Does naming the integers in $(\mathbb{R};+,\times)$ change the "definable topologies" significantly?

Question

This is motivated by this recent question of Gregory Nisbet:

For a structure $\mathcal{A}=(A;...)$ and a set $X\subseteq A$, let $\tau_X^\mathcal{A}$ be the topology on $A$ generated by sets definable in $\mathcal{A}$ with parameters from $X$. For example, $\tau^\mathcal{A}_A$ is always the discrete topology, although $\tau_\emptyset^\mathcal{A}$ need not be.

Now say that an expansion $\mathcal{B}$ of $\mathcal{A}$ is restrained iff for each $X\subseteq A$ there is some $Y\subseteq A$ with $\tau^\mathcal{B}_X=\tau^\mathcal{A}_Y$. Basically, a restrained expansion doesn't provide us with any new topological power except perhaps some "parameter juggling." I'm trying to get a sense of what the restrained expansions of a "tame" structure can look like, and the following seems a good starting point:

Is $\mathcal{R}_\mathbb{Z}=(\mathbb{R};+,\times,\mathbb{Z})$ a restrained expansion of $\mathcal{R}=(\mathbb{R};+,\times)$?

After some thought, my guess is that the answer is yes (so in particular a restrained expansion of an o-minimal structure need not be o-minimal). However, I don't immediately see how to prove this, the general obstacle being that $X$ is usually not definable in the expansion of $\mathcal{A}$ by constants naming the elements of $X$. In fact it's not immediately clear to me whether $\mathcal{R}$ has any non-o-minimal restrained expansions.

Answer 1

My intuition was way off and I missed a very easy argument.

Say that $A\subseteq\mathbb{R}$ is big iff it contains a transcendence basis, and small iff it doesn't. The point is that $\tau_\emptyset^{\mathcal{R}_\mathbb{Z}}$ is too small to be $\tau^\mathcal{R}_A$ if $A$ is big and too big to be $\tau^\mathcal{R}_A$ if $A$ is small; the first point is basically trivial, and the second point comes from considering the set $\mathbb{I}$ of irrationals (which is clearly definable in $\mathcal{R}_\mathbb{Z}$, hence open in $\tau^{\mathcal{R}_\mathbb{Z}}_\emptyset$). In detail:

Lemma 1: Suppose $A$ is uncountable. Then there is some real $\alpha\in A$ such that $\{\alpha\}$ is not open in $\tau_\emptyset^{\mathcal{R}_\mathbb{Z}}$, and so $\tau_\emptyset^{\mathcal{R}_\mathbb{Z}}\not=\tau^\mathcal{R}_A$.

Proof: Only countably many singletons can be open in $\tau^{\mathcal{R}_\mathbb{Z}}_\emptyset$, since the latter is second countable. $\quad\Box$

Lemma 2: Suppose $A$ is small. Then $\mathbb{I}$ is not open in $\tau_A^{\mathcal{R}}$, and so $\tau_\emptyset^{\mathcal{R}_\mathbb{Z}}\not=\tau_A^\mathcal{R}$.

Proof: By smallness, let $\alpha$ be a real number not in the algebraic closure of $A$. If $D\subseteq\mathbb{R}$ is definable in $\mathcal{R}$ using only parameters from $A$ and $\alpha\in D$, then by o-minimality of $\mathcal{R}$ and the transcendality assumption on $\alpha$ there must be some nontrivial open interval containing $\alpha$ which is a subset of $D$. But then $D\not\subseteq\mathbb{I}$. So $\mathbb{I}$ does not contain any of the basic opens of $\tau_A^\mathcal{R}$. $\quad\Box$

Since every big set is uncountable, we're done.

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Some final remarks:

• The above is a very "coarse" argument, and applies to a wide range of unary relations; in particular, if $A\subseteq\mathbb{R}$ is countable then in order for $\mathcal{R}_A$ to be a restrained expansion of $\mathcal{R}$ the set $A$ (and everything we can build from it) must be extremely sparse.

• On the other hand, there are countably infinite sets which yield restrained expansions. For example, the expansion $\mathcal{R}_{2^\mathbb{Z}}$ is d-minimal: every definable set is a union of an open set and finitely many discrete sets (see the bottom of page $2$ here). This means that for any set of parameters $A$ we have $$\tau_A^{\mathcal{R}_{2^\mathbb{Z}}}=\tau^\mathcal{R}_{\mathit{Def}_A(\mathcal{R}_{2^\mathbb{Z}})},$$ where $\mathit{Def}_X(\mathcal{Y})$ is the set of elements of $\mathcal{Y}$ definable using parameters from $X$.

• Finally, every o-minimal expansion is restrained. This is because definable sets in o-minimal expansions of $(\mathbb{R};<)$ are quite simple: if $\mathcal{A}=(\mathbb{R};<,...)$ is o-minimal, then for each $X\subseteq\mathbb{R}$ the sets definable in $\mathcal{A}$ with parameters from $X$ are exactly those definable in $(\mathbb{R};<)$ with parameters from the $\mathcal{A}$-definable closure of $X$ (think about endpoints).

The second and third bulletpoints above were pointed out to me by James Hanson. It would be interesting to fully determine the restrained expansions of $\mathcal{R}$ (at least by unary relations), and this does not seem to me impossible: right off the bat we can rule out "topologically complicated" expansions, so we do have some hope of getting actual results. If anyone has any thoughts on this I'd love to hear them!