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I have been studying basics of descriptive set theory lately. In the lecture notes I follow (sadly, the notes are written in Czech), there is the following definition:

Let X be a topological space. We say that a system $(G_\beta)_{\beta<\alpha}$ of open sets is a HK (Hausdorff-Kuratowski) scheme if $\alpha < \omega_1$, $G_\gamma\subseteq G_\beta$ whenever $\gamma\leq\beta < \alpha$, $G_\lambda =\bigcup\limits_{\beta < \lambda}G_\beta$ whenever $\lambda < \alpha$ is a limit ordinal.

My question is whether it's true that $X = \bigcup\limits_{\beta < \alpha}G_\beta$. I observed that for every $x\in\bigcup\limits_{\beta < \alpha}G_\beta$ there exists the smallest ordinal $\beta < \alpha$ such that $x\in G_\beta$ and this ordinal can't be a limit one. However, the notes claim that it follows that a HK system does cover the whole space. There is no restriction on $X$ in the definition but we may assume that $X$ is a polish space if needed.

My knowledge of set theory is pretty shallow (not speaking of intuition) so I might be overlooking something trivial, but I simply can't see why this should be true.

Thank you for any help!

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  • $\begingroup$ I can't see it either. What's to prevent $G_\beta=\emptyset$ for all $\beta$? $\endgroup$ – bof Jun 19 '16 at 18:20
  • $\begingroup$ I suspect that it’s supposed to be a strictly increasing sequence, so that $G_\gamma\subsetneqq G_\beta$ whenever $\gamma<\beta<\alpha$. This still doesn’t ensure that it covers $X$, however. Do the notes then go on to deal with something like $\bigcup_{\beta<\alpha,\beta\text{ even}}(G_{\beta+1}\setminus G_\beta)$? $\endgroup$ – Brian M. Scott Jun 19 '16 at 19:01
  • $\begingroup$ Good point - nothing prevets $G_\beta=\emptyset$ for all $\beta$. $\endgroup$ – user1321324 Jun 19 '16 at 19:18
  • $\begingroup$ Exactly, It's used to characterize $\mathbf{\Delta}_2^0$ sets in polish spaces. In that contruction, the system indeed covers the whole space. However, when proving that if there is a HK system such that... then $H\in\mathbf{\Delta}_2^0$. The argument goes like - $H$ is $\mathbf{\Sigma}_2^0$ as $H$ is union of G_{\beta +1}\setminus G_{\beta} where $\beta\in I$ ($I$ is some index subset). $\endgroup$ – user1321324 Jun 19 '16 at 19:24
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It seems that $X=\bigcup_{\beta<\alpha}G_\beta$ does not follow. Indeed, we can replace $X$ with $X\sqcup X$ and keep the $G_\beta$.

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  • $\begingroup$ Indeed. It seems the argument is wrong. $\endgroup$ – user1321324 Jun 19 '16 at 19:26
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Well, the claim must be false. Fortunately, I've gone carefully through the notes and the proofs that use that claim can be fixed. Thank you for pointing me out that the claim can't be true.

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