These are the completely ultrametrizable spaces.
Recall that a d:E2→[0,∞) is an ultrametric if
- d(x,y) = 0 ↔ x = y
- d(x,y) = d(y,x)
- d(x,z) ≤ max(d(x,y),d(y,z))
As usual, (E,d) is a complete ultrametric space if every Cauchy sequence converges.
Suppose E∞ is the inverse limit of the sequence En of discrete spaces, with fn:E∞→En being the limit maps. Then
d∞(x,y) = inf { 2-n : fn(x) = fn(y) }
is a complete ultrametric on E∞, which is compatible with the inverse limit topology.
Conversely, given a complete ultrametric space (E,d), the relation x ∼n y defined by d(x,y) ≤ 2-n is an equivalence relation. Let En be the quotient E/∼n, with the discrete topology. These spaces have obvious commuting maps between them, let E∞ be the inverse limit of this system. The map which sends each point of E to the sequence of its ∼n equivalence classes is a continuous map f:E→E∞. Because E is complete, this map f is a bijection. Moreover, a simple computation shows that this bijection is in fact a homeomorphism. Indeed, with d∞ defined as above, we have
d∞(f(x),f(y)) ≥ d(x,y) ≥ d∞(f(x),f(y))/2.
As Pete Clark pointed out in the comments, the above is an incomplete answer since the question does not assume that the inverse system is countable. However, the general case does admit a similar characterization in terms of uniformities. For the purposes of this answer, let us say that an ultrauniformity is a unformity with a fundamental system of entourages which consists of open (hence clopen) equivalence relations. The spaces in question are precisely the complete Hausdorff ultrauniform spaces.
Suppose E is the inverse limit of the discrete spaces Ei with limit maps fi:E→Ei. Without loss of generality, this is a directed system. Then the sets Ui = {(x,y): fi(x) = fi(y)} form a fundamental system of entourages for the topology on E, each of which is a clopen equivalence relation on E. The universal property of inverse limits guarantees that E is complete and Hausdorff. Indeed, every Cauchy filter on E defines a compatible sequence of points in the spaces Ei, which is the unique limit of this filter.
Conversely, suppose E is a complete Hausdorff ultrauniform space. If U is a fundamental entourage (so U is a clopen equivalence relation on E) then the quotient space E/U is a discrete space since the diagonal is clopen. In fact, E is the inverse limit of this directed system of quotients. It is a good exercise (for Pete's students) to show that completeness and Hausdorffness of E ensure that E satisfies the universal property of inverse limits.
Let $K$ be an infinite profinite group and let $K_n < K$ be a decreasing family of open subgroups with trivial intersection. Thus, $K$ acts continuously on the discrete space $X = \sqcup_n K/K_n$ and this in turn gives rise to a continuous action of $K$ on the free abelian group $\mathbb Z[X]$. Define $G$ to be the semidirect product $G = K \ltimes \mathbb Z[X]$.
$G$ is a locally compact group and if we denote $X_n = \sqcup_{m \geq n} K/K_m \subset X$ then since $K_n$ acts trivially on $X \setminus X_n$ we have that $K_n \ltimes \mathbb Z[X_n]$ is a family of open normal subgroups with trivial intersection in $G$. Also, $K < G$ is an open subgroup but contains no nontrivial normal subgroups since the action of $K$ on $X$ is faithful.
Best Answer
Every compact topological ring has "enough" open ideals and is thus profinite. See for example Sect. 5.1 in
Luis Ribes, Pavel Zalesski, Profinite Groups, Ergebnisse der Mathematik und ihrer Grenzgebiete, 3. Folge