# Which spaces are inverse limits of discrete spaces ?

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.

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Your guess is correct. I would guess that it is just the compactness that disappears but that is just a guess! I know there are results in this direction c'classically' but I would also point out that prodiscrete locales have been studied, i.e. don't take the inverse limit in the category of spaces, just look at the open set structure, if you see what I mean. –  Tim Porter Feb 16 '10 at 10:41
Good question. For some additional motivation, note that the "Borger-Carnahan etale fundamental group" of a scheme (see mathoverflow.net/questions/13176) is topologized in exactly this way: an inverse limit of not necessarily finite, discrete groups. –  Pete L. Clark Feb 16 '10 at 13:42
Continuing the previous thought, suppose we look only at inverse limits of discrete topological groups. In the case of finite groups, the inverse limit is compact, so by consideration of Haar measure, it is uncountable. What if the finiteness is dropped? The inverse limit need not be metrizable, but it seems like there is a good chance that it is complete in its (left, say) uniform structure. If so, that would force it not to be Q... –  Pete L. Clark Feb 16 '10 at 13:47
One consideration is that the limit functor is not exact on spaces, or on groups. This gives rise to the derived functors of systems of (Abelian) topological groups. These vanish if the topological groups are compact. I believe that you cannot retrieve the system from the limit unlike with the 'pro-finite' case, but I could not give chapter and verse on this. –  Tim Porter Feb 16 '10 at 15:39

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 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.

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Francois: This sounds good if the inverse system is countable. It cannot be correct in general, since e.g. an uncountable product of nontrivial discrete groups is not first countable, hence not metrizable. –  Pete L. Clark Feb 16 '10 at 15:32
"completely ultrametrizable" sounds like something my 7-year-old would say when describing something that I would call "good". –  Kevin Buzzard Feb 16 '10 at 15:32
Still, it is a very nice characterization of completely ultrametrizable spaces... –  Pete L. Clark Feb 16 '10 at 15:34
Pete, do you have a class full of Francoises, or what? –  Joel David Hamkins Feb 16 '10 at 17:59
In case it is not already clear, I think this is a fantastic answer, deserving of many more upvotes than it currently has. –  Pete L. Clark Feb 16 '10 at 22:59

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.

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The rationals Q are not an inverse limit of discrete spaces. Suppose that Q is the inverse limit of An, with a map from each An+1 to An. 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 2ω 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.)

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You can take \hat X the inverse limit given by system of cocients X/Y where Y is any open in X. First note that since X is compact and Y is open X/Y is finite. And if Y is contained in Y´, both open in X, then there is a canonic continuous map form X/Y into X/Y´.

Obviusly X with the proyections on open cocients is a co-cone over this system, then, there is a canonic function f fors nuoum X into \hat X. You must show that is continuous.

I hop it can help

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