"Clubiness" of projective sets of ordinals 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?
 A: 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 $0^\#$".
A: 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
