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.
General Topology – Totally Disconnected and Zero-Dimensional Spaces
gn.general-topologyreference-request
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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 $0$.
EDIT: I mentioned in the comments that Nik's list of definitions and theorems looks to me strikingly similar to Wallman's generalization of Stone duality.
After getting some feedback from Nik and taking some time to think about the problem more, the connection seems less direct than I previously suggested. However, I do still feel that there is a strong connection between Wallman's theory and Nik's, and I have edited my post to explain this as best I can.
This post does not completely answer Nik's question, but I hope it provides a useful partial answer.
Wallman's Construction:
The main difference between Wallman's theory and yours is that you seem to be thinking of lattice elements as points, whereas Wallman thought of them as closed sets.
I can't find a clear and thorough account of Wallman's results online except for his original paper (which is a bit lengthy, and uses some outdated terminology)
H. Wallman, ``Lattices and topological spaces," Annals of Mathematics vol. 39 (1938), pp. 112 - 126, available here.
I'll outline the construction here.
To every "compact, Hausdorff" lattice $L$ we may associate a topological space called $W(L)$. The points of $W(L)$ are not the members of $L$, but rather the ultrafilters on $L$. The idea is to put a topology on $W(L)$ in such a way that $L$ becomes (isomorphic to) the lattice of closed subsets of $W(L)$.
To put a topology on $W(L)$, let each $a \in L$ define a closed set $$C_a = \{p \in W(L) : a \in p\}.$$ In other words, $C_a$ is the set of all ultrafilters containing $a$. With your definition of a compact lattice, it's not hard to check that these sets constitute the closed subsets of a topology on $W(L)$.
(Sidebar: If you don't require that every subset of $L$ has a greatest lower bound, then this construction will still work. Instead of giving you a topology, the sets $C_a$ will then define a basis for the closed sets of a topology. To see this idea at work, I recommend the first part of section 1 of this paper of Dow and Hart.)
Given a closed set $C$ in $W(L)$, we obtain a "closed" subset of $L$ (in your terminology) by considering $\{a \in L : C \subseteq C_a\}$. Given a "closed" subset $C$ of $L$, we obtain a closed subset of $W(L)$ corresponding to it, namely $\bigcap_{a \in C}C_a$.
As Nik points out in the comments, the converse is not necessarily true: in general, the lattice of closed subsets of a compact Hausdorff space is not necessarily a "compact Hausdorff" lattice.
Why I think Wallman's idea is relevant here:
Using Wallman's construction and known facts from point-set topology, we can derive "near-miss" versions of some of Nik's theorems. For example, we can get a weaker version of Theorem 5, where we drop the continuity requirement but still require our map to be order-preserving:
Theorem: Suppose $L$ is a "compact Hausdorff" lattice and $a \not\leq b$ are in $L$. Then there is an order-preserving map from $L$ into $[0,1]$ taking $a$ to $1$ and $b$ to $0$.
Proof: Let $X = W(L)$ be the compact Hausdorff space whose lattice of closed sets is isomorphic to $L$. In $X$, $a$ and $b$ correspond to the nonempty closed sets $C_a$ and $C_b$, and the relation $a \not\leq b$ translates to $C_a \not\subseteq C_b$. Let $x$ be a point of $X$ that is in $C_a$ but not in $C_b$. Since $X$ is $T_{3\frac{1}{2}}$, there is a continuous function $f: X \rightarrow [0,1]$ such that $f(x) = 1$ and $f(C_b) = 0$. Let $\varphi: L \rightarrow [0,1]$ be defined by $$\varphi(d) = \sup \{f(z) : z \in C_d\}.$$ Clearly $\varphi(a) = 1$ and $\varphi(b) = 0$, and $\varphi$ is order-preserving. However, it is easy to come up with examples where $\varphi$ is does not preserve arbitrary meets and joins (hence the "near-miss").
Similarly, we can get a near-miss version of Theorem 6:
Theorem: If $L$ is a "compact Hausdorff" lattice, then there is an order-preserving injection from $L$ into a power of $[0,1]$.
Proof: Let $I = \{(a,b) \in L \times L : a \not\leq b\}$. For each $(a,b) = i \in I$, fix an order-preserving function $\varphi_i$ as in the previous theorem. Define $\varphi: L \rightarrow [0,1]^I$ by $\pi_i \circ \varphi(a) = \varphi_i(a)$. This function is clearly order-preserving. For injectivity, consider that for all $a \neq b \in L$, either $a \not\leq b$ or $b \not\leq a$.
We get a near-miss version of Theorem 4 in the same way (first prove a $0$-dimensional analogue of Theorem 5, stating that the function in question can be assumed to be two-valued).
Best Answer
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.
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.