[Math] Which spaces are inverse limits of discrete spaces

gn.general-topology

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.

Best Answer

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.

Related Question