Q: Is every formula of $L_\Omega$ equivalent to a formula of $L_1$ modulo $T_1$?
The notation comes from the following question: Is the following theory countably axiomatizable?
Edit: I mean $T_\Omega$.
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$\begingroup$ Okay, it looks like $T_0 \not= T_1$ and $T_1 = T_2 = T_\Omega$. $\endgroup$– David PokornyCommented Dec 14, 2016 at 2:44
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$\begingroup$ In what sense is $T_0\not=T_1$, if $T_1=T_2$? (Note that in my original answer where I said that the hierarchy collapses immediately in a sense, my indexing was wrong: $T_0$ is already "essentially the same as" $T_1$, in a precise sense; and I've edited that.) $\endgroup$– Noah SchweberCommented Dec 14, 2016 at 4:48
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$\begingroup$ $\exists v\,\forall x\, x\in v$ is not true in all models of $T_0$ but it is in all models of $T_1$. $\endgroup$– David PokornyCommented Dec 14, 2016 at 15:45
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2 Answers
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Noah is correct that over the theory $T_1$, we don't know about the new constants. But meanwhile, over the final theory $T_\Omega$, or what I would call $T_{\omega_1}$, every formula in the extended language is equivalent to a formula in the language of (class) set theory itself. You have added constants $c_\phi$ and axioms $\forall x\ (x\in c_\phi\iff \phi(x))$, where $\phi$ uses only earlier-mentioned constants. So by induction, any instance of $x\in c_\phi$ that appears in a formula can be systematically eliminated by this means, and so in fact we don't need any of these new constants.
So over your final theory, every formula is equivalant to an assertion in the second-order language of set theory.
answered Dec 14, 2016 at 2:00
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$\begingroup$ +1. Note for the OP that we don't have to wait until $T_\Omega$ to see this for a given sentence - each sentence at level $\alpha$ is immediately seen to be "redundant" by $T_\alpha$. $\endgroup$ Commented Dec 14, 2016 at 2:02
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$\begingroup$ Yes. But my main point is that they are all redundant all the way back to the original language. So the constants are not needed for any purpose---we could instead just assert the instances of comprehension. $\endgroup$ Commented Dec 14, 2016 at 2:05
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$\begingroup$ Indeed. I just wanted to point out to the OP that nothing "special" happens at $T_\Omega$. (By the way, in my original answer I meant to say that, but I got my indexing messed up - I fixed it now.) $\endgroup$ Commented Dec 14, 2016 at 2:09
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No, they are not. There are formulas in $L_2$, for example, involving constant symbols in $L_2$ but not in $L_1$. $T_1$ has nothing to say about those constant symbols, so no formula involving such a constant symbol (in a nontrivial manner) can be $T_1$-equivalent to an $L_1$ formula.
The sense in which $T_2$ is "no more than" $T_1$ is interpretability (and conservativity) - any model of $T_1$ interprets an expansion of itself which is a model of $T_2$.
answered Dec 14, 2016 at 1:56