# Basis for Open Ordinal Topology

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## Theorem

Let $\Gamma$ be a limit ordinal.

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

Consider the set $\BB$ of subsets of $\hointr 0 \Gamma$ of the form:

- $\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$

for $\alpha, \beta \in \hointr 0 \Gamma$.

Then $\BB$ forms a basis for $\hointr 0 \Gamma$.

## Proof

## 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