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