Prove $C$ and $S$ are $\Sigma^1_1$ and $\Pi^1_1$ respectively.

Question

Let $({}^{\omega_1}\omega_1,\tau)$ be a topological space where $\tau$ is generated by $\{N(f,\alpha)\mid f\in{}^{\omega_1}\omega_1, \alpha<\omega_1\}$ and $N(f,\alpha)=\{g:\omega_1\to\omega_1\mid g(\beta)=f(\beta)$ for $\beta<\alpha\}$. We say that $A\subset {}^{\omega_1}\omega_1$ is $\Pi^1_1$ if there is $B\subset{}^{\omega_1}\omega_1\times{}^{\omega_1}\omega_1$ open in the product topology with $$ f\in A\Longleftrightarrow(f,g)\in B, \text{ for every }g. $$ $A$ is $\Sigma^1_1$ if it is a complement of a $\Pi_1^1$. Let $$ C=\{\chi_A\mid A \text{ contains a club}\}, \ \ \ \ \ \ \ \ S=\{\chi_A\mid A\text{ is stationary}\}. $$ Prove $C$ is $\Sigma_1^1$ and $S$ is $\Pi_1^1$. (Here $\chi_A$ is the characteristic function of $A$, i.e. $\chi_A(a)=0$ for $a\in A$ and $1$ otherwise.)

Answer 1

For simplicity I'll think about $2^{\omega_1}$ and $2^{\omega_1}\times\omega_1^{\omega_1}$; it's a good exercise to check that if $A\subseteq\omega_1^{\omega_1}$ is $\Pi^1_1$ (resp. $\Sigma^1_1$) and $B\subseteq\omega_1^{\omega_1}$ is closed then $A\cap B$ is $\Pi_1^1$ (resp. $\Sigma^1_1$), so this is a benign modification.

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Here's an argument that $S$ is $\Pi^1_1$; the proof that $C$ is $\Sigma^1_1$ is similar.

Say that $g\in\omega_1^{\omega_1}$ enumerates a club iff there is some club $C$ such that $g(\alpha)$ is the $\alpha$th element of $C$ (enumerating $C$ in increasing order). We want to compute the complexity of the relation

If $g$ enumerates a club, then for some $\alpha\in ran(g)$ we have $f(\alpha)=0$.

(Note to readers that this isn't a typo; the OP's text is apparently inverting the usual definition of "characteristic function." Of course this doesn't actually make a difference.)

This can be rephrased as

Either $g$ does not enumerate a club, or $f(g(\beta))=0$ for some $\beta\in\omega_1$.

The second disjunct is an open condition on the pair $(f,g)\in 2^{\omega_1}\times {\omega_1}^{\omega_1}$, so we just need to show that the former is as well; that is, we need to show that if $g$ does not enumerate a club, this is detected at some countable stage.

This follows from the following "local" characterization of the property:

A function $g\in\omega_1^{\omega_1}$ enumerates a club iff $g$ is increasing and for all nonzero limit $\alpha<\omega_1$ we have $g(\alpha)=\sup_{\beta<\alpha}g(\beta)$. These are each closed properties, so we're done.