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Originally asked and bountied at MSE without success:

Say that a ring $R$ is spatial iff there is some topological space $\mathcal{X}$ such that $R\cong C(\mathcal{X})$, where $C(\mathcal{X})$ is the ring of continuous functions from $\mathcal{X}$ to $\mathbb{R}$.

My question is:

Is spatiality definable in $\mathsf{SOL}_{\infty,\infty}$?

Here $\mathsf{SOL}_{\infty,\infty}$ is the "large" infinitary version of second-order logic (with the standard semantics!), in which arbitrary set-sized Boolean combinations and homogeneous blocks of quantifiers are allowed. This is an incredibly powerful logic - already in $\mathsf{FOL}_{\infty,\infty}$ (defined here), each structure is characterized by a single sentence - but it still has limitations. The characterizations of spatiality I'm aware of (see Eric Wofsey's answer to the above-linked question) seem to require more than $\mathsf{SOL}_{\infty,\infty}$, but that logic is strong enough that I have no idea how to begin attacking a negative answer.

I've focused on $\mathsf{SOL}_{\infty,\infty}$ since - for me at least - it seems a plausible upper bound on what could count as an "algebraic" characterization of something.

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  • $\begingroup$ I don't quite see why the Gelfand duality style characterization cannot be formulated in second order logic. Could you clarify what part of the characterization doesn't work ? working with complex number for simplicity, I don't see why "there exists a norm making R a C* algebra" isn't working : Completeness for a fixed norm is definable in infinitary second order logic, the rest of the C* algebra axiom are just equation and "there exist a norm" can be phrased in terms of existence of a "unit ball" with certain property... What point am I messing ? $\endgroup$ Commented Jun 2, 2022 at 2:20
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    $\begingroup$ Oh. The point I am missing is that you want a characterization of rings of functions, and not of bounded functions. Sorry. got it. $\endgroup$ Commented Jun 2, 2022 at 2:26

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