Open Ordinal Space is not Compact in Closed Ordinal Space

Theorem
Let $\Gamma$ be a limit ordinal.

Let $\hointr 0 \Gamma$ denote the closed ordinal space on $\Gamma$.

Consider the compact subspace $\hointr 0 \Gamma$.

Then $\hointr 0 \Gamma$ is not compact in $\closedint 0 \Gamma$.

Proof
Consider the set:


 * $\set {\hointr 0 \Gamma: \alpha < \Gamma}$

This is an open cover of $\hointr 0 \Gamma$.

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

Hence the result by definition of compact.