Borel Hierarchy 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?
 A: 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.
