# Does this system of open sets have to cover the whole space?

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!

• I can't see it either. What's to prevent $G_\beta=\emptyset$ for all $\beta$? – bof Jun 19 '16 at 18:20
• 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 $\bigcup_{\beta<\alpha,\beta\text{ even}}(G_{\beta+1}\setminus G_\beta)$? – Brian M. Scott Jun 19 '16 at 19:01
• Good point - nothing prevets $G_\beta=\emptyset$ for all $\beta$. – user1321324 Jun 19 '16 at 19:18
• 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). – user1321324 Jun 19 '16 at 19:24

It seems that $X=\bigcup_{\beta<\alpha}G_\beta$ does not follow. Indeed, we can replace $X$ with $X\sqcup X$ and keep the $G_\beta$.