Open Ordinal Space is not Compact in Closed Ordinal Space
Jump to navigation
Jump to search
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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $41$. Closed Ordinal Space $[0, \Gamma] \ (\Gamma < \Omega)$: $7$