Closed Ordinal Space is Complete Order Space
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Theorem
Let $\Gamma$ be a limit ordinal.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a complete order space.
Proof
Let $H$ be a subset of an ordinal space.
Then $H$ has an infimum: its first element.
Let $H$ be a subset of $\closedint 0 \Gamma$.
Then $H$ has a supremum.
Therefore $\closedint 0 \Gamma$ is a complete order space.
$\blacksquare$
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 $[0, \Gamma] \ (\Gamma < \Omega)$: $6$