Basis for Open Ordinal Topology
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 $\alpha, \beta \in \left[{0 \,.\,.\, \Gamma}\right)$.
It's unclear from the source work used whether this is the basis for the Open Ordinal Topology or Closed Ordinal Topology.
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Then $\mathcal B$ forms a basis for $\left[{0 \,.\,.\, \Gamma}\right)$.
Proof
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Sources
- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology ... (previous) ... (next): $\text{II}: \ 40 - 43$
