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Problem below (kuratowski) is taken from Kechris's Classical Descriptive Set Theory, page $173.$


Definition: Given any sequence of sets $(A_n),$ $A_n\subseteq X,$ let $$\overline{\lim_n} \, A_n = \bigcap_n\bigcup_{m\geq n} A_m = \{x:x\text{ belongs to infinitely many } A_n\},$$ $$\underline{\lim}_n \, A_n = \bigcup_n\bigcap_{m\geq n} A_m = \{x:x\text{ belongs to all but finitely many } A_n\},$$ It is clear that $\underline{\lim}_n \, A_n \subseteq\overline{\lim_n} \, A_n.$ If they are equal, let $$\lim_n A_n= \overline{\lim_n} \, A_n = \underline{\lim}_n \, A_n .$$

Assume now that $X$ is metrizable, so that every closed set is a $G_\delta$ set. Let $\omega_1$ be the first uncountable ordinal, and for $1\leq \xi<\omega_1,$ define by transfinite recursion the classes $\Sigma_\xi^0,\Pi_\xi^0$ of subsets of $X$ as follows:

$$\Sigma_1^0=\{U\subseteq X: U \text{ is open}\},$$ $$\Pi_\xi^0 =\sim\Sigma_\xi^0,$$ $$\Sigma_\xi^0=\bigg\{ \bigcup_n A_n: A_n\in \Pi_{\xi_n}^0, \xi_n<\xi, n\in\mathbb{N} \bigg\}, \text{ if } \xi>1,$$ where $\sim$ refers to set complement. In addition, let $$\Delta_\xi^0 = \Sigma_\xi^0\cap \Pi_\xi^0.$$

For example, $\Sigma_1^0$ is a topology on $X$ where $\Pi_1^0$ is the collection of closed sets of $X.$ $\Sigma_2^0$ is the collection of all $F_\sigma$ sets while $\Pi_2^0$ is the collection of all $G_\delta$ sets. Note that $\Delta_1^0$ is the collection of all closed and open sets while $\Delta_2^0$ is the collection of all $F_\sigma$ and $G_\delta$ sets.


Question (Kuratowski) : Show that for $\xi>1,$ $$A \text{ is } \Delta_{\xi+1}^0 \iff A=\lim_nA_n\text{ for some sequence } (A_n) \text{ with } A_n\in\Delta_\xi^0.$$

For $(\Leftarrow)$ direction, by definition, $$A = \bigcap_n\bigcup_{m\geq n} A_m = \bigcup_n\bigcap_{m\geq n} A_m.$$ If $A_m\in\Delta_\xi^0 = \Sigma_\xi^0\cap \Pi_\xi^0,$ then $$\bigcup_{m\geq n}A_m\in \Sigma_{\xi}^0\quad \text{and}\quad \bigcap_{m\geq n}A_m\in\Pi_\xi^0.$$ Then by definition of $\Sigma_{\xi+1}^0$ and $\Pi_{\xi+1}^0,$ we have $$ \bigcap_n\bigcup_{m\geq n} A_m\in \Pi_{\xi+1}^0\quad\text{and}\quad \bigcup_n\bigcap_{m\geq n} A_m\in\Sigma_{\xi+1}^0.$$

I have no idea on how to start $(\Rightarrow)$ direction.

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  • $\begingroup$ You know, the lim inf and lim sup are very common and pretty easy to recognize, even when you are using a less common symbol for them. But the $\Delta, \sum, \prod$ notations you are using are far less common. I suggest explaining them instead of lim inf and lim sup. $\endgroup$ – Paul Sinclair May 19 '18 at 16:34
  • $\begingroup$ @PaulSinclair: Thanks for mentioning the unusual symbols. I have included their definitions in my question. $\endgroup$ – Idonknow May 19 '18 at 16:43
  • $\begingroup$ @Paul Sinclair Those notations are pretty common as well in descriptive set theory and in logic. $\endgroup$ – Régis Jun 6 '18 at 21:06
  • $\begingroup$ @Régis Do you know how to answer my questions? $\endgroup$ – Idonknow Jun 8 '18 at 6:26
  • $\begingroup$ I used to but I forgot. $\endgroup$ – Régis Jun 8 '18 at 8:03
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Note: This only holds if $\xi$ is successor ordinal.

To prove ($\Leftarrow$), first since $A \in \Delta_{\xi+1}^0, A\in\Sigma_{\xi+1}^0$ and $A\in\Pi_{\xi+1}^0$. So $$ A=\bigcup_nB_n=\bigcap_nC_n, \quad B_n\in \Delta_{\xi}^0,\: C_n\in\Delta_{\xi}^0 $$ Where $B_n\subset B_{n+1}\subset \cdots\subset A\subset \cdots\subset C_{n+1}\subset C_{n}$. Thus there is $A_n\in \Delta_{\xi}^0$ that $B_n\subset A_n\subset C_n$. Clearly $$ \overline{\lim_n} \, A_n = \bigcap_n\bigcup_{m\geq n} A_m \subset \bigcap_n\bigcup_{m\geq n} C_m =\bigcap_n C_n=A $$ And $$ \underline{\lim}_n \, A_n = \bigcup_n\bigcap_{m\geq n} A_m \supset \bigcup_n\bigcap_{m\geq n} B_m=\bigcup_n B_n =A $$ Thus $A=\lim_{n}A_n$.

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