Is there a space of all finite models?

Question

Let $\mathcal{L}=\{R_i\}_{i \in I}$ be a relational language. There is a natural construction of a compact space of countable $\mathcal{L}$-structures, namely, the space $$ \prod_{R_i \in \mathcal{L}} 2^{\mathbb{N}^{n_i}}$$ where $n_i$ is the arity of the relation symbol $R_i$.

Here the intuition behind the topology is that the more the relations of two structures agree on finite fragments of the underlying set (which is chosen as $\mathbb{N}$ without loss of generality) the closer these structures are. This idea will clearly break down if one considers finite $\mathcal{L}$-structures since a finite amount of data regarding its relations will enable us to distinguish a structure from all the others, so we should end up getting the discrete topology. Indeed, we do get this if $\mathcal{L}$ is finite and we replace $\mathbb{N}$ by $\{0,1,\dots,n\}$ and take the disjoint union of these spaces as $n$ ranges over $\mathbb{N}$.

Can one construct a topological space of all finite $\mathcal{L}$-structures other than the discrete space? Just to avoid any non-trivial topologies arising from the cardinality of $\mathcal{L}$, let us assume that $\mathcal{L}$ is finite; so we are looking for a countably infinite (preferably Hausdorff) topological space.

Obviously I'd prefer to have some kind of intuition behind the notion of closeness in this space. Given how much the space of countable structures has been studied, surely someone must have tried to construct such a space.

Answer 1

I was able to think of a way to topologize the set of finite $\mathcal{L}$-structures that resembles the idea behind the topology of the space of countable structures, namely, declaring the set of finite structures that have a common finite substructure as a basic open set. However, it produces a bad space in terms of separation axioms and whether or not this can become useful is another issue...

For each $n \in \mathbb{N}$, let $\text{Mod}_n(\mathcal{L})$ denote the set of $\mathcal{L}$-structures with universe $\{0,1,\dots,n\}$ and set $\text{Mod}_{<\omega}(\mathcal{L})=\bigcup_{n \in \mathbb{N}} \text{Mod}_n(\mathcal{L})$. For every $\mathcal{M} \in \text{Mod}_{<\omega}(\mathcal{L})$ set $$U_{\mathcal{M}}=\{\mathcal{N} \in \text{Mod}_{<\omega}(\mathcal{L}): \mathcal{M} \text{ is a substructure of } \mathcal{N}\}$$ Here, when I say "is a substructure of", I do not mean "contains a copy of" but I actually do want $\mathcal{M}$ to be physically contained in $\mathcal{N}$ as a substructure with its interpretations as the restrictions of those of $\mathcal{N}$.

Given $\mathcal{M}$ and $\mathcal{M}'$ in $\text{Mod}_{<\omega}(\mathcal{L})$ with domains $\{0,1,\dots,n\}$ and $\{0,1,\dots,n'\}$ respectively, if we have $\mathcal{N} \in U_{\mathcal{M}} \cap U_{\mathcal{M}'}$, then the substructure of $\mathcal{N}$ generated by $\{0,1,\dots,n\}$ is $\mathcal{M}$ and the substructure of $\mathcal{N}$ generated by $\{0,1,\dots,n'\}$ is $\mathcal{M}'$. It follows that if $U_{\mathcal{M}} \cap U_{\mathcal{M}'} \neq \emptyset$, then we have $U_{\mathcal{M}} \subseteq U_{\mathcal{M}'}$ or $U_{\mathcal{M}'} \subseteq U_{\mathcal{M}}$ depending on whether $n' \leq n$ or $n \leq n'$.

It follows that the set $\{U_{\mathcal{M}}: \mathcal{M} \in \text{Mod}_{<\omega}(\mathcal{L})\}$ is a basis for a topology. Let us endow $\text{Mod}_{<\omega}(\mathcal{L})$ with the topology generated by this basis. Then $\text{Mod}_{<\omega}(\mathcal{L})$ is a $T_0$-space since if $\mathcal{A} \neq \mathcal{B}$ and $|A| \leq |B|$, then $\mathcal{A} \notin U_{\mathcal{B}}$ and $\mathcal{B} \in U_{\mathcal{B}}$. This space is not $T_1$ since if $\mathcal{A}$ is a substructure of $\mathcal{B}$, then every open set containing $\mathcal{A}$ also contains $\mathcal{B}$.

As Eric Wofsey mentioned in the comments, convergence in this space is a bit unusual. If there is a sequence of finite $\mathcal{L}$-structures, then any finite $\mathcal{L}$-structure which is eventually a substructure of the elements of the sequence is a limit of this sequence. This is simply because every point $\mathcal{M}$ has a smallest open neighborhood, namely, $U_\mathcal{M}$.

The subspaces $\text{Mod}_n(\mathcal{L})$ are discrete since $U_{\mathcal{M}} \cap \text{Mod}_n(\mathcal{L})= \{\mathcal{M}\}$ whenever $\mathcal{M} \in \text{Mod}_n(\mathcal{L})$. But instead of taking the disjoint union of these discrete spaces which would result in a discrete space, we are "gluing" them at each finite structure in a way that I honestly don't understand.

Answer 2

This is not an answer, but rather an addendum to your answer that is far too long for comments. To try to start getting a sense of your topology, I looked at three simple cases.

If $\mathscr{L}$ is empty, you’re just getting $\omega$ with the topology whose non-empty sets are the tails; call this space $W$.

If $\mathscr{L}$ has a single constant symbol, we can identity $\operatorname{Mod}_{<\omega}(\mathscr{L})$ with

$$\{\langle n,m\rangle\in\omega\times\omega:m\le n\}\,:$$

$\langle n,m\rangle$ corresponds to the model $\mathscr{M}(n,m)$ on $\{0,\ldots,n\}$ with the constant interpreted as $m$. $U_{\mathscr{M}(n,m)}=\{\mathscr{M}(k,m):k\ge n\}$, so $\operatorname{Mod}_{<\omega}(\mathscr{L})$ is homeomorphic to $W\times\omega$, where $\omega$ has the discrete topology.

If $\mathscr{L}$ has a single unary relation symbol $R$, $\operatorname{Mod}_n(\mathscr{L})$ is a discrete set of power $2^{n+1}$. For each $s:\{0,\ldots,n\}\to\{0,1\}$ let $\mathscr{M}_s\in\operatorname{Mod}_n(\mathscr{L})$ be the model for which $s$ is the indicator function of the interpretation of $R$. If $m,n\in\omega$, $m\le n$, $\mathscr{M}_s\in\operatorname{Mod}_m(\mathscr{L})$, and $\mathscr{M}_t\in\operatorname{Mod}_n(\mathscr{L})$, then $\mathscr{M}_t\in U_{\mathscr{M}_s}$ if and only if $s=t\upharpoonright\{0,\ldots,n\}$.

Let $T=\bigcup_{n\in\omega}{^n}2$, so that $\langle T,\subseteq\rangle$ is the full binary tree of height $\omega$, and for $t\in T$ let $U_t=\{s\in T:t\subseteq s\}$; $\{U_t:t\in T\}$ is a base for a topology $\tau$ on $T$. The previous paragraph shows that in this case $\operatorname{Mod}_{<\omega}(\mathscr{L})$ is homeomorphic to $2\times T$, where $2$ has the discrete topology. It would be homeomorphic to $T$ itself if you included the empty model.