Take the language $\mathcal L(=,\in)_{\omega_1,\omega_1}$, if we restrict infinite quantification strings, i.e. of the forms $“(\forall v_i)_{i \in \omega}"; “(\exists v_i)_{i \in \omega}"$, to bounded forms, that is $“\forall v_0 \in x, \forall v_1 \in x,…"$ and $“\exists v_0 \in x , \exists v_1 \in x,…"$ abbreviated as $“(\forall v_i \in x)_{i \in \omega}"; “(\exists v_i \in x)_{i \in \omega}"$ respectively, or even if we further demand the bound to be a definable constant or the value of a function. Call the resulting language as "bounded–$\mathcal L(=,\in)_{\omega_1, \omega_1}$", and denote it by $\mathcal L(=,\in)_{\omega_1, [\omega_1]}$
How would $\mathcal L(=,\in)_{\omega_1, [\omega_1]}$ differ from $\mathcal L(=,\in)_{\omega_1, \omega_1}$ as regards basic properties of logic like Completeness, Compactness, etc..
As a clear example, if to $\mathcal L(=,\in)_{\omega_1, [\omega_1]}$ we add all axioms of $\sf ZFC$ written, as usual, in the fragment $\mathcal L(=,\in)_{\omega,\omega}$ and suppose we add an axiom of Foundation but in a bounded form, that is:
$\textbf{Foundation: } \\\forall \alpha \, (\forall v_i \in V_\alpha)_{i \in \omega} \, \exists x: \bigvee_{i \in \omega} (x=v_i) \land \bigwedge_{i \in \omega} (v_i \not \in x)$
Would that theory be different from theory written in $\mathcal L(=,\in)_{\omega_1,\omega_1}$, axiomatized by $\sf ZFC$ axioms in $\mathcal L(=,\in)_{\omega,\omega}$ , and Foundation written as above but unbounded.
Best Answer
Your new formulation of foundation is equivalent to the corresponding unbounded form. The reason is that every failing instance of the unbounded formulation will lead to a failing instance of the bounded formulation, since if there is a countable collection of $v_i$ with no $\in$-minimal element, then by the usual ZFC axioms it follows that $v_0$ is in some $V_\alpha$, and so the $v_i$ that appear within this $V_\alpha$ will fall under the bounded form, and an $\in$-minimal for that part of the family will be $\in$-minimal for the whole family since $V_\alpha$ is transitive.