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

Question

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!

Answer 1

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

Answer 2

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.