# Open Ordinal Space is not Compact in Closed Ordinal Space

## Theorem

Let $\Gamma$ be a limit ordinal.

Let $\hointr 0 \Gamma$ denote the open 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.

$\blacksquare$