# Basis for Open Ordinal Topology

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## Theorem

Let $\Gamma$ be a limit ordinal.

Let $\left[{0 \,.\,.\, \Gamma}\right)$ denote the open ordinal space on $\Gamma$.

Consider the set $\mathcal B$ of subsets of $\left[{0 \,.\,.\, \Gamma}\right)$ of the form:

- $\left({\alpha \,.\,.\, \beta + 1}\right) = \left({\alpha \,.\,.\, \beta}\right] = \left\{ {x \in \left[{0 \,.\,.\, \Gamma}\right): \alpha < x < \beta + 1}\right\}$

for $\alpha, \beta \in \left[{0 \,.\,.\, \Gamma}\right)$.

Then $\mathcal B$ forms a basis for $\left[{0 \,.\,.\, \Gamma}\right)$.

## Proof

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 40 - 43$