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For a countable admissible ordinal $\alpha$, let $\mathcal{L}_\alpha=\mathcal{L}_{\infty,\omega}\cap L_\alpha$ and let $\equiv_\alpha$ be the corresponding elementary equivalence relation. Say that such an $\alpha$ is locally Fraissean iff there is an $\mathcal{L}_\alpha$-sentence $\eta$ in the language $\{R,S,A,B\}$, with $R,S$ binary relation symbols and $A,B$ unary relation symbols, such that for every pair of $\{R\}$-structures $\mathcal{X},\mathcal{Y}\in L_\alpha$ we have $$\mathcal{X}\equiv_\alpha\mathcal{Y}\quad\iff\quad\exists\mathcal{M}\in L_\alpha[\mathcal{M}\models\eta\wedge (A^\mathcal{M}\cong\mathcal{X}\wedge B^\mathcal{M}\cong\mathcal{Y})^{L_\alpha}]$$ (the $L_\alpha$-superscript meaning that relevant isomorphisms have to exist within $L_\alpha$ itself).

For example, if $\equiv_\alpha$ coincides with $\cong$ on $\{R\}$-structures in $L_\alpha$, then $\alpha$ is locally Fraissean; in particular, if $L_\alpha\equiv L_{\omega_1}$ then $\alpha$ is locally Fraissean. But this seems like overkill:

What is the smallest locally Fraissean ordinal?

It's already not clear to me whether $\omega_1^{CK}$ is locally Fraissean - in particular, the argument Farmer S. gave that $\mathcal{L}_{\omega_1,\omega}$ doesn't satisfy the "global" version of the property above doesn't seem to trivially effectivize (although I could be missing something).

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  • $\begingroup$ About "$\mathcal{L}_\alpha=\mathcal{L}_{\infty,\omega}\cap L_\alpha$", what is $L_\alpha$ here, if it's the $\alpha$th level of the constructible hierarchy then how do we intersect it with a language? $\endgroup$
    – C7X
    Commented Apr 15, 2022 at 17:43
  • $\begingroup$ @C7X Remember that everything, in the ZFC context, is a set; this includes $\mathcal{L}_{\infty,\omega}$-formulas. You can find more details in Barwise's book Admissible sets and structures (this is a crucial notion throughout, e.g. in his completeness/compactness theorems). $\endgroup$
    – Noah Schweber
    Commented Apr 15, 2022 at 18:09
  • $\begingroup$ @C7X (Forgot to finish my comment!) So $\mathcal{L}_{\infty,\omega}\cap L_\alpha$ (with $L_\alpha$ being the $\alpha$th level of the constructible hierarchy as usual) does parse, although it looks weird at first. A big component of Barwise's research on infinitary logic is that, even though $\mathcal{L}_{\omega_1,\omega}$ itself is a bit wild, its "countable fragments" (the $\mathcal{L}_\alpha$s) are surprisingly tame. (Incidentally, there is a difference between $\mathcal{L}_{\infty,\omega}\cap L_\alpha$ and $\mathcal{L}_{\omega_1,\omega}\cap L_\alpha$ even when $\alpha$ is countable: (contd) $\endgroup$
    – Noah Schweber
    Commented Apr 16, 2022 at 4:57
  • $\begingroup$ formulas in the latter set are "explicitly countable," while formulas in the former are not. $L_\alpha$ may not be able to "see" that $\varphi\in\mathcal{L}_\alpha$ is in fact equivalent to an $\mathcal{L}_{\omega_1,\omega}$-formula. This usually isn't a big deal, but sometimes matters. See here.) $\endgroup$
    – Noah Schweber
    Commented Apr 16, 2022 at 5:00
  • $\begingroup$ I think we can always use $M$ such that $A^M=X$ and $B^M=Y$. Let $\mathcal{L}_\alpha(Z)$ be the set of $\mathcal{L}_\alpha$-formulae in the language $Z$, and for let $\equiv_\alpha$ be elementary equivalence for formulae in $\mathcal{L}_\alpha(\{R\})$. Then that means $\alpha$ is locally Fraissean iff there's a formula $\eta\in\mathcal{L}_\alpha(\{R,S,A,B\})$ such that $(M\models\eta)\iff(A^M\equiv_\alpha B^M)$ $\endgroup$
    – Racheline
    Commented Apr 17, 2022 at 7:32

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