ordinal spaces and order topology

Question

What is the difference between general order topology and ordinal space topology? I'm reading the first answer in Is every locally compact Hausdorff space paracompact?, and I don't understand why $E = [0,\omega) \times \{ \omega_1 \}$ and $F = \{ \omega \} \times [0,\omega_1)$ are closed. Every basis element containing $ \omega \times \omega_1 \in X - E $ must be the form of $(a,\omega] \times (c, \omega_1]$ in the order topology and since there must be an element of $[0,\omega)$ in $(a,\omega]$, $(a,\omega] \times (c, \omega_1]$ must intersect $E$. Then why is $ X - E$ open?

Answer 1

$X\setminus E=[0,\omega)\times[0,\omega_1]$; this is the product of an open set in $[0,\omega]$ and an open set in $[0,\omega_1]$, so it’s open in $[0,\omega]\times[0,\omega_1]$ and therefore open in the subspace $X$. $E$ is therefore the complement of an open subset of $X$, so $E$ is closed in $X$. The proof that $F$ is closed in $X$ is similar.

The point $\langle\omega,\omega_1\rangle$ is altogether irrelevant, since it’s not in $X$.