Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set

Theorem
Let $\Omega$ denote the first uncountable ordinal.

Let $\left[{0 \,.\,.\, \Omega}\right]$ denote the closed ordinal space on $\Omega$.

Then $\left\{ {\Omega}\right\}$ is a closed set of $\left[{0 \,.\,.\, \Omega}\right]$ but not a $G_\delta$ set.

Proof
The complement relative to $\left[{0 \,.\,.\, \Omega}\right]$ of $\left\{ {\Omega}\right\}$ is $\left[{0 \,.\,.\, \Omega}\right)$, which is open in $\left[{0 \,.\,.\, \Omega}\right]$.

Hence, by definition, $\left\{ {\Omega}\right\}$ is a closed set of $\left[{0 \,.\,.\, \Omega}\right]$.

Let $G_i$ be a countable set of open sets of $\left[{0 \,.\,.\, \Omega}\right]$ which contain $\Omega$.

Then we can find a set of basis elements of $\left[{0 \,.\,.\, \Omega}\right]$ of the form $\left({\alpha_i \,.\,.\, \Omega}\right] \subseteq G_i$ for each $i$.

Each $\left[{0 \,.\,.\, \alpha_i}\right)$ is countable.

We also have Countable Union of Countable Sets is Countable.

Thus the infimum of the $\alpha_i$ is an ordinal $\gamma$ which is less than $\Omega$.

Therefore:
 * $\displaystyle \bigcap G_i \supseteq \left({\gamma \,.\,.\, \Omega}\right] \ne \left\{ {\Omega}\right\}$

That is, $\left\{ {\Omega}\right\}$ cannot be expressed as a countable intersection of open sets of $\left[{0 \,.\,.\, \Omega}\right]$.