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10 votes
1 answer
255 views

How much of second-order arithmetic do you need for $\mathbf{\Sigma}^1_1$-determinacy to give you countable transitive models of $\mathsf{ZFC}$?

This is in some sense a follow-up to this question. The answer there says that over $\mathsf{Z}_2$ (second-order arithmetic), (boldface) $\mathbf{\Sigma}^1_1$-determinacy is enough to entail the ...
James E Hanson's user avatar
7 votes
1 answer
321 views

Strength of Borel determinacy

In this blog post by Gowers on Borel determinacy, Andres Caicedo says the following in a comment (slightly rephrased). Let $\mathsf{ZFC^-}$ be $\mathsf{ZFC}$ without power set and $\mathsf{ZC^-}$ be $...
new account's user avatar
13 votes
1 answer
524 views

How much determinacy do you need for second order arithmetic to be as strong as ZFC?

From Wikipedia (I couldn't find the original source): $\text{ZFC} + \{\text{there are $n$ Woodin cardinals: $n$ is a natural number}\}$ is conservative over $\text{Z}_2$ with projective determinacy. ...
Christopher King's user avatar
6 votes
0 answers
117 views

Reverse mathematics of Banach-Mazur games

Given $\mathcal{A}\subseteq\omega^\omega$, the Banach-Mazur game with payoff set $\mathcal{A}$ consists of players $1$ and $2$ alternately playing nonempty finite strings of naturals with player $1$ ...
Noah Schweber's user avatar