Does higher order arithmetic interpret the axiom of choice?

2012-08-03T02:55:58Z

By second order arithmetic I mean the axiomatic theory $Z_2$, that is Peano arithmetic extended by second order variables with the full comprehension axiom, and not defined semantically using power set in ZF. By third order arithmetic I mean that extended by third order variables and the comprehension axiom. And so on. Does each of these have an inner model which also satisfies the axiom of choice in each order, using constructibility? If not, do such inner models exist if we also extend induction to a higher order axiom? Is there a good reference on it?

Answer by Carl Mummert for Does higher order arithmetic interpret the axiom of choice?

2012-08-03T03:49:03Z

There is quite a bit of this in Simpson's book Subsystems of Second Order Arithmetic in the specific context of second-order arithmetic. Here are three relevant results:

Corollary VII.5.11 (conservation theorems). Let $T_0$ be any one of the $L_2$-theories $\Pi^1_\infty\text{-CA}_0$, $\Pi^1_{k+1}\text{-CA}_0$, $\Delta^1_{k+2}\text{-CA}_0$, $0 ≤ k < \infty$. Let $\phi$ be any $\Pi^1_4$ sentence. Suppose that $\phi$ is provable from $T_0$ plus $\exists X \forall Y (Y ∈ L(X ))$. Then $\phi$ is provable from $T_0$ alone.

Here $\Pi^1_\infty\text{-CA}_0$ has the full comprehension scheme for second order arithmetic, and hence also the full induction scheme.

Theorem VII.6.16 ($\Sigma^1_{k+3}$ choice schemes). The following is provable in $\text{ATR}_0$. Assume $\exists X \forall Y (Y ∈ L(X ))$. Then:

$\Sigma^1_{k+3}\text{-AC}_0$ is equivalent to $\Delta^1_{k+3}\text{-CA}_0$ .

$\Sigma^1_{k+3}\text{-DC}_0$ is equivalent to $\Delta^1_{k+3}\text{-CA}_0$ plus $\Sigma^1_{k+3}\text{-IND}$ .

Strong $\Sigma^1_{k+3}\text{-DC}_0$ is equivalent to $\Pi^1_{k+3}\text{-CA}_0$ .

$\Sigma^1_\infty \text{-DC}_0$ ( $=\bigcup_{k < \omega} \Sigma^1_k\text{-DC}_0$ ) is equivalent to $\Pi^1_\infty\text{-CA}_0$ .

and

Corollary IX.4.12 (conservation theorem). For all $k <\omega$, $\Sigma^1_{k+3}\text{-AC}_0$ (hence also $\Delta^1_{k+3}\text{-AC}_0$ ) is conservative over $\Pi^1_{k+2}\text{-CA}_0$ for $\Pi^1_4$ sentences.

Does higher order arithmetic interpret the axiom of choice? - MathOverflow

Does higher order arithmetic interpret the axiom of choice?

Answer by Carl Mummert for Does higher order arithmetic interpret the axiom of choice?