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.



totally disconnected and zero-dimensional spaces
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.


