Open Ordinal Space is not Compact in Closed Ordinal Space

Theorem
Let $\Gamma$ be a limit ordinal.

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

Consider the compact subspace $\left[{0 \,.\,.\, \Gamma}\right)$.

Then $\left[{0 \,.\,.\, \Gamma}\right)$ is not compact in $\left[{0 \,.\,.\, \Gamma}\right]$.

Proof
Consider the set:


 * $\left\{ {\left[{0 \,.\,.\, \alpha}\right): \alpha < \Gamma}\right\}$

This is an open cover of $\left[{0 \,.\,.\, \Gamma}\right)$.

But because $\Gamma$ is a limit ordinal, it has no finite subcover.

Hence the result by definition of compact.