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I'm a bit confused about the basics of the borel hierarchy. My question is this: if i have a closed set P and I make the set $\forall^\omega P$, is that $\Pi^0_3$? Similarly, if I have an open set P and i make the set $\exists^\omega$ P, is that $\Sigma^0_3$?

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1 Answer 1

This notations looks like it is from Moschovakis's book, Descriptive Set Theory. What you wrote is incorrect. If $P$ is closed, $(\forall^\omega)(P)$ is still closed and hence $\Pi_1^0$. Same thing for the other one.

To understand, $\forall^\omega P$, where $P$ is closed, means that $P$ is an closed subset of $X \times \omega$ where $X$ is a topological space (Polish space) and $\omega$ is the discrete space. So $\forall^\omega P = \{x \in X : (\forall n \in \omega)((x,n) \in P)\} = \bigcap_{n \in \omega} \pi^{-1}(n)$, where $\pi^{-1}$ is the continuous projection function onto the second coordinate. Since $\omega$ is discrete, $\{n\}$ is a closed set. Therefore, since $\pi$ is continuous, $\pi^{-1}(n)$ are all closed sets. Arbitrary intersection of closed sets are closed. Hence $(\forall^\omega)(P)$ is closed. By definition it is still $\Pi_1^0$. Similarly for the other one.

Another way of looking at this is that $(\forall^\omega)(P)$ is the countable intersection of the closed sets, $\pi^{-1}(n)$, which is still closed.

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