Uncountable Closed Ordinal Space is not Second-Countable

Theorem

Let $\Omega$ denote the first uncountable ordinal.

Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.

Then $\closedint 0 \Omega$ is not a second-countable space.

Proof

This theorem requires a proof. In particular: Show it does not have a countable basis 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: $43$. Closed Ordinal Space $[0, \Omega]$: $5$