4
$\begingroup$

i'm in trouble with an exercise on Kechris, Classical Descriptive Set Theory. The Theorem 22.4 shows $\Sigma_\xi^0(X)\neq\Pi_\xi^0(X)$ for each ordinal $\xi\lneq\omega_1$ and uncountable polish space $X$, using the existence of universal sets. The following exercise is: Show that if X is an uncountable polish space and $\lambda$ is a limit ordinal, then: $\bigcup_{\xi\lneq\lambda}\Sigma_\xi^0(X)\subsetneq\Delta_\lambda^0(X)$.

The inclusion is obvious. For the inequality, i would like to show that the set $A=\bigcup_{n\in\omega}A_n$, with $A_n$ taken in $\Sigma_{\xi_n}^0(X)\backslash\Pi_{\xi_n}^0(X)$ and $\xi_n\lneq\xi_{n+1}\lneq\dots\lneq\lambda$, is in $\Delta_\lambda^0(X)$ (clearly) but not in the first set. Is this a successfully way? Otherwise, what's the way?

$\endgroup$
  • $\begingroup$ I guess you mean to take the sequence $(\xi_n)_{n\in \omega}$ to be cofinal in $\lambda$? Then $A$ is in $\Sigma^0_\lambda$, but why should it be in $\Delta^0_\lambda$? That is, why is it in $\Pi^0_\lambda$? (You say this is clear, but I don't see it.) $\endgroup$ – Alex Kruckman Jan 23 '19 at 0:44
  • $\begingroup$ @NateEldredge Hmm... maybe I'm being dense here, but it seems like that observation only shows that $A$ is a union of sets in various $\Pi^0_\xi$ classes, so it's in $\Sigma^0_\lambda$. I'm asking why it's also in $\Pi^0_\lambda$. $\endgroup$ – Alex Kruckman Jan 23 '19 at 3:56
  • $\begingroup$ @AlexKruckman: Nope, it's me who's being dense. Thinking more carefully, I don't see how to prove it either. Sorry. $\endgroup$ – Nate Eldredge Jan 23 '19 at 4:07
  • $\begingroup$ The $\xi_n$ are cofinal in $\lambda$, so $A_n\in\Sigma_{\xi_n}^0\subseteq\Pi_{\xi_{n+1}}^0$ and $\xi_{n+1}<\lambda$. This was the idea for $A\in\Delta_\lambda^0$. I'm wrong? $\endgroup$ – Ajeje Jan 23 '19 at 8:40
  • $\begingroup$ @Ajeje: Yes, and $A$ is the countable union of the $A_n$, so that shows $A \in \mathbf\Sigma_\lambda^0$. To get it to be in $\mathbf\Delta_\lambda^0$, you have to also show it's in $\mathbf\Pi_\lambda^0$, and that means showing that $A^c$ can be written as a countable union of sets $B_n \in \mathbf\Sigma_{\xi_n}^0$. I don't see how you can show that. $\endgroup$ – Nate Eldredge Jan 23 '19 at 14:45
0
$\begingroup$

Your idea doesn't quite work as is; you will have to be more careful in choosing the $A_n$. For instance, as written, it could happen that $\xi_1 = 1$ and $A_1$ is an open ball, and all the remaining $A_n$ are contained inside $A_1$. Then $A$ is just $A_1$, which is certainly in $\bigcup_{\xi < \lambda} \mathbf\Sigma_\xi^0(X)$.

Also, I am unclear how you prove that $A \in \mathbf\Pi_\lambda^0(X)$ with your construction.

Here is a modification that will make it easier: "separate" the sets $A_n$. That is, choose a countable family $U_n$ of disjoint uncountable open subsets of $X$ (exercise: show that such a family exists), and let $A_n \in \mathbf\Pi_{\xi_n}^0(U_n) \setminus \mathbf\Sigma_{\xi_n}^0(U_n)$. Then letting $A = \bigcup_n A_n$ as before, it is not too hard to show that $A \in \mathbf{\Delta}_\lambda^0(X)$. (Hint: $A^c = (\bigcup_n U_n)^c \cup \bigcup_n (A_n^c \cap U_n)$). However, if $A \in \mathbf{\Sigma}_{\xi_n}^0(X)$ for some $n$, then $A_n = A \cap U_n \in \mathbf\Sigma_{\xi_n}^0(U_n)$ which is a contradiction.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, my idea doesn't work, i was thinking of something similar to your solution. Is there an easier way to solve the problem? When I first read it, i thought it was an easy consequence of the previous theorem $\endgroup$ – Ajeje Jan 24 '19 at 23:21
  • $\begingroup$ This seems relatively easy to me, and I don't see anything particularly easier. You need to construct a set $A$ that has at least as much "complexity" as each of the sets that make it up. If they can interact with each other, it can all fall apart, and this sort of separation seems to me like the most natural way to prevent that. $\endgroup$ – Nate Eldredge Jan 24 '19 at 23:33
  • $\begingroup$ @Ajeje: If you want to be a little more concrete, you can assume without loss of generality that $X$ is Cantor space, and it is clear that $X$ contains countably many disjoint (clopen) copies of Cantor space. $\endgroup$ – Nate Eldredge Jan 24 '19 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.