How are Topological Equivalences of Structures Homeomorphic?


Given a first-order vocabulary 𝜏, πΏπœ”πœ”(𝜏) is the set of first-order sentences of type 𝜏. The elementary topology on the class π‘†π‘‘πœ of first-order structures type 𝜏 is obtained by taking the family of elementary classes

π‘€π‘œπ‘‘(πœ‘) = {𝑀:𝑀 |= πœ‘}, πœ‘ ∈ πΏπœ”πœ”(𝜏)

as an open basis. Due to the presence of classical negation, this family is also a closed basis and thus the closed classes of π‘†π‘‘πœ are the first-order axiomatizable classes π‘€π‘œπ‘‘(𝑇), 𝑇 βŠ† πΏπœ”πœ”(𝜏). Possible foundational problems due to the fact that the topology is a class of classes may be settled observing that it is indexed by a set, namely the set of theories of type 𝜏.

The main facts of model theory are reflected by the topological properties of these spaces. Thus, the downward Löwenheim-Skolem theorem for sentences amounts to topological density of the subclass of countable structures. Łoś theorem on ultraproducts grants that U-limits exist for any ultrafilter π‘ˆ, condition well known to be equivalent to topological compactness, and to model theoretic compactness in this case.

These spaces are not Hausdorff or T1, but having a clopen basis they are regular; that is, closed classes and exterior points may be separated by disjoint open classes. All properties or regular compact spaces are then available: normality, complete regularity, uniformizability, the Baire property, etc.

Many model theoretic properties are related to the continuity of natural operations between classes of structures, where operations are seen to be continuous and play an important role in abstract model theory.

A topological space is regular if closed sets and exterior points may be separated by open sets. It is normal if disjoint closed sets may be separated by disjoint open sets. Thus, normality does not imply regularity here. However, a regular compact space is normal. Actually, a regular Lindelöf space is already normal

Consider the following equivalence relation in a space 𝑋: π‘₯ ≑ 𝑦 ⇔ 𝑐𝑙{π‘₯} = 𝑐𝑙{𝑦}

where 𝑐𝑙 denotes topological adherence. Clearly, π‘₯ ≑ 𝑦 iff π‘₯ and 𝑦 belong to the same closed (open) subsets (of a given basis). Let 𝑋/≑ be the quotient space and πœ‚ : 𝑋 β†’ 𝑋/≑ the natural projection. Then 𝑋/≑ is T0 by construction but not necessarily Hausdorff. The following claims thus follow:

a) πœ‚ : 𝑋 β†’ 𝑋/≑ induces an isomorphism between the respective lattices of BorelΒ subsets of 𝑋 and 𝑋/≑. In particular, it is open and closed, preserves disjointedness, preserves and reflects compactness and normality.

b) The assignment 𝑋 β†’ 𝑋/≑ is functorial, because ≑ is preserved by continuous functions and thus any continuous map 𝑓 : 𝑋 β†’ π‘Œ induces a continuous assignment 𝑓/≑ : 𝑋/≑ β†’ π‘Œ/≑ which commutes with composition.

c) 𝑋 β†’ 𝑋/≑ preserves products; that is, (𝛱𝑖𝑋𝑖)/≑ is canonically homeomorphic to 𝛱𝑖(𝑋𝑖/≑) with the product topology (monomorphisms are not preserved).

d) If 𝑋 is regular, the equivalence class of π‘₯ is 𝑐𝑙{π‘₯} (this may fail in the non-regular case).

e) If 𝑋 is regular, 𝑋/≑ is Hausdorff : if π‘₯ ≑̸ 𝑦 then π‘₯ ∉ 𝑐𝑙{𝑦} by (d); thus there are disjoint open sets π‘ˆ, 𝑉 in 𝑋 such that π‘₯ ∈ π‘ˆ, 𝑐𝑙{𝑦} βŠ† 𝑉, and their images under πœ‚ provide an open separation of πœ‚π‘₯ and πœ‚π‘¦ in 𝑋/≑ by (a).

f) If 𝐾1 and 𝐾2 are disjoint compact subsets of a regular topological space 𝑋 that cannot be separated by open sets there exist π‘₯𝑖 ∈ 𝐾𝑖, 𝑖 = 1, 2, such that π‘₯1 ≑ π‘₯2. Indeed, πœ‚πΎ1 and πœ‚πΎ2 are compact in 𝑋/≑ by continuity and thus closed because 𝑋/≑ is Hausdorff by (e). They can not be disjoint; otherwise, they would be separated by open sets whose inverse images would separate 𝐾1 and 𝐾2. Pick πœ‚π‘₯ = πœ‚π‘¦ ∈ πœ‚πΎ1 ∩ πœ‚πΎ2 with π‘₯ ∈ 𝐾1, 𝑦 ∈ 𝐾2.

Clearly then, for the elementary topology on π‘†π‘‘πœ, the relation ≑ coincides with elementary equivalence of structures and π‘†π‘‘πœ/≑ is homeomorphic to the Stone space of complete theories.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s