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


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