The space of all structures with countable universe and a given signature.

86 Views Asked by At

Let $L=\langle \{R_i\}\cup\{f_j\}\rangle , i \in I, j\in J$ be a signature and consider the space of all L structures with countable universe $\mathbb{N}$. This space is mentioned in the paper here on the connection between topological dynamics (specifically extreme amenability of topological groups), structural ramsey theory and Fraisse theory.)

$$X_L = \Pi_i 2^{\mathbb{n}^n_i} \times \Pi_j \mathbb{n}^{\mathbb{n^m_j}}$$

It is mentioned in the paper that if $L$ is relational, then $X_L$ is compact (homeomorphic to $2^{\mathbb{N}}$.

The thing is, in general I am struggling to understand what the topology on this space is. It is just a countable product space of spaces with discrete topology. Is this space significant in any other contexts?

2

There are 2 best solutions below

11
On BEST ANSWER

What you’ve written for $X_L$ isn’t quite right: it should be

$$X_L=\prod_{i\in I}2^{\Bbb N^{n_i}}\times\prod_{j\in J}\Bbb N^{\Bbb N^{m_j}}\;.$$

Saying that $L$ is relational is just saying that $J=\varnothing$, so that $X_L=\prod_{i\in I}2^{\Bbb N^{n_i}}$, which is the product of countably infinitely many discrete two-point spaces and therefore homeomorphic to $2^{\Bbb N}$. This is an extremely important space: it is homeomorphic to the (middle-thirds) Cantor set and in various guises crops up all over topology, real analysis, and set theory.

If $J\ne\varnothing$ and at least one $m_j>0$, $\prod_{j\in J}\Bbb N^{\Bbb N^{m_j}}$ is homeomorphic to $\Bbb N^{\Bbb N}$, the Cartesian product of countably infinitely many copies of $\Bbb N$ with the discrete topology, which in turn is homeomorphic to the space of irrational numbers with the topology that it inherits from the real line with its usual topology, a space that also crops up in a variety of places.

Topologically speaking, then, $X_L$ is homeomorphic to the Cantor set, the irrationals, or the Cartesian product of the two, and that product is actually homeomorphic to the irrationals, since $2^{\Bbb N}\times\Bbb N^{\Bbb N}=(2\times\Bbb N)^{\Bbb N}=\Bbb N^{\Bbb N}$, so it has a nice and very well understood topological structure.

Added: Andreas Blass reminds me that there are a couple of degenerate cases as well. When $L$ consists of nothing but a finite, non-empty set of constant symbols ($0$-ary function symbols), $X_L$ is homeomorphic to $\Bbb N$, a countably infinite discrete space, and when $L$ has no symbols at all, $X_L$ is just the one-point space.

1
On

It's not that alien a space: it's just the Cantor set (viewed as a subspace of $\mathbb{R}$ with the usual topology), or the set of infinite binary sequences with the topology generated by finite initial segments (the "space of paths through the binary tree"). I think much of the mystery around it comes from our "geometric" expectations, which bias us towards manifolds. Cantor space isn't really geometric (e.g. it's totally disconnected and homeomorphic to its square), but that shouldn't make it particularly mysterious.

Cantor space has some nice properties. It's usually first sighted when learning about cardinality (and in particular how you don't need "long" intervals to have "large" cardinality), but it has many more roles:

  • Compact metric spaces are all (up to homeomorphism) continuous images of Cantor space.

  • Cantor space is uniquely characterized up to homeomorphism as a compact totally disconnected metrizable space without isolated points. In particular, the topological space of $p$-adic integers is homeomorphic to Cantor space.

  • In general, Cantor space is a great source of counterexamples in topology and analysis:

  • Cantor space and its relatives (namely higher products of the two-element discrete space) show up everywhere in logic: e.g. the space of types over a structure, or the space of complete theories in the topological proof of compactness for propositional logic, etc.