $C_{\mathcal{A}}$ be the collection of all ideals $I\subseteq B$ such that if $R\in\mathcal{A}$ and $R\subseteq I$, then $\bigvee R\in I$ as well.

If $(B,\mathcal{A})$ is a Boolean admissibility system, then $\mathcal{V}(B,\mathcal{A}):=(C_{\mathcal{A}},\{\downarrow b|b\in B\})$ is a Boolean based frame. Similarly, if $(L,A)$ is a Boolean based frame, then $\mathcal{W}(L,A):=(A,\{R\subseteq A|\bigvee^{L}R\in A\})$ is a Boolean admissibility system. Furthermore, these correspondences give an equivalence between the category of Boolean admissibility systems and the category of Boolean based frames. Hence, we obtain a type of Stone-duality for zero-dimensional frames. If $(L,A)$ is a Boolean based frame, then $A=\mathfrak{B}(L)$ if and only if $\mathcal{W}(L,A)$ is subcomplete. Since $(L,\mathfrak{B}(L))$ is a Boolean based frame iff $L$ is a zero-dimensional frame, we conclude that the category of zero-dimensional frames is equivalent to the category of subcomplete Boolean admissibility systems.

If $(B,\mathcal{A})$ is a Boolean admissibility system, then $ \mathcal{V}(B,\mathcal{A})=(C_{\mathcal{A}},\{\downarrow b|b\in B\})$ is ultranormal with $\{\downarrow b|b\in B\}=\mathfrak{B}(C_{\mathcal{A}})$ if and only if whenever $I,J\in\mathcal{C}_{\mathcal{A}}$, then $\{a\vee b|a\in I,b\in J\}\in\mathcal{C}_{\mathcal{A}}$ as well. Hence, ultranormality simply means that the join of finitely many ideals in $\mathcal{C}_{\mathcal{A}}$ is an ideal in $\mathcal{C}_{\mathcal{A}}$.

If $(B,\mathcal{A})$ is a subcomplete Boolean admissibility system, then $\mathcal{V}(B,\mathcal{A})$ is ultraparacompact if and only if whenever $R\in\mathcal{A}$ there is some $S\in\mathcal{A}$ with $S\preceq R,\bigvee S=\bigvee R$ and $a\wedge b=0$ whenever $a,b\in S$ and $a\neq b$. Hence, the ultraparacompactness of frames translates into a version of ultraparacompactness in the dual subcomplete Boolean admissibility systems.

Answer 3

Here's just a reference showing that a Hausdorff locally compact totally disconnected space is zero-dimensional: Proposition 3.1.7 of Arhangel'skii and Tkachenko.

Answer 4

The answer to the secondary question is `no'. In this paper J. Terasawa constructed spaces of the form $\omega\cup\mathcal{A}$, where $\mathcal{A}$ is a maximal almost disjoint family, of arbitrary large covereing dimension, even infinite. Points in $\omega$ are isolated and basic neighbourhoods of $A\in\mathcal{A}$ are of the form $\lbrace A\rbrace\cup (A\setminus F)$, where $F$ is finite. This space is always Hausdorff, locally compact and zero-dimensional, but its covering dimension depends on $\mathcal{A}$.

Answer 5

To add yet another reference, I like the treatment in the book of Hurewicz and Wallman (Dimension Theory). In particular, in chapter II the theory of 0-dimensional sets is treated with quite a lot generality and examples.