Consider the open ordinal space $[0,\Omega)$, where $\Omega$ is the first uncountable ordinal. Can I say that every subset of $[0,\Omega)$ is $G_\delta$? If yes, does this imply that $[0,\Omega)$ is perfectly normal?
Thank you!
so, which version is the right one..
Note that this link, that @martini provided in the comments, shows that $[0,\Omega)$ is not perfectly normal (the countable limit ordinals are a closed subset of $[0, \Omega)$ that is not a $G_\delta$, as shown there by using the pressing down lemma.
The remarks from Counterexamples in Topology, e.g. item 5 in the posted picture, says that $[0, \Gamma)$ for $\Gamma < \Omega$ (!) is second countable (and thus perfectly normal etc.) Not for $\Gamma = \Omega$, as we saw above. So you misread what item 5. says.
Being an order topology, $[0, \Omega)$ is completely normal, which is also known as hereditarily normal: every subspace is normal, or equivalently, every two separated sets $A$ and $B$ (i.e. the closure of $A$ misses $B$ and vice versa) have disjoint open neighbourhoods.