Ordinal Space is Completely Normal

Theorem

Let $\Gamma$ denote a limit ordinal.

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

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

Then $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both completely normal.

Proof

By definition, $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both linearly ordered spaces.

The result follows from Linearly Ordered Space is Completely Normal.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $40 \text { - } 43$. Ordinal Space: $4$