# Necessity of omega-models in second order arithmetic

Are there examples of independence results over subsystems of true second order arithmetic that cannot be established using omega-models? To rule out trivial examples, let us assume that the base theory extends true first order arithemtic. A non example of such a statement would be Ramsey theorem for pairs since there is a computable coloring of pairs of integers into two colors without a computable (even from the halting problem, Jockusch) infinite homogeneous set.

An example of a use of non-omega models appears here - Also, see the introduction and question 6.1 in this paper.

I am not an expert in reverse mathematics so please feel free to offer any interesting known facts known to you here.

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– Andrés E. Caicedo May 18 '14 at 0:55
@Ashutosh: The base theory in your example from Chong–Slaman–Yang does not seem to extend "true first order arithmetic", unless you mean something different from what I understand. Could you clarify a bit on what you mean by "true first order arithmetic", please? – Lawrence Wong May 18 '14 at 9:05
I am not aware of any reverse math research in which the base theory includes true first-order arithmetic. But there is another point: the induction schemes in reverse mathematics all include set parameters. This is the case even for schemes like B$\Sigma^0_2$. Formulas of true first order arithmetic don't include set parameters. – Carl Mummert May 18 '14 at 10:54

For example, Hirst proved that the pigeonhole principle $\mathsf{RT}^1_{<\infty}$ is equivalent to $B\Sigma^0_2$, and thus is not provable in $\mathsf{RCA}_0$. But $B\Sigma^0_2$ holds in every $\omega$-model.
Similarly, there is a variation of $\mathsf{ACA}_0$, denoted $\mathsf{ACA}'_0$, which has an axiom asying that for all $n$, $0^{(n)}$ exists. In contrast, $\mathsf{ACA}_0$ only proves that $0^{(n)}$ exists for standard $n$. So every $\omega$-model of $\mathsf{ACA}_0$ is a model of $\mathsf{ACA}'_0$.
For an example with non-arithmetical induction: $\mathsf{ACA}_0$ plus $\Sigma^1_1$ induction proves the consistency of $\mathsf{ACA}_0$, so $\mathsf{ACA}_0$ does not prove $\Sigma^1_1$ induction.
Thanks a lot! Could you provide some references here? And maybe explain what $RT^{1}_{<\infty}$ stands for? Sorry for my ignornace. – Ashutosh May 18 '14 at 1:28