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I'm sure this is just my google-fu failing me, but: what are sufficient, non-overkill large cardinal axioms which guarantee "Every (boldface) $\Pi^1_n$ set of (real codes for) countable ordinals contains or is disjoint from a club subset of $\omega_1$"? (I asked this question on math.stackexchange a couple weeks ago https://math.stackexchange.com/questions/1539202/clubbiness-of-pi1-n-sets, and received some attention but no answer.)

To clarify, I'm asking about the strength over ZFC.

Here's a very silly upper bound: suppose $L(\mathbb{R})$ is a model of AD, and moreover every $\Pi^1_n$-sentence with real parameters is absolute between $L(\mathbb{R})$ and $V$ (actually, I think this is already a consequence of "$L(\mathbb{R})\models AD$," but I'm not sure). Then let $A\in V$ be a $\Pi^1_n$-set of countable ordinals, via the formula (with real parameters) $\varphi$. By the absoluteness assumption, $\varphi^{L(\mathbb{R})}=A$, so $A\in L(\mathbb{R})$. And since $L(\mathbb{R})\models AD$, $L(\mathbb{R})$ thinks $A$ contains or is disjoint from a club. But inner models compute club-ness correctly, so we're done.

This seems massively overkill to me, though - what is the right bound?

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    $\begingroup$ Any $\Pi^1_n$ sentence with real parameters is absolute between $L(\mathbb R)$ and $V$ because the quantifiers in the sentence range only over reals, and $L(\mathbb R)$ contains all of the reals. $\endgroup$
    – Andreas Blass
    Commented Dec 3, 2015 at 11:27
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    $\begingroup$ In view of my previous comment, it suffices to have enough determinacy to carry out, for $\Pi^1_n$ formulas, Solovay's proof that AD makes the club filter ultra. So you could check the complexity of the game that Solovay used, under the additional assumption that the target set is $\Pi^1_n$; presumably it will be be at worst a little higher than $\Pi^1_n$ in the projective hierarchy. So something in the neighborhood of $n$ Woodin cardinals would suffice. (I have no guess at the moment about whether significantly smaller cardinals will suffice; that's why this comment isn't an answer.) $\endgroup$
    – Andreas Blass
    Commented Dec 3, 2015 at 11:32
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    $\begingroup$ Expanding Andreas' post, if the target set is $\bf\Pi^1_n$ with $n\geq 2$, then $\bf\Sigma^1_{n+1}$ determinacy suffices. $\endgroup$
    – Cody Dance
    Commented Dec 3, 2015 at 22:10
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    $\begingroup$ What do we know about the strength of having a single club $C\subseteq\omega_1$ that decides (on a tail) every projectively definable set of countable ordinals? $\endgroup$
    – Joel David Hamkins
    Commented May 6, 2016 at 1:04

2 Answers 2

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An upper bound is an ineffable cardinal, which is weaker than $0^\#$. It is in the last few paragraphs of Harrington's paper "Analytic determinacy and $0^\#$".

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Meanwhile, let me provide a lower bound. Your situation is not provable in ZFC; it is false in $L$, and it is incompatible with having a projective well-ordering of the reals. Furthermore, it implies that $\omega_1$ is inaccessible to reals.

Theorem. If there is a projectively definable $\omega_1$-sequence of distinct reals (for example, if there is a projectively definable well-ordering of the reals), then there is a projectively definable set of countable ordinals $S\subseteq\omega_1$, of the same complexity as the sequence, that is both stationary and co-stationary. (Thus, it neither contains nor omits a club.)

Proof. Assume that there is a projectively definable $\omega_1$-sequence of distinct reals $\langle z_\alpha\mid\alpha<\omega_1\rangle$, where $z_\alpha\subseteq\omega$. This situation occurs, for example, under $V=L$ or indeed, if there is a projectively definable well-ordering of the reals, since we can let $z_\alpha$ be the $\alpha^{\rm th}$ real in that well-ordering.

Let $S_n=\{\alpha\mid n\in z_\alpha\}$. This is a projectively definable set of ordinals, of the same complexity as the sequence. If it is not both stationary and co-stationary, then there is a club $C_n$ that contains or omits $S_n$. Let $C=\bigcap_n C_n$, which is a club subset of $\omega_1$. Note that if $\alpha\in C$, then $n\in z_\alpha$ is determined by whether $C_n$ contained or omitted $S_n$. Thus, the $z_\alpha$ for $\alpha\in C$ must all agree, contradicting our assumption that the reals were distinct. So one of the sets $S_n$ must be both stationary and co-stationary. QED

Corollary. If every projectively definable set of countable ordinals contains or omits a club, then $\omega_1$ is inaccessible to reals.

Proof. If $\omega_1$ is not inaccessible to reals, then $\omega_1=\omega_1^{L[x]}$ for some real $x$. In this case, there is an projectively $x$-definable $\omega_1$-sequence of distinct reals, and so there will be a projectively $x$-definable set $S\subseteq\omega_1$ that is both stationary and co-stationary.QED

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