Which spaces are inverse limits of discrete spaces ?

Question

There is the following theorem:

"A space $X$ is the inverse limit of a system of discrete finite spaces, if and only if $X$ is totally disconnected, compact and Hausdorff."

A finite discrete space is totally disconnected, compact and Hausdorff and all those properties pass to inverse limits. I guess the other direction might be proved by taking the system of all decompositions of $X$ into disjoint clopen sets. The inverse limit should give $X$ back.

So what happens, if I dismiss the finiteness condition. As mentioned above every inverse limit of discrete spaces is totally disconnected, Hausdorff. So the question is:

"Which totally disconnected Hausdorff spaces are inverse limits of discrete spaces?"

For example I think it is impossible to write $\mathbb{Q}$ as an inverse limit of discrete spaces, but I don't have a proof.

Answer 1

These are the completely ultrametrizable spaces.

Recall that a d:E²→[0,∞) is an ultrametric if

1. d(x,y) = 0 ↔ x = y
2. d(x,y) = d(y,x)
3. 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 E_n of discrete spaces, with f_n:E_∞→E_n being the limit maps. Then

d_∞(x,y) = inf { 2^-n : f_n(x) = f_n(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 E_n 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.

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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 E_i with limit maps f_i:E→E_i. Without loss of generality, this is a directed system. Then the sets U_i = {(x,y): f_i(x) = f_i(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 E_i, 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.

Answer 2

The rational numbers are not the inverse limit of a countable sequence of discrete spaces, but the rational numbers are in fact the inverse limit of an uncountable collection of discrete spaces.

Another way to see that the rational numbers are not an inverse limit of a countable sequence of discrete spaces is to first take note that an inverse limit of a countable sequence of discrete spaces is metrizable by a complete metric. On the other hand, every completely metrizable subset of $\mathbb{R}$ is a $G_{\delta}$-set [DUG p. 307]. If $\mathbb{Q}$ were the intersection of countably many open sets $O_{n}$, then each $O_{n}$ would be dense making $\mathbb{Q}$ of second category. This is a contradiction. Therefore $\mathbb{Q}$ is not completely metrizable and not the inverse limit of a sequence of discrete spaces.

The rational numbers are in fact an inverse limit of discrete spaces. First take note that $\mathbb{Q}$ is Lindelof and regular, so $\mathbb{Q}$ is realcompact [WAL p. 41]. Another way to see that $\mathbb{Q}$ is realcompact is to take note that $\mathbb{Q}$ is paracompact and of cardinality below the first measurable cardinal. Furthermore, since $\mathbb{Q}$ is Lindelof and zero-dimensional. $\mathbb{Q}$ is strongly zero-dimensional[WAL p. 85]. Therefore since $\mathbb{Q}$ is realcompact and strongly zero-dimensional, $\mathbb{Q}$ is $\mathbb{N}$-compact[WAL p. 264]. Therefore since $\mathbb{Q}$ is $\mathbb{N}$-compact, $\mathbb{Q}$ is the inverse limit of discrete spaces[CHE].

The spaces which are inverse limits of discrete spaces are precisely the spaces with a compatible complete ultrauniformity as it was pointed out earlier. We shall call a topological space ultracomplete if it can be given a compatible complete ultrauniformity. Let $X$ be a zero-dimensional space. Then let $\mathcal{U}$ be the uniformity generated by equivalence relations $E$ such that each equivalence class in $E$ is a clopen set. Then we shall call $\mathcal{U}$ the fine ultrauniformity on the topological space $X$. One can show that a zero-dimensional space $X$ is ultracomplete if and only if $X$ is complete in the fine ultrauniformity.

On a different note, every inverse limit of discrete spaces is a closed subspace of a product of discrete spaces [DUG p. 429]. Furthermore, one can easily show that every closed subspace of a product of discrete spaces can be given a compatible complete ultrauniformity, and hence the closed subspaces of products of discrete spaces can be written as inverse limits of discrete spaces. Therefore the spaces representable as inverse limits of discrete spaces are the spaces representable as closed subspaces of products of discrete spaces. Furthermore, for spaces of cardinality below the first measurable cardinal, the N-compact spaces correspond with the spaces representable as inverse limits of discrete spaces.

[CHE] Chew, Kim-Peu. "N-compact Spaces as Limits of Inverse Systems of Discrete Spaces." Journal of the Australian Mathematical Society 14.04 (1972): 467.

[DUG] Dugundji, James. Topology. Boston: Allyn and Bacon, 1966.

[WAL] Walker, Russell C. The Stone-Cech Compactification. Berlin: Springer-Verlag, 1974.

Answer 3

The rationals Q are not an inverse limit of discrete spaces. Suppose that Q is the inverse limit of A_n, with a map from each A_n+1 to A_n. The inverse limit consists of infinite sequences that respect these maps. Thus, it is the set of infinite paths through the tree of finite coherent sequences. (We may restrict this tree to the finite sequences that actually lay on an infinite branch.) Since Q has no isolated points, this restricted tree is a splitting tree, and therefore has continuum 2^ω many infinite paths. But Q is countable, contradiction.

(We may assume the index set of the limit is ω, by passing to a cofinal ω sequence, since Q is countable and not discrete.)