Basis for Open Ordinal Topology

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

This article, or a section of it, needs explaining. In particular: It's unclear from the source work used whether this is the basis for the Open Ordinal Topology or Closed Ordinal Topology. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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

Proof

This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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